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== 9.3#h(0.6em)Solve Quadratic Equations Using the Quadratic Formula

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

Evaluate #math.equation(block: false, alt: "b squared minus 4 a b")[$b^(2) − 4 a b$] when #math.equation(block: false, alt: "a equals 3")[$a = 3$] and #math.equation(block: false, alt: "b equals −2 .")[$b = −2 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "28")[$28$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Simplify: #math.equation(block: false, alt: "the square root of 108 .")[$sqrt(108) .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "6 the square root of 3")[$6 sqrt(3)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Simplify: #math.equation(block: false, alt: "the square root of 50 .")[$sqrt(50) .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "5 the square root of 2")[$5 sqrt(2)$]
]
]


=== Solve Quadratic Equations Using the Quadratic Formula

When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes’. Mathematicians look for patterns when they do things over and over in order to make their work easier. In this section we will derive and use a formula to find the solution of a quadratic equation.

We have already seen how to solve a formula for a specific variable ‘in general’, so that we would do the algebraic steps only once, and then use the new formula to find the value of the specific variable. Now we will go through the steps of completing the square using the general form of a quadratic equation to solve a quadratic equation for #emph[x.]

We start with the standard form of a quadratic equation and solve it for #emph[x] by completing the square.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A quadratic equation, ax^2 + bx + c = 0, where 'a' cannot be zero, is shown in this image. It's a fundamental algebraic expression used to find the roots of a second-degree polynomial.], alt: "A quadratic equation, ax^2 + bx + c = 0, where 'a' cannot be zero, is shown in this image. It's a fundamental algebraic expression used to find the roots of a second-degree polynomial.", caption: none)]),
  [Isolate the variable terms on one side.], [#figure(figph[A mathematical equation displays 'ax^2 + bx = -c' in gray text against a plain white background, representing a quadratic equation.], alt: "A mathematical equation displays 'ax^2 + bx = -c' in gray text against a plain white background, representing a quadratic equation.", caption: none)],
  [Make the coefficient of #math.equation(block: false, alt: "x squared")[$x^(2)$] equal to 1, by #linebreak() dividing by #emph[a].], [#figure(figph[A mathematical equation showing the division of a quadratic equation by 'a', resulting in x^2 + (b/a)x = -c/a, a step in solving for x by completing the square.], alt: "A mathematical equation showing the division of a quadratic equation by 'a', resulting in x^2 + (b/a)x = -c/a, a step in solving for x by completing the square.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation is displayed, showing x squared plus b over a times x equals negative c over a, representing a step in solving a quadratic equation by completing the square.], alt: "A mathematical equation is displayed, showing x squared plus b over a times x equals negative c over a, representing a step in solving a quadratic equation by completing the square.", caption: none)],
  [To complete the square, find #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 times the fraction b over a close parenthesis squared")[$attach(( frac(1, 2) · frac(b, a) ), t: 2)$] and add it to both sides of the equation.], [],
