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== 9.8#h(0.6em)Solve Quadratic Inequalities

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Before you get started, take this readiness quiz.

Solve: #math.equation(block: false, alt: "2 x minus 3 equals 0 .")[$2 x − 3 = 0 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 3 over 2")[$x = frac(3, 2)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "2 y squared plus y equals 15")[$2 y^(2) + y = 15$]. #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "y equals minus 3 , y equals the fraction 5 over 2")[$y = − 3 , #h(0.2em) y = frac(5, 2)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "the fraction 1 over x squared plus 2 x minus 8 greater than 0")[$frac(1, x^(2) + 2 x − 8) > 0$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "minus ∞ , minus 4 union 2 , ∞")[$− ∞ , − 4 ∪ 2 , ∞$]
]
]

We have learned how to solve linear inequalities and rational inequalities previously. Some of the techniques we used to solve them were the same and some were different.

We will now learn to solve inequalities that have a quadratic expression. We will use some of the techniques from solving linear and rational inequalities as well as quadratic equations.

We will solve quadratic inequalities two ways—both graphically and algebraically.


=== Solve Quadratic Inequalities Graphically

A #strong[quadratic equation] is in standard form when written as #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0. If we replace the equal sign with an inequality sign, we have a #strong[quadratic inequality] in standard form.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Quadratic Inequality]

A #strong[quadratic inequality] is an inequality that contains a quadratic expression.

The standard form of a quadratic inequality is written:

#math.equation(block: true, alt: "a x squared plus b x plus c less than 0, a x squared plus b x plus c less than or equal to 0; a x squared plus b x plus c greater than 0, a x squared plus b x plus c greater than or equal to 0")[$a x^(2) + b x + c < 0 &  &  &  &  & a x^(2) + b x + c ≤ 0 \ a x^(2) + b x + c > 0 &  &  &  &  & a x^(2) + b x + c ≥ 0$]
]

The graph of a quadratic function #emph[f] (#emph[x]) = #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0 is a parabola. When we ask when is #emph[ax]#super[2] + #emph[bx] + #emph[c] \< 0, we are asking when is f(#emph[x]) \< 0. We want to know when the parabola is below the #emph[x]-axis.

When we ask when is #emph[ax]#super[2] + #emph[bx] + #emph[c] \> 0, we are asking when is #emph[f] (#emph[x]) \> 0. We want to know when the parabola is above the #emph[x]-axis.

#figure(figph[The first graph is an upward facing parabola, f of x, on an x y-coordinate plane. To the left of the function, f of x is greater than 0. Between the x-intercepts, f of x is less than 0. To the right of the function, f of x is greater than 0. The second graph is a downward-facing parabola, f of x, on an x y coordinate plane. To the left of the function, f of x is less than 0. Between the x-intercepts, f of x is greater than 0. To the right of the function, f of x is less than 0.], alt: "The first graph is an upward facing parabola, f of x, on an x y-coordinate plane. To the left of the function, f of x is greater than 0. Between the x-intercepts, f of x is less than 0. To the right of the function, f of x is greater than 0. The second graph is a downward-facing parabola, f of x, on an x y coordinate plane. To the left of the function, f of x is less than 0. Between the x-intercepts, f of x is greater than 0. To the right of the function, f of x is less than 0.", caption: none)

#examplebox("Example 1")[How to Solve a Quadratic Inequality Graphically][
Solve #math.equation(block: false, alt: "x squared minus 6 x plus 8 less than 0")[$x^(2) − 6 x + 8 < 0$] graphically. Write the solution in interval notation.

#solutionbox[
#figure(figph[The figure is a table with 3 columns. The first column is Step 1: Write the quadratic inequality in standard form. The second column says the inequality is in standard form. The third column says x squared minus 6 times x plus 8 less than 0.], alt: "The figure is a table with 3 columns. The first column is Step 1: Write the quadratic inequality in standard form. The second column says the inequality is in standard form. The third column says x squared minus 6 times x plus 8 less than 0.", caption: none)

#figure(figph[The figure is a table with 3 columns. The first column says Step 2-Graph the function f of x equals a times x squared plus b times x plus c using properties or transformations. The second column gives instructions and the third column shows the work for step 3 as follows. We will graph using properties. The function is f of x equals x squared minus 6 times x plus 8 where a equals 1, b equals negative 6, and c equals 8. Look at a in the function f of x equals x squared minus 6 times x plus 8. Since a is positive, the parabola opens upward. The equation of the axis of symmetry is the line x equals negative b divided by 2 times a, so x equals negative negative 6 divided by 2 times 1. X equals 3. The axis of symmetry is the line x equals 3. The vertex is on the axis of symmetry. Substitute x equals 3 into the function, so f of 3 equals 3 squared minus 6 times 3 plus 8. F of 3 equals negative 1, so the vertex is (3, negative 1). We find f of 0 in order to find the y-intercept, so f of 0 equals 0 squared minus 6 times 0 plus 8. F of 0 equals 8, so the y intercept is (0, 8). We use the axis of symmetry to find a point symmetric to the y-intercept. The y-intercept is 3 units left of the axis of symmetry, x equals 3. A point 3 units to the right of the axis of symmetry has x equals 6. Point symmetric to y-intercept is (6, 8). We solve f of x equals 0 in order to find the x-intercepts. We can solve this quadratic equation by factoring. 0 equals x squared minus 6 times x plus 8, 0 equals the quantity x minus 2 times the quantity x minus 4, x equals 2 or x equals 4. The x-intercepts are (2, 0) and (4, 0). We graph the vertex, intercepts, and the point symmetric to the y-intercept. We connect these 5 points to sketch the parabola shown that is upward-facing with the points found through this process.], alt: "The figure is a table with 3 columns. The first column says Step 2-Graph the function f of x equals a times x squared plus b times x plus c using properties or transformations. The second column gives instructions and the third column shows the work for step 3 as follows. We will graph using properties. The function is f of x equals x squared minus 6 times x plus 8 where a equals 1, b equals negative 6, and c equals 8. Look at a in the function f of x equals x squared minus 6 times x plus 8. Since a is positive, the parabola opens upward. The equation of the axis of symmetry is the line x equals negative b divided by 2 times a, so x equals negative negative 6 divided by 2 times 1. X equals 3. The axis of symmetry is the line x equals 3. The vertex is on the axis of symmetry. Substitute x equals 3 into the function, so f of 3 equals 3 squared minus 6 times 3 plus 8. F of 3 equals negative 1, so the vertex is (3, negative 1). We find f of 0 in order to find the y-intercept, so f of 0 equals 0 squared minus 6 times 0 plus 8. F of 0 equals 8, so the y intercept is (0, 8). We use the axis of symmetry to find a point symmetric to the y-intercept. The y-intercept is 3 units left of the axis of symmetry, x equals 3. A point 3 units to the right of the axis of symmetry has x equals 6. Point symmetric to y-intercept is (6, 8). We solve f of x equals 0 in order to find the x-intercepts. We can solve this quadratic equation by factoring. 0 equals x squared minus 6 times x plus 8, 0 equals the quantity x minus 2 times the quantity x minus 4, x equals 2 or x equals 4. The x-intercepts are (2, 0) and (4, 0). We graph the vertex, intercepts, and the point symmetric to the y-intercept. We connect these 5 points to sketch the parabola shown that is upward-facing with the points found through this process.", caption: none)

#figure(figph[The figure is a table with 3 columns. The first column says Step 3- Determine the solution from the graph. The second column gives instructions. X squared minus 6 x plus 8 less than 0. The inequality asks for the values of x which make the function less than 0. Which values of x make the parabola below the x-axis. We do not include the values 2, 4 as the inequality is strictly less than. The third column says The solution, in interval notation, is (2, 4).], alt: "The figure is a table with 3 columns. The first column says Step 3- Determine the solution from the graph. The second column gives instructions. X squared minus 6 x plus 8 less than 0. The inequality asks for the values of x which make the function less than 0. Which values of x make the parabola below the x-axis. We do not include the values 2, 4 as the inequality is strictly less than. The third column says The solution, in interval notation, is (2, 4).", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Solve #math.equation(block: false, alt: "x squared plus 2 x minus 8 less than 0")[$x^(2) + 2 x − 8 < 0$] graphically and ⓑ write the solution in interval notation.

#solutionbox[
ⓐ  #linebreak() 
#figure(figph[This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 2, negative 9), y-intercept of (0, 8), and axis of symmetry shown at x equals negative 2.], alt: "This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 2, negative 9), y-intercept of (0, 8), and axis of symmetry shown at x equals negative 2.", caption: none)

 #linebreak() 
ⓑ #math.equation(block: false, alt: "open parenthesis −4 , 2 close parenthesis")[$( −4 , 2 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Solve #math.equation(block: false, alt: "x squared minus 8 x plus 12 greater than or equal to 0")[$x^(2) − 8 x + 12 ≥ 0$] graphically and ⓑ write the solution in interval notation.

#solutionbox[
ⓐ  #linebreak() #figure(figph[This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (4, negative 4) and x-intercepts of (2, 0) and (6, 0).], alt: "This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (4, negative 4) and x-intercepts of (2, 0) and (6, 0).", caption: none)

 #linebreak() 
ⓑ #math.equation(block: false, alt: "open parenthesis − ∞ , 2 ] union [ 6 , ∞ close parenthesis")[$( "−" ∞ , 2 ] ∪ [ 6 , ∞ )$]
]
]

We list the steps to take to solve a quadratic inequality graphically.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Solve a quadratic inequality graphically.]

+ Write the quadratic inequality in standard form.
+ Graph the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a x squared plus b x plus c .")[$f ( x ) = a x^(2) + b x + c .$]
+ Determine the solution from the graph.
]

In the last example, the parabola opened upward and in the next example, it opens downward. In both cases, we are looking for the part of the parabola that is below the #emph[x]-axis but note how the position of the parabola affects the solution.

#examplebox("Example 2")[][
Solve #math.equation(block: false, alt: "− x squared minus 8 x minus 12 less than or equal to 0")[$"−" x^(2) − 8 x − 12 ≤ 0$] graphically. Write the solution in interval notation.