  [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 the fraction b over a close parenthesis squared equals the fraction b squared over 4 a squared")[$attach(( frac(1, 2) frac(b, a) ), t: 2) = frac(b^(2), 4 a^(2))$]], [#figure(figph[A mathematical equation demonstrating a step in completing the square: x^2 + (b/a)x + (b^2)/(4a^2) = -c/a + (b^2)/(4a^2), with the term b^2/(4a^2) highlighted in red on both sides.], alt: "A mathematical equation demonstrating a step in completing the square: x^2 + (b/a)x + (b^2)/(4a^2) = -c/a + (b^2)/(4a^2), with the term b^2/(4a^2) highlighted in red on both sides.", caption: none)],
  [The left side is a perfect square, factor it.], [#figure(figph[The mathematical equation (x + b/(2a))^2 = -c/a + b^2/(4a^2) is shown on a white background. It represents an intermediate step in solving quadratic equations.], alt: "The mathematical equation (x + b/(2a))^2 = -c/a + b^2/(4a^2) is shown on a white background. It represents an intermediate step in solving quadratic equations.", caption: none)],
  [Find the common denominator of the right #linebreak() side and write equivalent fractions with #linebreak() the common denominator.], [#figure(figph[A mathematical equation illustrating a step in completing the square, showing (x + b/(2a))^2 equals b^2/(4a^2) minus (c/a) multiplied by (4a/4a).], alt: "A mathematical equation illustrating a step in completing the square, showing (x + b/(2a))^2 equals b^2/(4a^2) minus (c/a) multiplied by (4a/4a).", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation illustrating a step in solving for x using the completing the square method, which is commonly used to derive the quadratic formula.], alt: "A mathematical equation illustrating a step in solving for x using the completing the square method, which is commonly used to derive the quadratic formula.", caption: none)],
  [Combine to one fraction.], [#figure(figph[A mathematical equation shows (x + b/(2a))^2 = (b^2 - 4ac)/(4a^2), representing an intermediate step in deriving the quadratic formula by completing the square.], alt: "A mathematical equation shows (x + b/(2a))^2 = (b^2 - 4ac)/(4a^2), representing an intermediate step in deriving the quadratic formula by completing the square.", caption: none)],
  [Use the square root property.], [#figure(figph[A mathematical equation shows x plus b over 2a equals plus or minus the square root of b squared minus 4ac, all divided by 4a squared, illustrating a step in deriving the quadratic formula.], alt: "A mathematical equation shows x plus b over 2a equals plus or minus the square root of b squared minus 4ac, all divided by 4a squared, illustrating a step in deriving the quadratic formula.", caption: none)],
  [Simplify the radical.], [#figure(figph[A mathematical equation shown as x + b/(2a) = +/- sqrt(b^2 - 4ac) / (2a), a step in deriving the quadratic formula, isolated on a white background.], alt: "A mathematical equation shown as x + b/(2a) = +/- sqrt(b^2 - 4ac) / (2a), a step in deriving the quadratic formula, isolated on a white background.", caption: none)],
  [Add #math.equation(block: false, alt: "minus the fraction b over 2 a")[$− frac(b, 2 a)$] to both sides of the equation.], [#figure(figph[The quadratic formula, showing the solutions for x are equal to negative b divided by two a, plus or minus the square root of b squared minus four a c, all divided by two a.], alt: "The quadratic formula, showing the solutions for x are equal to negative b divided by two a, plus or minus the square root of b squared minus four a c, all divided by two a.", caption: none)],
  [Combine the terms on the right side.], [#figure(figph[The quadratic formula, x = (-b ×1×4b^2 - 4ac) / 2a, is displayed, used to find the roots of a quadratic equation. This fundamental algebraic equation is a key tool in mathematics for solving polynomials of degree 2.], alt: "The quadratic formula, x = (-b ×1×4b^2 - 4ac) / 2a, is displayed, used to find the roots of a quadratic equation. This fundamental algebraic equation is a key tool in mathematics for solving polynomials of degree 2.", caption: none)],
  [], [This equation is the Quadratic Formula.],
))

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

The solutions to a #strong[quadratic equation] of the form #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0, where #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$] are given by the formula:

#math.equation(block: true, alt: "x equals the fraction − b plus or minus the square root of b squared minus 4 a c over 2 a")[$x = frac("−" b ± sqrt(b^(2) − 4 a c), 2 a)$]
]

To use the #strong[Quadratic Formula], we substitute the values of #emph[a], #emph[b], and #emph[c] from the standard form into the expression on the right side of the formula. Then we simplify the expression. The result is the pair of solutions to the quadratic equation.

Notice the formula is an equation. Make sure you use both sides of the equation.

#examplebox("Example 1")[How to Solve a Quadratic Equation Using the Quadratic Formula][
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "2 x squared plus 9 x minus 5 equals 0 .")[$2 x^(2) + 9 x − 5 = 0 .$]

#solutionbox[
#figure(figph[Step 1 is to write the quadratic equation in standard form, a times x squared plus b x plus c equals zero, and identify the values a, b, and c. The equation 2 x squared plus 9 x minus 5 equals zero is in standard form. A equals 2, b equals 9, and c equals negative 5.], alt: "Step 1 is to write the quadratic equation in standard form, a times x squared plus b x plus c equals zero, and identify the values a, b, and c. The equation 2 x squared plus 9 x minus 5 equals zero is in standard form. A equals 2, b equals 9, and c equals negative 5.", caption: none)