#solutionbox[
#figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([The quadratic inequality in standard form.], [], [#math.equation(block: false, alt: "minus x squared minus 8 x minus 12 less than or equal to 0")[$− x^(2) − 8 x − 12 ≤ 0$]]),
  [Graph the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals − x squared minus 8 x minus 12")[$f ( x ) = "−" x^(2) − 8 x − 12$].], [], [The parabola opens downward. #linebreak() #figure(figph[A red U-shaped arrow pointing downwards on both ends, suggesting a flow or return, against a plain white background.], alt: "A red U-shaped arrow pointing downwards on both ends, suggesting a flow or return, against a plain white background.", caption: none)],
  [Find the line of symmetry.], [], [#math.equation(block: false, alt: "x equals minus the fraction b over 2 a")[$#h(1.8em) x = − frac(b, 2 a)$] #linebreak() 
#math.equation(block: false, alt: "x equals minus the fraction minus 8 over 2 open parenthesis −1 close parenthesis")[$#h(1.8em) x = − frac(− 8, 2 ( −1 ))$] #linebreak() 
#math.equation(block: false, alt: "x equals −4")[$#h(1.8em) x = −4$]],
  [Find the vertex.], [], [#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals − x squared minus 8 x minus 12")[$#h(0.65em) f ( x ) = "−" x^(2) − 8 x − 12$] #linebreak() 
#math.equation(block: false, alt: "f open parenthesis −4 close parenthesis equals − open parenthesis −4 close parenthesis squared minus 8 open parenthesis −4 close parenthesis minus 12")[$f ( −4 ) = "−" attach(( −4 ), t: 2) − 8 ( −4 ) − 12$] #linebreak() 
#math.equation(block: false, alt: "f open parenthesis −4 close parenthesis equals −16 plus 32 minus 12")[$f ( −4 ) = −16 + 32 − 12$] #linebreak() 
#math.equation(block: false, alt: "f open parenthesis −4 close parenthesis equals 4")[$f ( −4 ) = 4$] #linebreak() 
Vertex #math.equation(block: false, alt: "open parenthesis −4 , 4 close parenthesis")[$( −4 , 4 )$]],
  [Find the #emph[x]-intercepts. Let #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0")[$f ( x ) = 0$].], [], [#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals − x squared minus 8 x minus 12")[$#h(0.75em) f ( x ) = "−" x^(2) − 8 x − 12$] #linebreak() 
#math.equation(block: false, alt: "0 equals − x squared minus 8 x minus 12")[$#h(1.95em) 0 = "−" x^(2) − 8 x − 12$]],
  [Factor. #linebreak() 
Use the Zero Product Property.], [], [#math.equation(block: false, alt: "0 equals −1 open parenthesis x plus 6 close parenthesis open parenthesis x plus 2 close parenthesis")[$#h(1.95em) 0 = −1 ( x + 6 ) ( x + 2 )$] #linebreak() 
#math.equation(block: false, alt: "x equals −6 x equals −2")[$#h(2em) x = −6 #h(1.5em) x = −2$]],
  [Graph the parabola.], [], [#emph[x]-intercepts #math.equation(block: false, alt: "open parenthesis −6 , 0 close parenthesis , open parenthesis −2 , 0 close parenthesis")[$( −6 , 0 ) , ( −2 , 0 )$] #linebreak() #figure(figph[A graph shows a downward-opening parabola on a Cartesian coordinate system. The parabola has its vertex at (-4, 4) and intersects the x-axis at (-6, 0) and (-2, 0).], alt: "A graph shows a downward-opening parabola on a Cartesian coordinate system. The parabola has its vertex at (-4, 4) and intersects the x-axis at (-6, 0) and (-2, 0).", caption: none)],
  [Determine the solution from the graph. #linebreak() 
We include the #emph[x]-intercepts as the inequality #linebreak() is “less than or equal to.”], [], [#math.equation(block: false, alt: "open parenthesis − ∞ , − 6 ] union [ − 2 , ∞ close parenthesis")[$( "−" ∞ , #h(0.2em) "−" 6 ] ∪ [ "−" 2 , #h(0.2em) ∞ )$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Solve #math.equation(block: false, alt: "− x squared minus 6 x minus 5 greater than 0")[$"−" x^(2) − 6 x − 5 > 0$] graphically and ⓑ write the solution in interval notation.

#solutionbox[
ⓐ  #linebreak() 
#figure(figph[A downward-facing parabola on the x y-coordinate plane. It has a vertex of (negative 3, 4), a y-intercept at (0, negative 5), and an axis of symmetry shown at x equals negative 3.], alt: "A downward-facing parabola on the x y-coordinate plane. It has a vertex of (negative 3, 4), a y-intercept at (0, negative 5), and an axis of symmetry shown at x equals negative 3.", caption: none)


 #linebreak() 
ⓑ #math.equation(block: false, alt: "open parenthesis −5 , −1 close parenthesis")[$( −5 , −1 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
ⓐ Solve #math.equation(block: false, alt: "− x squared plus 10 x minus 16 less than or equal to 0")[$"−" x^(2) + 10 x − 16 ≤ 0$] graphically and ⓑ write the solution in interval notation.

#solutionbox[
ⓐ  #linebreak() 
#figure(figph[A downward-facing parabola on the x y-coordinate plane. It has a vertex of (5, 9), a y-intercept at (0, negative 16), and an axis of symmetry of x equals 5.], alt: "A downward-facing parabola on the x y-coordinate plane. It has a vertex of (5, 9), a y-intercept at (0, negative 16), and an axis of symmetry of x equals 5.", caption: none)

 #linebreak() 
ⓑ #math.equation(block: false, alt: "open parenthesis − ∞ , 2 ] union [ 8 , ∞ close parenthesis")[$( "−" ∞ , 2 ] ∪ [ 8 , ∞ )$]
]
]


=== Solve Quadratic Inequalities Algebraically

The algebraic method we will use is very similar to the method we used to solve rational inequalities. We will find the zero partition numbers for the inequality, which will be the solutions to the related quadratic equation. Remember a polynomial expression can change signs only where the expression is zero.

We will use the #strong[zero partition numbers] to divide the number line into intervals and then determine whether the quadratic expression will be positive or negative in the interval. We then determine the solution for the inequality.

#examplebox("Example 3")[How To Solve Quadratic Inequalities Algebraically][
Solve #math.equation(block: false, alt: "x squared minus x minus 12 greater than or equal to 0")[$x^(2) − x − 12 ≥ 0$] algebraically. Write the solution in interval notation.

#solutionbox[
#figure(figph[This figure is a table giving the instructions for solving x squared minus x minus 12 greater than or equal to 0 algebraically. It consists of 3 columns where the instructions are given in the first column, the explanation in the second, and the work in the third. Step 1 is to write the quadratic inequality in standard form. The quadratic inequality in already in standard form, so x squared minus x minus 12 greater than or equal to 0.], alt: "This figure is a table giving the instructions for solving x squared minus x minus 12 greater than or equal to 0 algebraically. It consists of 3 columns where the instructions are given in the first column, the explanation in the second, and the work in the third. Step 1 is to write the quadratic inequality in standard form. The quadratic inequality in already in standard form, so x squared minus x minus 12 greater than or equal to 0.", caption: none)

#figure(figph[Step 2 is to determine the zero partition number -- the solutions to the related quadratic equation. To do this, change the inequality sign to an equal sign and then solve the equation. x squared minus x minus 12 equals 0 factors to the quantity x plus 3 times the quantity x minus 4 equals 0. Then, x plus 3 equals 0 and x minus 4 equals 0 to give x equals negative 3 and x equals 4.], alt: "Step 2 is to determine the zero partition number -- the solutions to the related quadratic equation. To do this, change the inequality sign to an equal sign and then solve the equation. x squared minus x minus 12 equals 0 factors to the quantity x plus 3 times the quantity x minus 4 equals 0. Then, x plus 3 equals 0 and x minus 4 equals 0 to give x equals negative 3 and x equals 4.", caption: none)

#figure(figph[Step 3 is to use the zero partition numbers to divide the number line into intervals. Use negative 3 and 4 to divide the number line into intervals. A number line is shown that includes from left to right the values of negative 3, 0, and 4, with dotted lines on negative 3 and 4.], alt: "Step 3 is to use the zero partition numbers to divide the number line into intervals. Use negative 3 and 4 to divide the number line into intervals. A number line is shown that includes from left to right the values of negative 3, 0, and 4, with dotted lines on negative 3 and 4.", caption: none)

#figure(figph[Step 4 says above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality. X equals negative 5, x equals 0, and x equals 5 are chosen to test. The expression negative x squared minus x minus 12 is given with negative 5 squared minus negative 5 minus 12 underneath, which gives 18. The expression negative x squared minus x minus 12 is given with 0 squared minus 0 minus 12 underneath, which gives 12. The expression negative x squared minus x minus 12 is given with 5 squared minus 5 minus 12 underneath, which gives 8.], alt: "Step 4 says above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality. X equals negative 5, x equals 0, and x equals 5 are chosen to test. The expression negative x squared minus x minus 12 is given with negative 5 squared minus negative 5 minus 12 underneath, which gives 18. The expression negative x squared minus x minus 12 is given with 0 squared minus 0 minus 12 underneath, which gives 12. The expression negative x squared minus x minus 12 is given with 5 squared minus 5 minus 12 underneath, which gives 8.", caption: none)

#figure(figph[For Step 5, determine the intervals where the inequality is correct. Write the solution in interval notation. x squared minus x minus 12 greater than or equal to 0 is shown. The inequality is positive in the first and last intervals and equals 0 at the points negative 4, 3 . The solution, in interval notation, is (negative infinity, negative 3\] in union with \[4, infinity).], alt: "For Step 5, determine the intervals where the inequality is correct. Write the solution in interval notation. x squared minus x minus 12 greater than or equal to 0 is shown. The inequality is positive in the first and last intervals and equals 0 at the points negative 4, 3 . The solution, in interval notation, is (negative infinity, negative 3] in union with [4, infinity).", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "x squared plus 2 x minus 8 greater than or equal to 0")[$x^(2) + 2 x − 8 ≥ 0$] algebraically. Write the solution in interval notation.

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , −4 ] union [ 2 , ∞ close parenthesis")[$( "−" ∞ , −4 ] ∪ [ 2 , ∞ )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "x squared minus 2 x minus 15 less than or equal to 0")[$x^(2) − 2 x − 15 ≤ 0$] algebraically. Write the solution in interval notation.

#solutionbox[
#math.equation(block: true, alt: "[ −3 , 5 ]")[$[ −3 , 5 ]$]
]
]

In this example, since the expression #math.equation(block: false, alt: "x squared minus x minus 12")[$x^(2) − x − 12$] factors nicely, we can also find the sign in each interval much like we did when we solved rational inequalities. We find the sign of each of the factors, and then the sign of the product. Our number line would like this:

#figure(figph[The figure shows the expression x squared minus x minus 12 factored to the quantity of x plus 3 times the quantity of x minus 4. The image shows a number line showing dotted lines on negative 3 and 4. It shows the signs of the quantity x plus 3 to be negative, positive, positive, and the signs of the quantity x minus 4 to be negative, negative, positive. Under the number line, it shows the quantity x plus 3 times the quantity x minus 4 with the signs positive, negative, positive.], alt: "The figure shows the expression x squared minus x minus 12 factored to the quantity of x plus 3 times the quantity of x minus 4. The image shows a number line showing dotted lines on negative 3 and 4. It shows the signs of the quantity x plus 3 to be negative, positive, positive, and the signs of the quantity x minus 4 to be negative, negative, positive. Under the number line, it shows the quantity x plus 3 times the quantity x minus 4 with the signs positive, negative, positive.", caption: none)

The result is the same as we found using the other method.

We summarize the steps here.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Solve a quadratic inequality algebraically.]

+ Write the quadratic inequality in standard form.
+ Determine the zero partition numbers—the solutions to the related quadratic equation.
+ Use the zero partition numbers to divide the number line into intervals.
+ Above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality.
+ Determine the intervals where the inequality is correct. Write the solution in interval notation.
]

#examplebox("Example 4")[][
Solve #math.equation(block: false, alt: "minus x squared plus 6 x minus 7 greater than or equal to 0")[$attach(− x, t: 2) + 6 x − 7 ≥ 0$] algebraically. Write the solution in interval notation.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Write the quadratic inequality in standard form.], [#math.equation(block: false, alt: "minus x squared plus 6 x minus 7 greater than or equal to 0")[$− x^(2) + 6 x − 7 ≥ 0$]]),
  [Multiply both sides of the inequality by #math.equation(block: false, alt: "−1")[$−1$]. #linebreak() 
Remember to reverse the inequality sign.], [#math.equation(block: false, alt: "x squared minus 6 x plus 7 less than or equal to 0")[$#h(0.7em) x^(2) − 6 x + 7 ≤ 0$]],
  [Determine the zero partition numbers by solving #linebreak() the related quadratic equation.], [#math.equation(block: false, alt: "x squared minus 6 x plus 7 equals 0")[$#h(0.7em) x^(2) − 6 x + 7 = 0$]],
  [Write the Quadratic Formula.], [#math.equation(block: false, alt: "x equals the fraction minus b plus or minus the square root of b squared minus 4 a c over 2 a")[$#h(4.9em) x = frac(− b ± sqrt(b^(2) − 4 a c), 2 a)$]],
  [Then substitute in the values of #math.equation(block: false, alt: "a , b , c")[$a , b , c$].], [#math.equation(block: false, alt: "x equals the fraction minus open parenthesis −6 close parenthesis plus or minus the square root of open parenthesis −6 close parenthesis squared minus 4 times 1 times open parenthesis 7 close parenthesis over 2 times 1")[$#h(4.9em) x = frac(− ( −6 ) ± sqrt(attach(( −6 ), t: 2) − 4 ⋅ 1 ⋅ ( 7 )), 2 ⋅ 1)$]],
  [Simplify.], [#math.equation(block: false, alt: "x equals the fraction 6 plus or minus the square root of 8 over 2")[$#h(4.9em) x = frac(6 ± sqrt(8), 2)$]],
  [Simplify the radical.], [#math.equation(block: false, alt: "x equals the fraction 6 plus or minus 2 the square root of 2 over 2")[$#h(4.9em) x = frac(6 ± 2 sqrt(2), 2)$]],
  [Remove the common factor, 2.], [#math.equation(block: false, alt: "x equals the fraction 2 open parenthesis 3 plus or minus the square root of 2 close parenthesis over 2")[$#h(4.9em) x = frac(2 ( 3 ± sqrt(2) ), 2)$] #linebreak() 
#math.equation(block: false, alt: "x equals 3 plus or minus the square root of 2")[$#h(4.9em) x = 3 ± sqrt(2)$] #linebreak() 
#math.equation(block: false, alt: "x equals 3 plus the square root of 2 x equals 3 minus the square root of 2")[$#h(4.9em) x = 3 + sqrt(2) #h(2em) x = 3 − sqrt(2)$] #linebreak() 
#math.equation(block: false, alt: "x approximately equals 1.6 x approximately equals 4.4")[$#h(4.9em) x ≈ 1.6 #h(3.35em) x ≈ 4.4$]],
  [Use the zero partition numbers to divide the #linebreak() number line into intervals. #linebreak() 
Test numbers from each interval #linebreak() in the original inequality.], [#figure(figph[Sign analysis of the quadratic expression -x^2 + 6x - 7 on a number line, showing it's positive between 3-sqrt(2) (approx 1.6) and 3+sqrt(2) (approx 4.4), and negative elsewhere.], alt: "Sign analysis of the quadratic expression -x^2 + 6x - 7 on a number line, showing it's positive between 3-sqrt(2) (approx 1.6) and 3+sqrt(2) (approx 4.4), and negative elsewhere.", caption: none)],
  [Determine the intervals where the #linebreak() inequality is correct. Write the solution #linebreak() in interval notation.], [#math.equation(block: false, alt: "− x squared plus 6 x minus 7 greater than or equal to 0")[$"−" x^(2) + 6 x − 7 ≥ 0$] in the middle interval #linebreak() 
#math.equation(block: false, alt: "[ 3 minus the square root of 2 , 3 plus the square root of 2 ]")[$[ 3 − sqrt(2) , #h(0.5em) 3 + sqrt(2) ]$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "− x squared plus 2 x plus 1 greater than or equal to 0")[$"−" x^(2) + 2 x + 1 ≥ 0$] algebraically. Write the solution in interval notation.

#solutionbox[
#math.equation(block: true, alt: "[ −1 minus the square root of 2 , −1 plus the square root of 2 ]")[$[ −1 − sqrt(2) , −1 + sqrt(2) ]$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "− x squared plus 8 x minus 14 less than 0")[$"−" x^(2) + 8 x − 14 < 0$] algebraically. Write the solution in interval notation.

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , 4 minus the square root of 2 close parenthesis union open parenthesis 4 plus the square root of 2 , ∞ close parenthesis")[$( "−" ∞ , 4 − sqrt(2) ) ∪ ( 4 + sqrt(2) , ∞ )$]
]
]

The solutions of the quadratic inequalities in each of the previous examples, were either an interval or the union of two intervals. This resulted from the fact that, in each case we found two solutions to the corresponding quadratic equation #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0. These two solutions then gave us either the two #emph[x-]intercepts for the graph or the two zero partition numbers to divide the number line into intervals.

This correlates to our previous discussion of the number and type of solutions to a quadratic equation using the discriminant.

For a quadratic equation of the form #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0, #math.equation(block: false, alt: "a not equal to 0 .")[$a ≠ 0 .$]

#figure(figph[The figure is a table with 3 columns. Column 1 is labeled discriminant, column 2 is Number/Type of solution, and column 3 is Typical Graph. Reading across the columns, if b squared minus 4 times a times c is greater than 0, there will be 2 real solutions because there are 2 x-intercepts on the graph. The image of a typical graph an upward or downward parabola with 2 x-intercepts. If the discriminant b squared minus 4 times a times c is equals to 0, then there is 1 real solution because there is 1 x-intercept on the graph. The image of the typical graph is an upward- or downward-facing parabola that has a vertex on the x-axis instead of crossing through it. If the discriminant b squared minus 4 times a times c is less than 0, there are 2 complex solutions because there is no x-intercept. The image of the typical graph shows an upward- or downward-facing parabola that does not cross the x-axis.], alt: "The figure is a table with 3 columns. Column 1 is labeled discriminant, column 2 is Number/Type of solution, and column 3 is Typical Graph. Reading across the columns, if b squared minus 4 times a times c is greater than 0, there will be 2 real solutions because there are 2 x-intercepts on the graph. The image of a typical graph an upward or downward parabola with 2 x-intercepts. If the discriminant b squared minus 4 times a times c is equals to 0, then there is 1 real solution because there is 1 x-intercept on the graph. The image of the typical graph is an upward- or downward-facing parabola that has a vertex on the x-axis instead of crossing through it. If the discriminant b squared minus 4 times a times c is less than 0, there are 2 complex solutions because there is no x-intercept. The image of the typical graph shows an upward- or downward-facing parabola that does not cross the x-axis.", caption: none)

The last row of the table shows us when the parabolas never intersect the #emph[x]-axis. Using the Quadratic Formula to solve the quadratic equation, the radicand is a negative. We get two complex solutions.

In the next example, the quadratic inequality solutions will result from the solution of the quadratic equation being complex.

#examplebox("Example 5")[][
Solve, writing any solution in interval notation:

ⓐ #math.equation(block: false, alt: "x squared minus 3 x plus 4 greater than 0")[$x^(2) − 3 x + 4 > 0$] ⓑ #math.equation(block: false, alt: "x squared minus 3 x plus 4 less than or equal to 0")[$x^(2) − 3 x + 4 ≤ 0$]

#solutionbox[
ⓐ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Write the quadratic inequality in standard form.], [#math.equation(block: false, alt: "x squared minus 3 x plus 4 greater than 0")[$#h(0.8em) x^(2) − 3 x + 4 > 0$]]),
  [Determine the zero partition numbers by solving #linebreak() the related quadratic equation.], [#math.equation(block: false, alt: "x squared minus 3 x plus 4 equals 0")[$#h(1.5em) x^(2) − 3 x + 4 = 0$]],
  [Write the Quadratic Formula.], [#math.equation(block: false, alt: "x equals the fraction minus b plus or minus the square root of b squared minus 4 a c over 2 a")[$x = frac(− b ± sqrt(b^(2) − 4 a c), 2 a)$]],
  [Then substitute in the values of #math.equation(block: false, alt: "a , b , c")[$a , b , c$].], [#math.equation(block: false, alt: "x equals the fraction minus open parenthesis − 3 close parenthesis plus or minus the square root of open parenthesis − 3 close parenthesis squared minus 4 times 1 times open parenthesis 4 close parenthesis over 2 times 1")[$x = frac(− ( "−" 3 ) ± sqrt(attach(( "−" 3 ), t: 2) − 4 ⋅ 1 ⋅ ( 4 )), 2 ⋅ 1)$]],
  [Simplify.], [#math.equation(block: false, alt: "x equals the fraction 3 plus or minus the square root of minus 7 over 2")[$x = frac(3 ± sqrt(− 7), 2)$]],
  [Simplify the radicand.], [#math.equation(block: false, alt: "x equals the fraction 3 plus or minus the square root of 7 i over 2")[$x = frac(3 ± sqrt(7) i, 2)$]],
  [The complex solutions tell us the #linebreak() parabola does not intercept the #emph[x]-axis. #linebreak() Also, the parabola opens upward. This #linebreak() tells us that the parabola is completely above the #emph[x]-axis.], [Complex solutions #linebreak() #figure(figph[A white background features a horizontal line with arrows pointing left and right. Above the center of this line, a curved arrow forms an upward-pointing U-shape, with both ends curving upwards and terminating in arrowheads.], alt: "A white background features a horizontal line with arrows pointing left and right. Above the center of this line, a curved arrow forms an upward-pointing U-shape, with both ends curving upwards and terminating in arrowheads.", caption: none)],
))

We are to find the solution to #math.equation(block: false, alt: "x squared minus 3 x plus 4 greater than 0 .")[$x^(2) − 3 x + 4 > 0 .$] Since for all values of #math.equation(block: false, alt: "x")[$x$] the graph is above the #emph[x]-axis, all values of #emph[x] make the inequality true. In interval notation we write #math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis .")[$( "−" ∞ , ∞ ) .$]

ⓑ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Write the quadratic inequality in standard form.], [#math.equation(block: false, alt: "x squared minus 3 x plus 4 less than or equal to 0")[$x^(2) − 3 x + 4 ≤ 0$]]),
  [Determine the zero partition numbers by solving the related quadratic equation], [#math.equation(block: false, alt: "x squared minus 3 x plus 4 equals 0")[$x^(2) − 3 x + 4 = 0$]],
))

Since the corresponding quadratic equation is the same as in part (a), the parabola will be the same. The parabola opens upward and is completely above the #emph[x]-axis—no part of it is below the #emph[x]-axis.