#figure(figph[Step 2. Write the quadratic formula. Then substitute the values of a, b, and c. Substitute a equals 2, b equals 9, and c equals negative 5 into the equation x equals the quotient negative b plus or minus the square root of the difference b squared minus 4 a c divided by 2 a. So x equals the quotient negative 9 plus or minus the square root of the difference 9 squared minus the product 4 times 2 times negative 5 divided by the product 2 times 2.], alt: "Step 2. Write the quadratic formula. Then substitute the values of a, b, and c. Substitute a equals 2, b equals 9, and c equals negative 5 into the equation x equals the quotient negative b plus or minus the square root of the difference b squared minus 4 a c divided by 2 a. So x equals the quotient negative 9 plus or minus the square root of the difference 9 squared minus the product 4 times 2 times negative 5 divided by the product 2 times 2.", caption: none)

#figure(figph[In step 3, simplify the fraction and solve for x. x equals the quotient negative 9 plus or minus the square root of the difference 81 minus negative 40 divided by 4. Simplify the radicand. x equals the quotient negative 9 plus or minus the square root of 121 divided by 4. Simplify the square root. x equals the quotient negative 9 plus or minus 11 divided by 4. Separate into two equations. The first equation is x equals the quotient negative 9 plus 11 divided by 4 which simplifies to 2 divided by 4. The first solution is x equals one half. The second equation is x equals the quotient negative 9 minus 11 divided by 4 which simplifies to negative 20 divided by 4. The second solution is x equals negative 5.], alt: "In step 3, simplify the fraction and solve for x. x equals the quotient negative 9 plus or minus the square root of the difference 81 minus negative 40 divided by 4. Simplify the radicand. x equals the quotient negative 9 plus or minus the square root of 121 divided by 4. Simplify the square root. x equals the quotient negative 9 plus or minus 11 divided by 4. Separate into two equations. The first equation is x equals the quotient negative 9 plus 11 divided by 4 which simplifies to 2 divided by 4. The first solution is x equals one half. The second equation is x equals the quotient negative 9 minus 11 divided by 4 which simplifies to negative 20 divided by 4. The second solution is x equals negative 5.", caption: none)

#figure(figph[The fourth, and final, step is to check the solution. Put each answer into the original equation to check. First, substitute x equals one half into the original equation, 2 x squared plus 9 x minus 5 equals 0. This yields 2 times the square of one half plus nine times one half minus 5. We need to show that this expression equals 0. Simplify the square. 2 times one fourth plus nine times one half minus 5 equals one half plus 9 halves minus 5, or 10 halves minus 5. 5 minus 5 equals 0, so x equals one half is indeed a solution. Next substitute x = negative 5 into the equation 2 x squared plus 9 x minus 5 equals 0. This yields 2 times the square of negative 5 plus 9 times negative 5 minus 5. We need to show that this expression equals 0. Simplify the square. 2 times 25 plus nine times negative 5 minus 5 equals 50 minus 45 minus 5, or 0. x equals negative 5 is a solution as well.], alt: "The fourth, and final, step is to check the solution. Put each answer into the original equation to check. First, substitute x equals one half into the original equation, 2 x squared plus 9 x minus 5 equals 0. This yields 2 times the square of one half plus nine times one half minus 5. We need to show that this expression equals 0. Simplify the square. 2 times one fourth plus nine times one half minus 5 equals one half plus 9 halves minus 5, or 10 halves minus 5. 5 minus 5 equals 0, so x equals one half is indeed a solution. Next substitute x = negative 5 into the equation 2 x squared plus 9 x minus 5 equals 0. This yields 2 times the square of negative 5 plus 9 times negative 5 minus 5. We need to show that this expression equals 0. Simplify the square. 2 times 25 plus nine times negative 5 minus 5 equals 50 minus 45 minus 5, or 0. x equals negative 5 is a solution as well.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "3 y squared minus 5 y plus 2 equals 0")[$3 y^(2) − 5 y + 2 = 0$].