We are to find the solution to #math.equation(block: false, alt: "x squared minus 3 x plus 4 less than or equal to 0 .")[$x^(2) − 3 x + 4 ≤ 0 .$] Since for all values of #emph[x] the graph is never below the #emph[x]-axis, no values of #emph[x] make the inequality true. There is no solution to the inequality.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve and write any solution in interval notation: #linebreak() ⓐ #math.equation(block: false, alt: "− x squared plus 2 x minus 4 less than or equal to 0")[$"−" x^(2) + 2 x − 4 ≤ 0$] ⓑ #math.equation(block: false, alt: "− x squared plus 2 x minus 4 greater than or equal to 0")[$"−" x^(2) + 2 x − 4 ≥ 0$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis")[$( "−" ∞ , ∞ )$] #linebreak() ⓑ no solution
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve and write any solution in interval notation: #linebreak() ⓐ #math.equation(block: false, alt: "x squared plus 3 x plus 3 less than 0")[$x^(2) + 3 x + 3 < 0$] ⓑ #math.equation(block: false, alt: "x squared plus 3 x plus 3 greater than 0")[$x^(2) + 3 x + 3 > 0$]

#solutionbox[
ⓐ no solution #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis − ∞ , ∞ close parenthesis")[$( "−" ∞ , ∞ )$]
]
]


=== Key Concepts

- Solve a Quadratic Inequality Graphically
+ Write the quadratic inequality in standard form.
+ Graph the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a x squared plus b x plus c")[$f ( x ) = a x^(2) + b x + c$] using properties or transformations.
+ Determine the solution from the graph.
- How to Solve a Quadratic Inequality Algebraically
+ Write the quadratic inequality in standard form.
+ Determine the zero partition numbers -- the solutions to the related quadratic equation.
+ Use the zero partition numbers to divide the number line into intervals.
+ Above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality.
+ Determine the intervals where the inequality is correct. Write the solution in interval notation.


=== Section Exercises


==== Practice Makes Perfect

#strong[Solve Quadratic Inequalities Graphically]

In the following exercises, ⓐ solve graphically and ⓑ write the solution in interval notation.

#math.equation(block: true, alt: "x squared plus 6 x plus 5 greater than 0")[$x^(2) + 6 x + 5 > 0$]

#solutionbox[
ⓐ  #linebreak() 
#figure(figph[The graph shown is an upward-facing parabola with vertex (negative 3, negative 4) and y-intercept (0,5).], alt: "The graph shown is an upward-facing parabola with vertex (negative 3, negative 4) and y-intercept (0,5).", caption: none)

 #linebreak() 
ⓑ #math.equation(block: false, alt: "open parenthesis − ∞ , −5 close parenthesis union open parenthesis −1 , ∞ close parenthesis")[$( "−" ∞ , −5 ) ∪ ( −1 , ∞ )$]
]

#math.equation(block: true, alt: "x squared plus 4 x minus 12 less than 0")[$x^(2) + 4 x − 12 < 0$]

#math.equation(block: true, alt: "x squared plus 4 x plus 3 less than or equal to 0")[$x^(2) + 4 x + 3 ≤ 0$]

#solutionbox[
ⓐ  #linebreak() 
#figure(figph[The graph shown is an upward facing parabola with vertex (negative 2, negative 1) and y-intercept (0,3).], alt: "The graph shown is an upward facing parabola with vertex (negative 2, negative 1) and y-intercept (0,3).", caption: none)

 #linebreak() 
ⓑ #math.equation(block: false, alt: "[ −3 , −1 ]")[$[ −3 , −1 ]$]
]

#math.equation(block: true, alt: "x squared minus 6 x plus 8 greater than or equal to 0")[$x^(2) − 6 x + 8 ≥ 0$]

#math.equation(block: true, alt: "− x squared minus 3 x plus 18 less than or equal to 0")[$"−" x^(2) − 3 x + 18 ≤ 0$]

#solutionbox[
ⓐ  #linebreak() 
#figure(figph[The graph shown is a downward-facing parabola with vertex (negative 1 and 5 tenths, 20) and y-intercept (0, 18).], alt: "The graph shown is a downward-facing parabola with vertex (negative 1 and 5 tenths, 20) and y-intercept (0, 18).", caption: none)

 #linebreak() 
ⓑ #math.equation(block: false, alt: "open parenthesis − ∞ , −6 ] union [ 3 , ∞ close parenthesis")[$( "−" ∞ , −6 ] ∪ [ 3 , ∞ )$]
]

#math.equation(block: true, alt: "− x squared plus 2 x plus 24 less than 0")[$"−" x^(2) + 2 x + 24 < 0$]

#math.equation(block: true, alt: "− x squared plus x plus 12 greater than or equal to 0")[$"−" x^(2) + x + 12 ≥ 0$]

#solutionbox[
ⓐ  #linebreak() 
#figure(figph[The graph shown is a downward facing parabola with a y-intercept of (0, 12) and x-intercepts (negative 3, 0) and (4, 0).], alt: "The graph shown is a downward facing parabola with a y-intercept of (0, 12) and x-intercepts (negative 3, 0) and (4, 0).", caption: none)

 #linebreak() 
ⓑ #math.equation(block: false, alt: "[ −3 , 4 ]")[$[ −3 , 4 ]$]
]

#math.equation(block: true, alt: "− x squared plus 2 x plus 15 greater than 0")[$"−" x^(2) + 2 x + 15 > 0$]

In the following exercises, solve each inequality algebraically and write any solution in interval notation.

#math.equation(block: true, alt: "x squared plus 3 x minus 4 greater than or equal to 0")[$x^(2) + 3 x − 4 ≥ 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , −4 ] union [ 1 , ∞ close parenthesis")[$( "−" ∞ , −4 ] ∪ [ 1 , ∞ )$]
]

#math.equation(block: true, alt: "x squared plus x minus 6 less than or equal to 0")[$x^(2) + x − 6 ≤ 0$]

#math.equation(block: true, alt: "x squared minus 7 x plus 10 less than 0")[$x^(2) − 7 x + 10 < 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis 2 , 5 close parenthesis")[$( 2 , 5 )$]
]

#math.equation(block: true, alt: "x squared minus 4 x plus 3 greater than 0")[$x^(2) − 4 x + 3 > 0$]

#math.equation(block: true, alt: "x squared plus 8 x greater than minus 15")[$x^(2) + 8 x > − 15$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , −5 close parenthesis union open parenthesis −3 , ∞ close parenthesis")[$( "−" ∞ , −5 ) ∪ ( −3 , ∞ )$]
]

#math.equation(block: true, alt: "x squared plus 8 x less than minus 12")[$x^(2) + 8 x < − 12$]

#math.equation(block: true, alt: "x squared minus 4 x plus 2 less than or equal to 0")[$x^(2) − 4 x + 2 ≤ 0$]

#solutionbox[
#math.equation(block: true, alt: "[ 2 minus the square root of 2 , 2 plus the square root of 2 ]")[$[ 2 − sqrt(2) , 2 + sqrt(2) ]$]
]

#math.equation(block: true, alt: "− x squared plus 8 x minus 11 less than 0")[$"−" x^(2) + 8 x − 11 < 0$]

#math.equation(block: true, alt: "x squared minus 10 x greater than minus 19")[$x^(2) − 10 x > − 19$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , 5 minus the square root of 6 close parenthesis union open parenthesis 5 plus the square root of 6 , ∞ close parenthesis")[$( "−" ∞ , 5 − sqrt(6) ) ∪ ( 5 + sqrt(6) , ∞ )$]
]

#math.equation(block: true, alt: "x squared plus 6 x less than minus 3")[$x^(2) + 6 x < − 3$]

#math.equation(block: true, alt: "−6 x squared plus 19 x minus 10 greater than or equal to 0")[$−6 x^(2) + 19 x − 10 ≥ 0$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 2 over 3 , the fraction 5 over 2")[$frac(2, 3) , " " frac(5, 2)$]
]

#math.equation(block: true, alt: "−3 x squared minus 4 x plus 4 less than or equal to 0")[$−3 x^(2) − 4 x + 4 ≤ 0$]

#math.equation(block: true, alt: "−2 x squared plus 7 x plus 4 greater than or equal to 0")[$−2 x^(2) + 7 x + 4 ≥ 0$]

#solutionbox[
#math.equation(block: true, alt: "[ − the fraction 1 over 2 , 4 ]")[$[ "−" frac(1, 2) , 4 ]$]
]

#math.equation(block: true, alt: "2 x squared plus 5 x minus 12 greater than 0")[$2 x^(2) + 5 x − 12 > 0$]

#math.equation(block: true, alt: "x squared plus 3 x plus 5 greater than 0")[$x^(2) + 3 x + 5 > 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , ∞ close parenthesis .")[$( "−" ∞ , ∞ ) .$]
]

#math.equation(block: true, alt: "x squared minus 3 x plus 6 less than or equal to 0")[$x^(2) − 3 x + 6 ≤ 0$]

#math.equation(block: true, alt: "− x squared plus x minus 7 greater than 0")[$"−" x^(2) + x − 7 > 0$]

#solutionbox[
no solution
]

#math.equation(block: true, alt: "− x squared minus 4 x minus 5 less than 0")[$"−" x^(2) − 4 x − 5 < 0$]

#math.equation(block: true, alt: "−2 x squared plus 8 x minus 10 less than 0")[$−2 x^(2) + 8 x − 10 < 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , ∞ close parenthesis .")[$( "−" ∞ , ∞ ) .$]
]

#math.equation(block: true, alt: "− x squared plus 2 x minus 7 greater than or equal to 0")[$"−" x^(2) + 2 x − 7 ≥ 0$]


==== Writing Exercises

Explain zero partition numbers and how they are used to solve quadratic inequalities algebraically.

#solutionbox[
Answers will vary.
]

Solve #math.equation(block: false, alt: "x squared plus 2 x greater than or equal to 8")[$x^(2) + 2 x ≥ 8$] both graphically and algebraically. Which method do you prefer, and why?

Describe the steps needed to solve a quadratic inequality graphically.

#solutionbox[
Answers will vary.
]

Describe the steps needed to solve a quadratic inequality algebraically.


==== Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

#figure(figph[This figure is a list to assess your understanding of the concepts presented in this section. It has 4 columns labeled I can…, Confidently, With some help, and No-I don’t get it! Below I can…, there is solve quadratic inequalities graphically and solve quadratic inequalities algebraically. The other columns are left blank for you to check you understanding.], alt: "This figure is a list to assess your understanding of the concepts presented in this section. It has 4 columns labeled I can…, Confidently, With some help, and No-I don’t get it! Below I can…, there is solve quadratic inequalities graphically and solve quadratic inequalities algebraically. The other columns are left blank for you to check you understanding.", caption: none)

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?


=== Chapter Review Exercises


==== Solve Quadratic Equations Using the Square Root Property

#strong[Solve Quadratic Equations of the form #emph[ax]#super[2] = #emph[k] Using the Square Root Property]

In the following exercises, solve using the Square Root Property.

#math.equation(block: true, alt: "y squared equals 144")[$y^(2) = 144$]

#solutionbox[
#math.equation(block: true, alt: "y equals plus or minus 12")[$y = ± 12$]
]

#math.equation(block: true, alt: "n squared minus 80 equals 0")[$n^(2) − 80 = 0$]

#math.equation(block: true, alt: "4 a squared equals 100")[$4 a^(2) = 100$]

#solutionbox[
#math.equation(block: true, alt: "a equals plus or minus 5")[$a = ± 5$]
]

#math.equation(block: true, alt: "2 b squared equals 72")[$2 b^(2) = 72$]

#math.equation(block: true, alt: "r squared plus 32 equals 0")[$r^(2) + 32 = 0$]

#solutionbox[
#math.equation(block: true, alt: "r equals plus or minus 4 the square root of 2 i")[$r = ± 4 sqrt(2) i$]
]

#math.equation(block: true, alt: "t squared plus 18 equals 0")[$t^(2) + 18 = 0$]

#math.equation(block: true, alt: "the fraction 2 over 3 w squared minus 20 equals 30")[$frac(2, 3) w^(2) − 20 = 30$]

#solutionbox[
#math.equation(block: true, alt: "w equals plus or minus 5 the square root of 3")[$w = ± 5 sqrt(3)$]
]

#math.equation(block: true, alt: "5 c squared plus 3 equals 19")[$5 c^(2) + 3 = 19$]

#strong[Solve Quadratic Equations of the Form #math.equation(block: false, alt: "a open parenthesis x minus h close parenthesis squared equals k")[$a attach(( x − h ), t: 2) = k$] Using the Square Root Property]

In the following exercises, solve using the Square Root Property.

#math.equation(block: true, alt: "open parenthesis p minus 5 close parenthesis squared plus 3 equals 19")[$attach(( p − 5 ), t: 2) + 3 = 19$]

#solutionbox[
#math.equation(block: true, alt: "p equals 1 , p equals 9")[$p = 1 , #h(0.2em) p = 9$]
]

#math.equation(block: true, alt: "open parenthesis u plus 1 close parenthesis squared equals 45")[$attach(( u + 1 ), t: 2) = 45$]

#math.equation(block: true, alt: "open parenthesis x minus the fraction 1 over 4 close parenthesis squared equals the fraction 3 over 16")[$attach(( x − frac(1, 4) ), t: 2) = frac(3, 16)$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 1 over 4 plus or minus the fraction the square root of 3 over 4")[$x = frac(1, 4) ± frac(sqrt(3), 4)$]
]

#math.equation(block: true, alt: "open parenthesis y minus the fraction 2 over 3 close parenthesis squared equals the fraction 2 over 9")[$attach(( y − frac(2, 3) ), t: 2) = frac(2, 9)$]

#math.equation(block: true, alt: "open parenthesis n minus 4 close parenthesis squared minus 50 equals 150")[$attach(( n − 4 ), t: 2) − 50 = 150$]

#solutionbox[
#math.equation(block: true, alt: "n equals 4 plus or minus 10 the square root of 2")[$n = 4 ± 10 sqrt(2)$]
]

#math.equation(block: true, alt: "open parenthesis 4 c minus 1 close parenthesis squared equals −18")[$attach(( 4 c − 1 ), t: 2) = −18$]

#math.equation(block: true, alt: "n squared plus 10 n plus 25 equals 12")[$n^(2) + 10 n + 25 = 12$]

#solutionbox[
#math.equation(block: true, alt: "n equals −5 plus or minus 2 the square root of 3")[$n = −5 ± 2 sqrt(3)$]
]

#math.equation(block: true, alt: "64 a squared plus 48 a plus 9 equals 81")[$64 a^(2) + 48 a + 9 = 81$]


==== Solve Quadratic Equations by Completing the Square

#strong[Solve Quadratic Equations Using Completing the Square]

In the following exercises, complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

#math.equation(block: true, alt: "x squared plus 22 x")[$x^(2) + 22 x$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x plus 11 close parenthesis squared")[$attach(( x + 11 ), t: 2)$]
]

#math.equation(block: true, alt: "m squared minus 8 m")[$m^(2) − 8 m$]

#math.equation(block: true, alt: "a squared minus 3 a")[$a^(2) − 3 a$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis a minus the fraction 3 over 2 close parenthesis squared")[$attach(( a − frac(3, 2) ), t: 2)$]
]

#math.equation(block: true, alt: "b squared plus 13 b")[$b^(2) + 13 b$]

In the following exercises, solve by completing the square.

#math.equation(block: true, alt: "d squared plus 14 d equals −13")[$d^(2) + 14 d = −13$]

#solutionbox[
#math.equation(block: true, alt: "d equals −13 , −1")[$d = −13 , −1$]
]

#math.equation(block: true, alt: "y squared minus 6 y equals 36")[$y^(2) − 6 y = 36$]

#math.equation(block: true, alt: "m squared plus 6 m equals −109")[$m^(2) + 6 m = −109$]

#solutionbox[
#math.equation(block: true, alt: "m equals −3 plus or minus 10 i")[$m = −3 ± 10 i$]
]

#math.equation(block: true, alt: "t squared minus 12 t equals −40")[$t^(2) − 12 t = −40$]

#math.equation(block: true, alt: "v squared minus 14 v equals −31")[$v^(2) − 14 v = −31$]

#solutionbox[
#math.equation(block: true, alt: "v equals 7 plus or minus 3 the square root of 2")[$v = 7 ± 3 sqrt(2)$]
]

#math.equation(block: true, alt: "w squared minus 20 w equals 100")[$w^(2) − 20 w = 100$]

#math.equation(block: true, alt: "m squared plus 10 m minus 4 equals −13")[$m^(2) + 10 m − 4 = −13$]

#solutionbox[
#math.equation(block: true, alt: "m equals −9 , −1")[$m = −9 , −1$]
]

#math.equation(block: true, alt: "n squared minus 6 n plus 11 equals 34")[$n^(2) − 6 n + 11 = 34$]

#math.equation(block: true, alt: "a squared equals 3 a plus 8")[$a^(2) = 3 a + 8$]

#solutionbox[
#math.equation(block: true, alt: "a equals the fraction 3 over 2 plus or minus the fraction the square root of 41 over 2")[$a = frac(3, 2) ± frac(sqrt(41), 2)$]
]

#math.equation(block: true, alt: "b squared equals 11 b minus 5")[$b^(2) = 11 b − 5$]

#math.equation(block: true, alt: "open parenthesis u plus 8 close parenthesis open parenthesis u plus 4 close parenthesis equals 14")[$( u + 8 ) ( u + 4 ) = 14$]

#solutionbox[
#math.equation(block: true, alt: "u equals −6 plus or minus 3 the square root of 2")[$u = −6 ± 3 sqrt(2)$]
]

#math.equation(block: true, alt: "open parenthesis z minus 10 close parenthesis open parenthesis z plus 2 close parenthesis equals 28")[$( z − 10 ) ( z + 2 ) = 28$]

#strong[Solve Quadratic Equations of the form #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0 by Completing the Square]

In the following exercises, solve by completing the square.

#math.equation(block: true, alt: "3 p squared minus 18 p plus 15 equals 15")[$3 p^(2) − 18 p + 15 = 15$]

#solutionbox[
#math.equation(block: true, alt: "p equals 0 , 6")[$p = 0 , 6$]
]

#math.equation(block: true, alt: "5 q squared plus 70 q plus 20 equals 0")[$5 q^(2) + 70 q + 20 = 0$]

#math.equation(block: true, alt: "4 y squared minus 6 y equals 4")[$4 y^(2) − 6 y = 4$]

#solutionbox[
#math.equation(block: true, alt: "y equals minus the fraction 1 over 2 , 2")[$y = − frac(1, 2) , 2$]
]

#math.equation(block: true, alt: "2 x squared plus 2 x equals 4")[$2 x^(2) + 2 x = 4$]

#math.equation(block: true, alt: "3 c squared plus 2 c equals 9")[$3 c^(2) + 2 c = 9$]

#solutionbox[
#math.equation(block: true, alt: "c equals minus the fraction 1 over 3 plus or minus the fraction 2 the square root of 7 over 3")[$c = − frac(1, 3) ± frac(2 sqrt(7), 3)$]
]

#math.equation(block: true, alt: "4 d squared minus 2 d equals 8")[$4 d^(2) − 2 d = 8$]

#math.equation(block: true, alt: "2 x squared plus 6 x equals −5")[$2 x^(2) + 6 x = −5$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 3 over 2 plus or minus the fraction 1 over 2 i")[$x = frac(3, 2) ± frac(1, 2) i$]
]

#math.equation(block: true, alt: "2 x squared plus 4 x equals −5")[$2 x^(2) + 4 x = −5$]


==== Solve Quadratic Equations Using the Quadratic Formula

In the following exercises, solve by using the Quadratic Formula.