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

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "4 z squared plus 2 z minus 6 equals 0")[$4 z^(2) + 2 z − 6 = 0$].

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

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Solve a quadratic equation using the quadratic formula.]

+ Write the quadratic equation in standard form, #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0. Identify the values of #emph[a], #emph[b], and #emph[c].
+ Write the Quadratic Formula. Then substitute in the values of #emph[a], #emph[b], and #emph[c].
+ Simplify.
+ Check the solutions.
]

If you say the formula as you write it in each problem, you’ll have it memorized in no time! And remember, the Quadratic Formula is an EQUATION. Be sure you start with “#emph[x] =”.

#examplebox("Example 2")[][
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "x squared minus 6 x equals −5 .")[$x^(2) − 6 x = −5 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical equation is displayed, showing 'x squared minus 6x equals minus 5' against a white background.], alt: "A mathematical equation is displayed, showing 'x squared minus 6x equals minus 5' against a white background.", caption: none)]),
  [Write the equation in standard form by adding #linebreak() 5 to each side.], [#figure(figph[A clear, focused image displays the quadratic equation x^2 - 6x + 5 = 0 presented in black text against a plain white background.], alt: "A clear, focused image displays the quadratic equation x^2 - 6x + 5 = 0 presented in black text against a plain white background.", caption: none)],
  [This equation is now in standard form.], [#figure(figph[Two quadratic equations are displayed: the general form ax^2 + bx + c = 0 in red, and a specific example x^2 - 6x + 5 = 0 in black.], alt: "Two quadratic equations are displayed: the general form ax^2 + bx + c = 0 in red, and a specific example x^2 - 6x + 5 = 0 in black.", caption: none)],
  [Identify the values of #math.equation(block: false, alt: "a , b , c .")[$a , " " " " b , " " " " c .$]], [#figure(figph[The image displays mathematical coefficients a=1 in light blue, b=-6 in red, and c=5 in light green, set against a plain white background.], alt: "The image displays mathematical coefficients a=1 in light blue, b=-6 in red, and c=5 in light green, set against a plain white background.", caption: none)],
  [Write the Quadratic Formula.], [#figure(figph[The quadratic formula, used to find the solutions for x in a quadratic equation of the form ax^2 + bx + c = 0.], alt: "The quadratic formula, used to find the solutions for x in a quadratic equation of the form ax^2 + bx + c = 0.", caption: none)],
  [Then substitute in the values of #math.equation(block: false, alt: "a , b , c .")[$a , " " " " b , " " " " c .$]], [#figure(figph[The quadratic formula, x = -(-6) ×177 sqrt((-6)^2 - 4 \* 1 \* 5) / (2 \* 1), demonstrates the substitution of values for 'a', 'b', and 'c' to solve a quadratic equation.], alt: "The quadratic formula, x = -(-6) ×177 sqrt((-6)^2 - 4 * 1 * 5) / (2 * 1), demonstrates the substitution of values for 'a', 'b', and 'c' to solve a quadratic equation.", caption: none)],
  [Simplify.], [#figure(figph[A quadratic formula step showing x equals six plus or minus the square root of thirty-six minus twenty, all divided by two.], alt: "A quadratic formula step showing x equals six plus or minus the square root of thirty-six minus twenty, all divided by two.", caption: none)

 #linebreak() #figure(figph[A mathematical equation shows x equals the fraction of (6 plus or minus the square root of 16) all over 2. This is a step in solving a quadratic equation.], alt: "A mathematical equation shows x equals the fraction of (6 plus or minus the square root of 16) all over 2. This is a step in solving a quadratic equation.", caption: none)