#math.equation(block: true, alt: "4 x squared minus 5 x plus 1 equals 0")[$4 x^(2) − 5 x + 1 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 1 over 4 , 1")[$x = frac(1, 4) , 1$]
]

#math.equation(block: true, alt: "7 y squared plus 4 y minus 3 equals 0")[$7 y^(2) + 4 y − 3 = 0$]

#math.equation(block: true, alt: "r squared minus r minus 42 equals 0")[$r^(2) − r − 42 = 0$]

#solutionbox[
#math.equation(block: true, alt: "r equals −6 , 7")[$r = −6 , 7$]
]

#math.equation(block: true, alt: "t squared plus 13 t plus 22 equals 0")[$t^(2) + 13 t + 22 = 0$]

#math.equation(block: true, alt: "4 v squared plus v minus 5 equals 0")[$4 v^(2) + v − 5 = 0$]

#solutionbox[
#math.equation(block: true, alt: "v equals minus the fraction 5 over 4 , 1")[$v = − frac(5, 4) , #h(0.2em) 1$]
]

#math.equation(block: true, alt: "2 w squared plus 9 w plus 2 equals 0")[$2 w^(2) + 9 w + 2 = 0$]

#math.equation(block: true, alt: "3 m squared plus 8 m plus 2 equals 0")[$3 m^(2) + 8 m + 2 = 0$]

#solutionbox[
#math.equation(block: true, alt: "m equals the fraction −4 plus or minus the square root of 10 over 3")[$m = frac(−4 ± sqrt(10), 3)$]
]

#math.equation(block: true, alt: "5 n squared plus 2 n minus 1 equals 0")[$5 n^(2) + 2 n − 1 = 0$]

#math.equation(block: true, alt: "6 a squared minus 5 a plus 2 equals 0")[$6 a^(2) − 5 a + 2 = 0$]

#solutionbox[
#math.equation(block: true, alt: "a equals the fraction 5 over 12 plus or minus the fraction the square root of 23 over 12 i")[$a = frac(5, 12) ± frac(sqrt(23), 12) i$]
]

#math.equation(block: true, alt: "4 b squared minus b plus 8 equals 0")[$4 b^(2) − b + 8 = 0$]

#math.equation(block: true, alt: "u open parenthesis u minus 10 close parenthesis plus 3 equals 0")[$u ( u − 10 ) + 3 = 0$]

#solutionbox[
#math.equation(block: true, alt: "u equals 5 plus or minus the square root of 22")[$u = 5 ± sqrt(22)$]
]

#math.equation(block: true, alt: "5 z open parenthesis z minus 2 close parenthesis equals 3")[$5 z ( z − 2 ) = 3$]

#math.equation(block: true, alt: "the fraction 1 over 8 p squared minus the fraction 1 over 5 p equals minus the fraction 1 over 20")[$frac(1, 8) p^(2) − frac(1, 5) p = − frac(1, 20)$]

#solutionbox[
#math.equation(block: true, alt: "p equals the fraction 4 plus or minus the square root of 6 over 5")[$p = frac(4 ± sqrt(6), 5)$]
]

#math.equation(block: true, alt: "the fraction 2 over 5 q squared plus the fraction 3 over 10 q equals the fraction 1 over 10")[$frac(2, 5) q^(2) + frac(3, 10) q = frac(1, 10)$]

#math.equation(block: true, alt: "4 c squared plus 4 c plus 1 equals 0")[$4 c^(2) + 4 c + 1 = 0$]

#solutionbox[
#math.equation(block: true, alt: "c equals minus the fraction 1 over 2")[$c = − frac(1, 2)$]
]

#math.equation(block: true, alt: "9 d squared minus 12 d equals −4")[$9 d^(2) − 12 d = −4$]

#strong[Use the Discriminant to Predict the Number of Solutions of a Quadratic Equation]

In the following exercises, determine the number of solutions for each quadratic equation.

ⓐ #math.equation(block: false, alt: "9 x squared minus 6 x plus 1 equals 0")[$9 x^(2) − 6 x + 1 = 0$] #linebreak() ⓑ #math.equation(block: false, alt: "3 y squared minus 8 y plus 1 equals 0")[$3 y^(2) − 8 y + 1 = 0$] #linebreak() ⓒ #math.equation(block: false, alt: "7 m squared plus 12 m plus 4 equals 0")[$7 m^(2) + 12 m + 4 = 0$] #linebreak() ⓓ #math.equation(block: false, alt: "5 n squared minus n plus 1 equals 0")[$5 n^(2) − n + 1 = 0$]

#solutionbox[
ⓐ 1 ⓑ 2 ⓒ 2 ⓓ 0
]

ⓐ #math.equation(block: false, alt: "5 x squared minus 7 x minus 8 equals 0")[$5 x^(2) − 7 x − 8 = 0$] #linebreak() ⓑ #math.equation(block: false, alt: "7 x squared minus 10 x plus 5 equals 0")[$7 x^(2) − 10 x + 5 = 0$] #linebreak() ⓒ #math.equation(block: false, alt: "25 x squared minus 90 x plus 81 equals 0")[$25 x^(2) − 90 x + 81 = 0$] #linebreak() ⓓ #math.equation(block: false, alt: "15 x squared minus 8 x plus 4 equals 0")[$15 x^(2) − 8 x + 4 = 0$]

#strong[Identify the Most Appropriate Method to Use to Solve a Quadratic Equation]

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve.

ⓐ #math.equation(block: false, alt: "16 r squared minus 8 r plus 1 equals 0")[$16 r^(2) − 8 r + 1 = 0$] #linebreak() ⓑ #math.equation(block: false, alt: "5 t squared minus 8 t plus 3 equals 9")[$5 t^(2) − 8 t + 3 = 9$] #linebreak() ⓒ #math.equation(block: false, alt: "3 open parenthesis c plus 2 close parenthesis squared equals 15")[$3 attach(( c + 2 ), t: 2) = 15$]

#solutionbox[
ⓐ factor ⓑ Quadratic Formula ⓒ square root
]

ⓐ #math.equation(block: false, alt: "4 d squared plus 10 d minus 5 equals 21")[$4 d^(2) + 10 d − 5 = 21$] #linebreak() ⓑ #math.equation(block: false, alt: "25 x squared minus 60 x plus 36 equals 0")[$25 x^(2) − 60 x + 36 = 0$] #linebreak() ⓒ #math.equation(block: false, alt: "6 open parenthesis 5 v minus 7 close parenthesis squared equals 150")[$6 attach(( 5 v − 7 ), t: 2) = 150$]


==== Solve Equations in Quadratic Form

#strong[Solve Equations in Quadratic Form]

In the following exercises, solve.

#math.equation(block: true, alt: "x to the power 4 minus 14 x squared plus 24 equals 0")[$x^(4) − 14 x^(2) + 24 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals plus or minus the square root of 2 , plus or minus 2 the square root of 3")[$x = ± sqrt(2) , #h(0.2em) ± 2 sqrt(3)$]
]

#math.equation(block: true, alt: "x to the power 4 plus 4 x squared minus 32 equals 0")[$x^(4) + 4 x^(2) − 32 = 0$]

#math.equation(block: true, alt: "4 x to the power 4 minus 5 x squared plus 1 equals 0")[$4 x^(4) − 5 x^(2) + 1 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals plus or minus 1 , plus or minus the fraction 1 over 2")[$x = ± 1 , #h(0.2em) ± frac(1, 2)$]
]

#math.equation(block: true, alt: "open parenthesis 2 y plus 3 close parenthesis squared plus 3 open parenthesis 2 y plus 3 close parenthesis minus 28 equals 0")[$attach(( 2 y + 3 ), t: 2) + 3 ( 2 y + 3 ) − 28 = 0$]

#math.equation(block: true, alt: "x plus 3 the square root of x minus 28 equals 0")[$x + 3 sqrt(x) − 28 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals 16")[$x = 16$]
]

#math.equation(block: true, alt: "6 x plus 5 the square root of x minus 6 equals 0")[$6 x + 5 sqrt(x) − 6 = 0$]

#math.equation(block: true, alt: "x to the power the fraction 2 over 3 minus 10 x to the power the fraction 1 over 3 plus 24 equals 0")[$x^(frac(2, 3)) − 10 x^(frac(1, 3)) + 24 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals 64 , 216")[$x = 64 , #h(0.2em) 216$]
]

#math.equation(block: true, alt: "x plus 7 x to the power the fraction 1 over 2 plus 6 equals 0")[$x + 7 x^(frac(1, 2)) + 6 = 0$]

#math.equation(block: true, alt: "8 x to the power −2 minus 2 x to the power −1 minus 3 equals 0")[$8 x^(−2) − 2 x^(−1) − 3 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals −2 , the fraction 4 over 3")[$x = −2 , #h(0.2em) frac(4, 3)$]
]


==== Solve Applications of Quadratic Equations

#strong[Solve Applications Modeled by Quadratic Equations]

In the following exercises, solve by using the method of factoring, the square root principle, or the Quadratic Formula. Round your answers to the nearest tenth, if needed.

Find two consecutive odd numbers whose product is 323.

Find two consecutive even numbers whose product is 624.

#solutionbox[
Two consecutive even numbers whose product is 624 are 24 and 26, and −24 and −26.
]

A triangular banner has an area of 351 square centimeters. The length of the base is two centimeters longer than four times the height. Find the height and length of the base.

Julius built a triangular display case for his coin collection. The height of the display case is six inches less than twice the width of the base. The area of the of the back of the case is 70 square inches. Find the height and width of the case.

#solutionbox[
The height is 14 inches and the width is 10 inches.
]

A tile mosaic in the shape of a right triangle is used as the corner of a rectangular pathway. The hypotenuse of the mosaic is 5 feet. One side of the mosaic is twice as long as the other side. What are the lengths of the sides? Round to the nearest tenth.

#figure(figph[A rectangle is shown is a right triangle in the corner. The hypotenuse of the triangle is 5 feet, the longer leg is 2 times s and the shorter leg is s.], alt: "A rectangle is shown is a right triangle in the corner. The hypotenuse of the triangle is 5 feet, the longer leg is 2 times s and the shorter leg is s.", caption: none)

A rectangular piece of plywood has a diagonal which measures two feet more than the width. The length of the plywood is twice the width. What is the length of the plywood’s diagonal? Round to the nearest tenth.

#solutionbox[
The length of the diagonal is 3.6 feet.
]

The front walk from the street to Pam’s house has an area of 250 square feet. Its length is two less than four times its width. Find the length and width of the sidewalk. Round to the nearest tenth.

For Sophia’s graduation party, several tables of the same width will be arranged end to end to give serving table with a total area of 75 square feet. The total length of the tables will be two more than three times the width. Find the length and width of the serving table so Sophia can purchase the correct size tablecloth . Round answer to the nearest tenth.

#solutionbox[
The width of the serving table is 4.7 feet and the length is 16.1 feet. #linebreak() 
#figure(figph[Four tables arranged end-to-end are shown. Together, they have an area of 75 feet. The short side measures w and the long side measures 3 times w plus 2.], alt: "Four tables arranged end-to-end are shown. Together, they have an area of 75 feet. The short side measures w and the long side measures 3 times w plus 2.", caption: none)
]

A ball is thrown vertically in the air with a velocity of 160 ft/sec. Use the formula #emph[h] = −16#emph[t]#super[2] + #emph[v]#sub[0]#emph[t] to determine when the ball will be 384 feet from the ground. Round to the nearest tenth.

The couple took a small airplane for a quick flight up to the wine country for a romantic dinner and then returned home. The plane flew a total of 5 hours and each way the trip was 360 miles. If the plane was flying at 150 mph, what was the speed of the wind that affected the plane?

#solutionbox[
The speed of the wind was 30 mph.
]

Ezra kayaked up the river and then back in a total time of 6 hours. The trip was 4 miles each way and the current was difficult. If Roy kayaked at a speed of 5 mph, what was the speed of the current?

Two handymen can do a home repair in 2 hours if they work together. One of the men takes 3 hours more than the other man to finish the job by himself. How long does it take for each handyman to do the home repair individually?

#solutionbox[
One man takes 3 hours and the other man 6 hours to finish the repair alone.
]


==== Graph Quadratic Functions Using Properties

#strong[Recognize the Graph of a Quadratic Function]

In the following exercises, graph by plotting point.

Graph #math.equation(block: false, alt: "y equals x squared minus 2")[$y = x^(2) − 2$]

Graph #math.equation(block: false, alt: "y equals − x squared plus 3")[$y = "−" x^(2) + 3$]

#solutionbox[
#figure(figph[This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (3, 0) and other points of (negative 2, negative 1) and (2, negative 1).], alt: "This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (3, 0) and other points of (negative 2, negative 1) and (2, negative 1).", caption: none)
]

In the following exercises, determine if the following parabolas open up or down.

ⓐ #math.equation(block: false, alt: "y equals −3 x squared plus 3 x minus 1")[$y = −3 x^(2) + 3 x − 1$] #linebreak() ⓑ #math.equation(block: false, alt: "y equals 5 x squared plus 6 x plus 3")[$y = 5 x^(2) + 6 x + 3$]

ⓐ #math.equation(block: false, alt: "y equals x squared plus 8 x minus 1")[$y = x^(2) + 8 x − 1$] #linebreak() ⓑ #math.equation(block: false, alt: "y equals −4 x squared minus 7 x plus 1")[$y = −4 x^(2) − 7 x + 1$]

#solutionbox[
ⓐ up ⓑ down
]

#strong[Find the Axis of Symmetry and Vertex of a Parabola]

In the following exercises, find ⓐ the equation of the axis of symmetry and ⓑ the vertex.

#math.equation(block: true, alt: "y equals − x squared plus 6 x plus 8")[$y = "−" x^(2) + 6 x + 8$]

#math.equation(block: true, alt: "y equals 2 x squared minus 8 x plus 1")[$y = 2 x^(2) − 8 x + 1$]

#solutionbox[
#math.equation(block: true, alt: "x equals 2 ; open parenthesis 2 , −7 close parenthesis")[$x = 2 ; ( 2 , −7 )$]
]

#strong[Find the Intercepts of a Parabola]

In the following exercises, find the #emph[x]- and #emph[y]-intercepts.

#math.equation(block: true, alt: "y equals x squared minus 4 x minus 5")[$y = x^(2) − 4 x − 5$]

#math.equation(block: true, alt: "y equals x squared minus 8 x plus 15")[$y = x^(2) − 8 x + 15$]

#solutionbox[
#math.equation(block: true, alt: "y : open parenthesis 0 , 15 close parenthesis; x : open parenthesis 3 , 0 close parenthesis , open parenthesis 5 , 0 close parenthesis")[$y : ( 0 , 15 ) \ x : ( 3 , 0 ) , ( 5 , 0 )$]
]

#math.equation(block: true, alt: "y equals x squared minus 4 x plus 10")[$y = x^(2) − 4 x + 10$]

#math.equation(block: true, alt: "y equals −5 x squared minus 30 x minus 46")[$y = −5 x^(2) − 30 x − 46$]

#solutionbox[
#math.equation(block: true, alt: "y : open parenthesis 0 , −46 close parenthesis; x : none")[$y : ( 0 , −46 ) \ x : "none"$]
]

#math.equation(block: true, alt: "y equals 16 x squared minus 8 x plus 1")[$y = 16 x^(2) − 8 x + 1$]

#math.equation(block: true, alt: "y equals x squared plus 16 x plus 64")[$y = x^(2) + 16 x + 64$]

#solutionbox[
#math.equation(block: true, alt: "y : open parenthesis 0 , 64 close parenthesis; x : open parenthesis −8 , 0 close parenthesis")[$y : ( 0 , 64 ) \ x : ( −8 , 0 )$]
]

#strong[Graph Quadratic Functions Using Properties]

In the following exercises, graph by using its properties.

#math.equation(block: true, alt: "y equals x squared plus 8 x plus 15")[$y = x^(2) + 8 x + 15$]

#math.equation(block: true, alt: "y equals x squared minus 2 x minus 3")[$y = x^(2) − 2 x − 3$]

#solutionbox[
#figure(figph[This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (1, negative 4) and a y-intercept of (0, negative 3).], alt: "This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (1, negative 4) and a y-intercept of (0, negative 3).", caption: none)
]

#math.equation(block: true, alt: "y equals − x squared plus 8 x minus 16")[$y = "−" x^(2) + 8 x − 16$]

#math.equation(block: true, alt: "y equals 4 x squared minus 4 x plus 1")[$y = 4 x^(2) − 4 x + 1$]

#solutionbox[
#figure(figph[This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (one-half, 0) and a y-intercept of (0, 1).], alt: "This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (one-half, 0) and a y-intercept of (0, 1).", caption: none)
]

#math.equation(block: true, alt: "y equals x squared plus 6 x plus 13")[$y = x^(2) + 6 x + 13$]

#math.equation(block: true, alt: "y equals −2 x squared minus 8 x minus 12")[$y = −2 x^(2) − 8 x − 12$]

#solutionbox[
#figure(figph[This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 2, negative 4) and a y-intercept of (0, negative 12).], alt: "This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 2, negative 4) and a y-intercept of (0, negative 12).", caption: none)
]

#strong[Solve Maximum and Minimum Applications]

In the following exercises, find the minimum or maximum value.

#math.equation(block: true, alt: "y equals 7 x squared plus 14 x plus 6")[$y = 7 x^(2) + 14 x + 6$]

#math.equation(block: true, alt: "y equals −3 x squared plus 12 x minus 10")[$y = −3 x^(2) + 12 x − 10$]

#solutionbox[
The maximum value is 2 when #emph[x] = 2.
]

In the following exercises, solve. Rounding answers to the nearest tenth.

A ball is thrown upward from the ground with an initial velocity of 112 ft/sec. Use the quadratic equation #emph[h] = −16#emph[t]#super[2] + 112#emph[t] to find how long it will take the ball to reach maximum height, and then find the maximum height.

A daycare facility is enclosing a rectangular area along the side of their building for the children to play outdoors. They need to maximize the area using 180 feet of fencing on three sides of the yard. The quadratic equation #math.equation(block: false, alt: "A open parenthesis x close parenthesis equals x 90 minus the fraction x over 2")[$A ( x ) = x 90 − frac(x, 2)$] gives the area, #emph[A], of the yard for the length, #emph[x], of the building that will border the yard. Find the length of the building that should border the yard to maximize the area, and then find the maximum area.

#figure(figph[An odd-shaped figure is given. 3 sides of a rectangle are attached to the right side of the figure.], alt: "An odd-shaped figure is given. 3 sides of a rectangle are attached to the right side of the figure.", caption: none)

#solutionbox[
The length adjacent to the building is 90 feet giving a maximum area of 4,050 square feet.
]


==== Graph Quadratic Functions Using Transformations

#strong[Graph Quadratic Functions of the form #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared plus k")[$f ( x ) = x^(2) + k$]]

In the following exercises, graph each function using a vertical shift.

#math.equation(block: true, alt: "g open parenthesis x close parenthesis equals x squared plus 4")[$g ( x ) = x^(2) + 4$]

#math.equation(block: true, alt: "h open parenthesis x close parenthesis equals x squared minus 3")[$h ( x ) = x^(2) − 3$]

#solutionbox[
#figure(figph[This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 3, 0) and other points of (negative 1, negative 2) and (1, negative 2).], alt: "This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 3, 0) and other points of (negative 1, negative 2) and (1, negative 2).", caption: none)
]

In the following exercises, graph each function using a horizontal shift.

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis x plus 1 close parenthesis squared")[$f ( x ) = attach(( x + 1 ), t: 2)$]

#math.equation(block: true, alt: "g open parenthesis x close parenthesis equals open parenthesis x minus 3 close parenthesis squared")[$g ( x ) = attach(( x − 3 ), t: 2)$]

#solutionbox[
#figure(figph[This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (3, 0) and other points of (2, 1) and (4,1).], alt: "This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (3, 0) and other points of (2, 1) and (4,1).", caption: none)
]

In the following exercises, graph each function using transformations.

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis x plus 2 close parenthesis squared plus 3")[$f ( x ) = attach(( x + 2 ), t: 2) + 3$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis x plus 3 close parenthesis squared minus 2")[$f ( x ) = attach(( x + 3 ), t: 2) − 2$]

#solutionbox[
#figure(figph[This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 3, negative 2) and other points of (negative 5, 2) and (negative 1, 2).], alt: "This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 3, negative 2) and other points of (negative 5, 2) and (negative 1, 2).", caption: none)
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis x minus 1 close parenthesis squared plus 4")[$f ( x ) = attach(( x − 1 ), t: 2) + 4$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis x minus 4 close parenthesis squared minus 3")[$f ( x ) = attach(( x − 4 ), t: 2) − 3$]

#solutionbox[
#figure(figph[This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (4, negative 3) and other points of (3, negative 2) and (5, negative 2).], alt: "This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (4, negative 3) and other points of (3, negative 2) and (5, negative 2).", caption: none)
]

#strong[Graph Quadratic Functions of the form #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a x squared")[$f ( x ) = a x^(2)$]]

In the following exercises, graph each function.

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 x squared")[$f ( x ) = 2 x^(2)$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals − x squared")[$f ( x ) = "−" x^(2)$]

#solutionbox[
#figure(figph[This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (0, 0) and other points of (negative 1, negative 1) and (1, negative 1).], alt: "This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (0, 0) and other points of (negative 1, negative 1) and (1, negative 1).", caption: none)
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction 1 over 2 x squared")[$f ( x ) = frac(1, 2) x^(2)$]

#strong[Graph Quadratic Functions Using Transformations]

In the following exercises, rewrite each function in the #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a open parenthesis x minus h close parenthesis squared plus k")[$f ( x ) = a attach(( x − h ), t: 2) + k$] form by completing the square.

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 x squared minus 4 x minus 4")[$f ( x ) = 2 x^(2) − 4 x − 4$]

#solutionbox[
#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 open parenthesis x minus 1 close parenthesis squared minus 6")[$f ( x ) = 2 attach(( x − 1 ), t: 2) − 6$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 3 x squared plus 12 x plus 8")[$f ( x ) = 3 x^(2) + 12 x + 8$]

In the following exercises, ⓐ rewrite each function in #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a open parenthesis x minus h close parenthesis squared plus k")[$f ( x ) = a attach(( x − h ), t: 2) + k$] form and ⓑ graph it by using transformations.