 #linebreak() #figure(figph[A mathematical equation displays x equals 6 plus or minus 4, all divided by 2.], alt: "A mathematical equation displays x equals 6 plus or minus 4, all divided by 2.", caption: none)],
  [Rewrite to show two solutions.], [#figure(figph[Two mathematical expressions for 'x' are displayed: x = (6+4)/2 and x = (6-4)/2, representing two possible solutions in a mathematical context.], alt: "Two mathematical expressions for 'x' are displayed: x = (6+4)/2 and x = (6-4)/2, representing two possible solutions in a mathematical context.", caption: none)],
  [Simplify.], [#figure(figph[Two mathematical equations are displayed horizontally: x = 10/2, and x = 2/2.], alt: "Two mathematical equations are displayed horizontally: x = 10/2, and x = 2/2.", caption: none)],
  [], [#figure(figph[The image shows two mathematical expressions on a white background: 'x = 5,' and 'x = 1'.], alt: "The image shows two mathematical expressions on a white background: 'x = 5,' and 'x = 1'.", caption: none)],
  [Check: #linebreak()  #linebreak() 
#figure(figph[Verification of solutions for the quadratic equation x^2 - 6x + 5 = 0, demonstrating that both x=5 and x=1 satisfy the equation, resulting in 0=0.], alt: "Verification of solutions for the quadratic equation x^2 - 6x + 5 = 0, demonstrating that both x=5 and x=1 satisfy the equation, resulting in 0=0.", caption: none)], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "a squared minus 2 a equals 15")[$a^(2) − 2 a = 15$].

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

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "b squared plus 24 equals −10 b")[$b^(2) + 24 = −10 b$].

#solutionbox[
#math.equation(block: true, alt: "b equals −6 , b equals −4")[$b = −6 , b = −4$]
]
]

When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the #strong[Quadratic Formula]. If we get a #strong[radical] as a solution, the final answer must have the radical in its simplified form.

#examplebox("Example 3")[][
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "2 x squared plus 10 x plus 11 equals 0 .")[$2 x^(2) + 10 x + 11 = 0 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A quadratic equation is displayed, reading '2x^2 + 10x + 11 = 0'.], alt: "A quadratic equation is displayed, reading '2x^2 + 10x + 11 = 0'.", caption: none)]),
  [This equation is in standard form.], [#figure(figph[The general form of a quadratic equation, ax^2 + bx + c = 0, is shown above a specific example: 2x^2 + 10x + 11 = 0.], alt: "The general form of a quadratic equation, ax^2 + bx + c = 0, is shown above a specific example: 2x^2 + 10x + 11 = 0.", caption: none)],
  [Identify the values of #emph[a], #emph[b], and #emph[c].], [#figure(figph[The image displays the values of three variables: a = 2, b = 10, and c = 11, written in a horizontal line with different colors for each variable and its value.], alt: "The image displays the values of three variables: a = 2, b = 10, and c = 11, written in a horizontal line with different colors for each variable and its value.", caption: none)],
  [Write the Quadratic Formula.], [#figure(figph[The image displays the quadratic formula, an equation used to solve quadratic equations for the variable x. It shows x equals negative b, plus or minus the square root of b squared minus 4ac, all divided by 2a.], alt: "The image displays the quadratic formula, an equation used to solve quadratic equations for the variable x. It shows x equals negative b, plus or minus the square root of b squared minus 4ac, all divided by 2a.", caption: none)],
  [Then substitute in the values of #emph[a], #emph[b], and #emph[c].], [#figure(figph[The quadratic formula is shown with values substituted for solving an equation: x = (- (10) +/- sqrt((10)^2 - 4 \* 2 \* (11))) / (2 \* 2). The numbers 10, 2, and 11 are color-coded in red, cyan, and yellow, respectively.], alt: "The quadratic formula is shown with values substituted for solving an equation: x = (- (10) +/- sqrt((10)^2 - 4 * 2 * (11))) / (2 * 2). The numbers 10, 2, and 11 are color-coded in red, cyan, and yellow, respectively.", caption: none)],
  [Simplify.], [#figure(figph[A quadratic formula calculation shows x equals negative 10 plus or minus the square root of 100 minus 88, all divided by 4.], alt: "A quadratic formula calculation shows x equals negative 10 plus or minus the square root of 100 minus 88, all divided by 4.", caption: none)],
  [], [#figure(figph[A mathematical equation showing x equals the fraction with a numerator of -10 plus or minus the square root of 12, all divided by 4. This is a common step in the quadratic formula.], alt: "A mathematical equation showing x equals the fraction with a numerator of -10 plus or minus the square root of 12, all divided by 4. This is a common step in the quadratic formula.", caption: none)],
  [Simplify the radical.], [#figure(figph[The equation shows x equals the fraction with numerator -10 plus or minus 2 times the square root of 3, and denominator 4.], alt: "The equation shows x equals the fraction with numerator -10 plus or minus 2 times the square root of 3, and denominator 4.", caption: none)],
  [Factor out the common factor in the numerator.], [#figure(figph[The equation x equals the fraction 2 multiplied by the quantity -5 plus or minus the square root of 3, all divided by 4, is displayed on a white background.], alt: "The equation x equals the fraction 2 multiplied by the quantity -5 plus or minus the square root of 3, all divided by 4, is displayed on a white background.", caption: none)],
  [Remove the common factors.], [#figure(figph[Equation showing x equals negative five plus or minus the square root of three, all divided by two.], alt: "Equation showing x equals negative five plus or minus the square root of three, all divided by two.", caption: none)],
  [Rewrite to show two solutions.], [#figure(figph[Two mathematical expressions for x are displayed. The first is x = (-5 + sqrt(3))/2, and the second is x = (-5 - sqrt(3))/2, representing solutions to a quadratic equation.], alt: "Two mathematical expressions for x are displayed. The first is x = (-5 + sqrt(3))/2, and the second is x = (-5 - sqrt(3))/2, representing solutions to a quadratic equation.", caption: none)],
  [Check: #linebreak() 
We leave the check for you!], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "3 m squared plus 12 m plus 7 equals 0")[$3 m^(2) + 12 m + 7 = 0$].

#solutionbox[
#math.equation(block: true, alt: "m equals the fraction −6 plus the square root of 15 over 3 , m equals the fraction −6 minus the square root of 15 over 3")[$m = frac(−6 + sqrt(15), 3) , m = frac(−6 − sqrt(15), 3)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "5 n squared plus 4 n minus 4 equals 0")[$5 n^(2) + 4 n − 4 = 0$].

#solutionbox[
#math.equation(block: true, alt: "n equals the fraction −2 plus 2 the square root of 6 over 5 , n equals the fraction −2 minus 2 the square root of 6 over 5")[$n = frac(−2 + 2 sqrt(6), 5) , n = frac(−2 − 2 sqrt(6), 5)$]
]
]

When we substitute #emph[a], #emph[b], and #emph[c] into the Quadratic Formula and the #strong[radicand] is negative, the quadratic equation will have imaginary or complex solutions. We will see this in the next example.

#examplebox("Example 4")[][
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "3 p squared plus 2 p plus 9 equals 0 .")[$3 p^(2) + 2 p + 9 = 0 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A quadratic equation displayed as 3p^2 + 2p + 9 = 0.], alt: "A quadratic equation displayed as 3p^2 + 2p + 9 = 0.", caption: none)]),
  [This equation is in standard form], [#figure(figph[Two quadratic equations are displayed: the general form ax^2 + bx + c = 0 in red and a specific example 3p^2 + 2p + 9 = 0 in black.], alt: "Two quadratic equations are displayed: the general form ax^2 + bx + c = 0 in red and a specific example 3p^2 + 2p + 9 = 0 in black.", caption: none)],
  [Identify the values of #math.equation(block: false, alt: "a , b , c .")[$a , b , c .$]], [#figure(figph[The image displays the values of three variables: a=3, b=2, and c=9, written in different colors on a white background.], alt: "The image displays the values of three variables: a=3, b=2, and c=9, written in different colors on a white background.", caption: none)],
  [Write the Quadratic Formula.], [#figure(figph[The quadratic formula, an algebraic formula that provides the solution(s) for a quadratic equation, where p equals negative b plus or minus the square root of b squared minus 4ac, all divided by 2a.], alt: "The quadratic formula, an algebraic formula that provides the solution(s) for a quadratic equation, where p equals negative b plus or minus the square root of b squared minus 4ac, all divided by 2a.", caption: none)],
  [Then substitute in the values of #math.equation(block: false, alt: "a , b , c")[$a , b , c$].], [#figure(figph[A mathematical expression for 'p' using the quadratic formula, featuring numbers 2, 3, and 9. It shows p = \[-2 ×1 sqrt(2^2 - 4 \* 3 \* 9)\] / (2 \* 3).], alt: "A mathematical expression for 'p' using the quadratic formula, featuring numbers 2, 3, and 9. It shows p = [-2 ×1 sqrt(2^2 - 4 * 3 * 9)] / (2 * 3).", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation shows p equals a fraction where the numerator is -2 plus or minus the square root of (4 minus 108), and the denominator is 6.], alt: "A mathematical equation shows p equals a fraction where the numerator is -2 plus or minus the square root of (4 minus 108), and the denominator is 6.", caption: none)],
  [], [#figure(figph[A mathematical equation showing the variable p is equal to a fraction where the numerator is -2 plus or minus the square root of -104, and the denominator is 6.], alt: "A mathematical equation showing the variable p is equal to a fraction where the numerator is -2 plus or minus the square root of -104, and the denominator is 6.", caption: none)],
  [Simplify the radical using complex numbers.], [#figure(figph[The image displays a mathematical equation for the variable 'p', showing it as a complex number: p = (-2  A plus-minus sign   sqrt(104) i) / 6.], alt: "The image displays a mathematical equation for the variable 'p', showing it as a complex number: p = (-2 A plus-minus sign sqrt(104) i) / 6.", caption: none)],
  [Simplify the radical.], [#figure(figph[The image shows the mathematical equation for p, which is equal to a fraction with -2 plus or minus 2 times the square root of 26i in the numerator, and 6 in the denominator.], alt: "The image shows the mathematical equation for p, which is equal to a fraction with -2 plus or minus 2 times the square root of 26i in the numerator, and 6 in the denominator.", caption: none)],
  [Factor the common factor in the numerator.], [#figure(figph[A mathematical equation displaying 'p' equals 2 multiplied by the quantity of negative 1 plus or minus the square root of 26i, all divided by 6.], alt: "A mathematical equation displaying 'p' equals 2 multiplied by the quantity of negative 1 plus or minus the square root of 26i, all divided by 6.", caption: none)],
  [Remove the common factors.], [#figure(figph[A mathematical equation shows p equals the fraction with a numerator of -1 plus or minus the square root of 26 multiplied by i, and a denominator of 3.], alt: "A mathematical equation shows p equals the fraction with a numerator of -1 plus or minus the square root of 26 multiplied by i, and a denominator of 3.", caption: none)],
  [Rewrite in standard #math.equation(block: false, alt: "a plus b i")[$a + b i$] form.], [#figure(figph[The image shows the complex solution for p, expressed as p = -1/3 plus or minus (square root of 26 multiplied by i) all divided by 3, written in a clear, standard mathematical notation.], alt: "The image shows the complex solution for p, expressed as p = -1/3 plus or minus (square root of 26 multiplied by i) all divided by 3, written in a clear, standard mathematical notation.", caption: none)],
  [Write as two solutions.], [#figure(figph[The image displays the two complex conjugate values for 'p': p = -1/3 + (sqrt(26)i)/3 and p = -1/3 - (sqrt(26)i)/3.], alt: "The image displays the two complex conjugate values for 'p': p = -1/3 + (sqrt(26)i)/3 and p = -1/3 - (sqrt(26)i)/3.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "4 a squared minus 2 a plus 8 equals 0")[$4 a^(2) − 2 a + 8 = 0$].

#solutionbox[
#math.equation(block: true, alt: "a equals the fraction 1 over 4 plus the fraction the square root of 31 over 4 i , a equals the fraction 1 over 4 minus the fraction the square root of 31 over 4 i")[$a = frac(1, 4) + frac(sqrt(31), 4) i , #h(0.2em) #h(0.2em) a = frac(1, 4) − frac(sqrt(31), 4) i$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "5 b squared plus 2 b plus 4 equals 0")[$5 b^(2) + 2 b + 4 = 0$].