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 3 x squared minus 6 x minus 1")[$f ( x ) = 3 x^(2) − 6 x − 1$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 open parenthesis x minus 1 close parenthesis squared minus 4")[$f ( x ) = 3 attach(( x − 1 ), t: 2) − 4$] #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (1, negative 4) and other points of (0, negative 1) and (2, negative 1).], alt: "This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (1, negative 4) and other points of (0, negative 1) and (2, negative 1).", caption: none)
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals −2 x squared minus 12 x minus 5")[$f ( x ) = −2 x^(2) − 12 x − 5$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 x squared plus 4 x plus 6")[$f ( x ) = 2 x^(2) + 4 x + 6$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 open parenthesis x plus 1 close parenthesis squared plus 4")[$f ( x ) = 2 attach(( x + 1 ), t: 2) + 4$] #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 1, 4) and other points of (negative 2, 6) and (0, 6).], alt: "This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 1, 4) and other points of (negative 2, 6) and (0, 6).", caption: none)
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 3 x squared minus 12 x plus 7")[$f ( x ) = 3 x^(2) − 12 x + 7$]

In the following exercises, ⓐ rewrite each function in #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a open parenthesis x minus h close parenthesis squared plus k")[$f ( x ) = a attach(( x − h ), t: 2) + k$] form and ⓑ graph it using properties.

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals −3 x squared minus 12 x minus 5")[$f ( x ) = −3 x^(2) − 12 x − 5$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals −3 open parenthesis x plus 2 close parenthesis squared plus 7")[$f ( x ) = −3 attach(( x + 2 ), t: 2) + 7$] #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 2, 7) and other points of (negative 4, negative 5) and (0, negative 5).], alt: "This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 2, 7) and other points of (negative 4, negative 5) and (0, negative 5).", caption: none)
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 x squared minus 12 x plus 7")[$f ( x ) = 2 x^(2) − 12 x + 7$]

#strong[Find a Quadratic Function from its Graph]

In the following exercises, write the quadratic function in #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a open parenthesis x minus h close parenthesis squared plus k")[$f ( x ) = a attach(( x − h ), t: 2) + k$] form.

#figure(figph[This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 1, negative 1) and other points of (negative 2, negative 4) and (0, negative 4).], alt: "This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 1, negative 1) and other points of (negative 2, negative 4) and (0, negative 4).", caption: none)

#solutionbox[
#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis x plus 1 close parenthesis squared minus 5")[$f ( x ) = attach(( x + 1 ), t: 2) − 5$]
]

#figure(figph[This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (2, 4) and other points of (0, 8) and (4, 8).], alt: "This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (2, 4) and other points of (0, 8) and (4, 8).", caption: none)


==== Solve Quadratic Inequalities

#strong[Solve Quadratic Inequalities Graphically]

In the following exercises, solve graphically and write the solution in interval notation.

#math.equation(block: true, alt: "x squared minus x minus 6 greater than 0")[$x^(2) − x − 6 > 0$]

#solutionbox[
ⓐ  #linebreak() #figure(figph[This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (one-half, negative 6 and one-fourth) and other points of (0, negative 6) and (1, negative 6).], alt: "This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (one-half, negative 6 and one-fourth) and other points of (0, negative 6) and (1, negative 6).", caption: none)

 #linebreak() 
ⓑ #math.equation(block: false, alt: "open parenthesis minus ∞ , −2 close parenthesis union open parenthesis 3 , ∞ close parenthesis")[$( − ∞ , −2 ) ∪ ( 3 , ∞ )$]
]

#math.equation(block: true, alt: "x squared plus 4 x plus 3 less than or equal to 0")[$x^(2) + 4 x + 3 ≤ 0$]

#math.equation(block: true, alt: "− x squared minus x plus 2 greater than or equal to 0")[$"−" x^(2) − x + 2 ≥ 0$]

#solutionbox[
ⓐ  #linebreak() #figure(figph[This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (negative one-half, 2 and one-fourth) and other points of (negative 2, 0) and (1, 0).], alt: "This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (negative one-half, 2 and one-fourth) and other points of (negative 2, 0) and (1, 0).", caption: none)

 #linebreak() 
ⓑ #math.equation(block: false, alt: "[ −2 , 1 ]")[$[ −2 , 1 ]$]
]

#math.equation(block: true, alt: "− x squared plus 2 x plus 3 less than 0")[$"−" x^(2) + 2 x + 3 < 0$]

In the following exercises, solve each inequality algebraically and write any solution in interval notation.

#math.equation(block: true, alt: "x squared minus 6 x plus 8 less than 0")[$x^(2) − 6 x + 8 < 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis 2 , 4 close parenthesis")[$( 2 , 4 )$]
]

#math.equation(block: true, alt: "x squared plus x greater than 12")[$x^(2) + x > 12$]

#math.equation(block: true, alt: "x squared minus 6 x plus 4 less than or equal to 0")[$x^(2) − 6 x + 4 ≤ 0$]

#solutionbox[
#math.equation(block: true, alt: "[ 3 minus the square root of 5 , 3 plus the square root of 5 ]")[$[ 3 − sqrt(5) , 3 + sqrt(5) ]$]
]

#math.equation(block: true, alt: "2 x squared plus 7 x minus 4 greater than 0")[$2 x^(2) + 7 x − 4 > 0$]

#math.equation(block: true, alt: "− x squared plus x minus 6 greater than 0")[$"−" x^(2) + x − 6 > 0$]

#solutionbox[
no solution
]

#math.equation(block: true, alt: "x squared minus 2 x plus 4 greater than or equal to 0")[$x^(2) − 2 x + 4 ≥ 0$]


=== Practice Test

Use the Square Root Property to solve the quadratic equation #math.equation(block: false, alt: "3 open parenthesis w plus 5 close parenthesis squared equals 27 .")[$3 attach(( w + 5 ), t: 2) = 27 .$]

#solutionbox[
#math.equation(block: true, alt: "w equals −2 , w equals −8")[$w = −2 , w = −8$]
]

Use Completing the Square to solve the quadratic equation #math.equation(block: false, alt: "a squared minus 8 a plus 7 equals 23 .")[$a^(2) − 8 a + 7 = 23 .$]

Use the Quadratic Formula to solve the quadratic equation #math.equation(block: false, alt: "2 m squared minus 5 m plus 3 equals 0 .")[$2 m^(2) − 5 m + 3 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "m equals 1 , m equals the fraction 3 over 2")[$m = 1 , m = frac(3, 2)$]
]

Solve the following quadratic equations. Use any method.

#math.equation(block: true, alt: "2 x open parenthesis 3 x minus 2 close parenthesis minus 1 equals 0")[$2 x ( 3 x − 2 ) − 1 = 0$]

#math.equation(block: true, alt: "the fraction 9 over 4 y squared minus 3 y plus 1 equals 0")[$frac(9, 4) y^(2) − 3 y + 1 = 0$]

#solutionbox[
#math.equation(block: true, alt: "y equals the fraction 2 over 3")[$y = frac(2, 3)$]
]

Use the discriminant to determine the number and type of solutions of each quadratic equation.

#math.equation(block: true, alt: "6 p squared minus 13 p plus 7 equals 0")[$6 p^(2) − 13 p + 7 = 0$]

#math.equation(block: true, alt: "3 q squared minus 10 q plus 12 equals 0")[$3 q^(2) − 10 q + 12 = 0$]

#solutionbox[
2 complex
]

Solve each equation.

#math.equation(block: true, alt: "4 x to the power 4 minus 17 x squared plus 4 equals 0")[$4 x^(4) − 17 x^(2) + 4 = 0$]

#math.equation(block: true, alt: "y to the power the fraction 2 over 3 plus 2 y to the power the fraction 1 over 3 minus 3 equals 0")[$y^(frac(2, 3)) + 2 y^(frac(1, 3)) − 3 = 0$]

#solutionbox[
#math.equation(block: true, alt: "y equals 1 , y equals −27")[$y = 1 , y = −27$]
]

For each parabola, find ⓐ which direction it opens, ⓑ the equation of the axis of symmetry, ⓒ the vertex, ⓓ the #emph[x-] and #emph[y]-intercepts, and e) the maximum or minimum value.

#math.equation(block: true, alt: "y equals 3 x squared plus 6 x plus 8")[$y = 3 x^(2) + 6 x + 8$]

#math.equation(block: true, alt: "y equals minus x squared minus 8 x minus 16")[$y = − x^(2) − 8 x − 16$]

#solutionbox[
ⓐ down ⓑ #math.equation(block: false, alt: "x equals −4")[$x = −4$] #linebreak() ⓒ #math.equation(block: false, alt: "open parenthesis −4 , 0 close parenthesis")[$( −4 , 0 )$] ⓓ #math.equation(block: false, alt: "y : open parenthesis 0 , −16 close parenthesis ; x : open parenthesis −4 , 0 close parenthesis")[$y ":" #h(0.2em) ( 0 , −16 ) ";" #h(0.2em) x ":" #h(0.2em) ( −4 , 0 )$] #linebreak() ⓔ maximum value of #math.equation(block: false, alt: "0")[$0$] when #math.equation(block: false, alt: "x equals −4")[$x = −4$]
]

Graph each quadratic function using intercepts, the vertex, and the equation of the axis of symmetry.

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x squared plus 6 x plus 9")[$f ( x ) = x^(2) + 6 x + 9$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals −2 x squared plus 8 x plus 4")[$f ( x ) = −2 x^(2) + 8 x + 4$]

#solutionbox[
#figure(figph[This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (2, 12) and other points of (0, 4) and (4, 4).], alt: "This figure shows a downward-opening parabola on the x y-coordinate plane. It has a vertex of (2, 12) and other points of (0, 4) and (4, 4).", caption: none)
]

In the following exercises, graph each function using transformations.

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis x plus 3 close parenthesis squared plus 2")[$f ( x ) = attach(( x + 3 ), t: 2) + 2$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x squared minus 4 x minus 1")[$f ( x ) = x^(2) − 4 x − 1$]

#solutionbox[
#figure(figph[This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (2, negative 5) and other points of (0, negative 1) and (4, negative 1).], alt: "This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (2, negative 5) and other points of (0, negative 1) and (4, negative 1).", caption: none)

 #linebreak() 
#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 open parenthesis x minus 1 close parenthesis squared minus 6")[$f ( x ) = 2 attach(( x − 1 ), t: 2) − 6$]
]

In the following exercises, solve each inequality algebraically and write any solution in interval notation.

#math.equation(block: true, alt: "x squared minus 6 x minus 8 less than or equal to 0")[$x^(2) − 6 x − 8 ≤ 0$]

#math.equation(block: true, alt: "2 x squared plus x minus 10 greater than 0")[$2 x^(2) + x − 10 > 0$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis − ∞ , minus the fraction 5 over 2 close parenthesis union open parenthesis 2 , ∞ close parenthesis")[$( "−" ∞ , − frac(5, 2) ) ∪ ( 2 , ∞ )$]
]

Model the situation with a quadratic equation and solve by any method.

Find two consecutive even numbers whose product is 360.

The length of a diagonal of a rectangle is three more than the width. The length of the rectangle is three times the width. Find the length of the diagonal. (Round to the nearest tenth.)

#solutionbox[
The diagonal is 4.4 units long.
]

A water balloon is launched upward at the rate of 86 ft/sec. Using the formula #emph[h] = −16#emph[t]#super[2] + 86#emph[t] find how long it will take the balloon to reach the maximum height, and then find the maximum height. Round to the nearest tenth.