#solutionbox[
#math.equation(block: true, alt: "b equals minus the fraction 1 over 5 plus the fraction the square root of 19 over 5 i , b equals minus the fraction 1 over 5 minus the fraction the square root of 19 over 5 i")[$b = − frac(1, 5) + frac(sqrt(19), 5) i , #h(0.2em) #h(0.2em) b = − frac(1, 5) − frac(sqrt(19), 5) i$]
]
]

Remember, to use the Quadratic Formula, the equation must be written in standard form, #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0. Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

#examplebox("Example 5")[][
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "x open parenthesis x plus 6 close parenthesis plus 4 equals 0 .")[$x ( x + 6 ) + 4 = 0 .$]

#solutionbox[
Our first step is to get the equation in standard form.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical equation is displayed, showing x multiplied by the quantity (x plus 6), plus 4, which equals 0. The equation is x(x+6) + 4 = 0.], alt: "A mathematical equation is displayed, showing x multiplied by the quantity (x plus 6), plus 4, which equals 0. The equation is x(x+6) + 4 = 0.", caption: none)]),
  [Distribute to get the equation in standard form.], [#figure(figph[A mathematical equation is displayed, showing a quadratic expression set equal to zero: x^2 + 6x + 4 = 0. The equation is rendered in a clear, digital font on a plain white background.], alt: "A mathematical equation is displayed, showing a quadratic expression set equal to zero: x^2 + 6x + 4 = 0. The equation is rendered in a clear, digital font on a plain white background.", caption: none)],
  [This equation is now in standard form], [#figure(figph[Two quadratic equations are displayed: 'ax^2 + bx + c = 0' in red text, followed by 'x^2 + 6x + 4 = 0' in black text below it.], alt: "Two quadratic equations are displayed: 'ax^2 + bx + c = 0' in red text, followed by 'x^2 + 6x + 4 = 0' in black text below it.", caption: none)],
  [Identify the values of #math.equation(block: false, alt: "a , b , c .")[$a , b , c .$]], [#figure(figph[The image displays the values of three variables: a=1, b=6, and c=4, with each variable and its value presented in a distinct color, suggesting a mathematical or programming context.], alt: "The image displays the values of three variables: a=1, b=6, and c=4, with each variable and its value presented in a distinct color, suggesting a mathematical or programming context.", caption: none)],
  [Write the Quadratic Formula.], [#figure(figph[The quadratic formula: x equals negative b, plus or minus the square root of b squared minus 4ac, all divided by 2a.], alt: "The quadratic formula: x equals negative b, plus or minus the square root of b squared minus 4ac, all divided by 2a.", caption: none)],
  [Then substitute in the values of #math.equation(block: false, alt: "a , b , c")[$a , b , c$].], [#figure(figph[Mathematical equation showing the quadratic formula with 'b=6' (red), 'a=1' (light blue), and 'c=4' (yellow) color-coded and substituted into the expression x = -(6) ×plusminus ×sqrt{(6)^2 - 4 \* 1 \* 4} / (2 \* 1).], alt: "Mathematical equation showing the quadratic formula with 'b=6' (red), 'a=1' (light blue), and 'c=4' (yellow) color-coded and substituted into the expression x = -(6) ×plusminus ×sqrt{(6)^2 - 4 * 1 * 4} / (2 * 1).", caption: none)],
  [Simplify.], [#figure(figph[The image shows a step in solving a quadratic equation, with x equal to a fraction where the numerator is -6 plus or minus the square root of (36 minus 16), and the denominator is 2.], alt: "The image shows a step in solving a quadratic equation, with x equal to a fraction where the numerator is -6 plus or minus the square root of (36 minus 16), and the denominator is 2.", caption: none)],
  [], [#figure(figph[A mathematical equation is displayed: x equals the fraction with a numerator of -6 plus or minus the square root of 20, and a denominator of 2.], alt: "A mathematical equation is displayed: x equals the fraction with a numerator of -6 plus or minus the square root of 20, and a denominator of 2.", caption: none)],
  [Simplify the radical.], [#figure(figph[A mathematical equation shows x equals the fraction with a numerator of -6 plus or minus 2 times the square root of 5, all divided by 2.], alt: "A mathematical equation shows x equals the fraction with a numerator of -6 plus or minus 2 times the square root of 5, all divided by 2.", caption: none)],
  [Factor the common factor in the numerator.], [#figure(figph[A mathematical equation shows x equals 2 times the quantity of negative 3 plus or minus the square root of 5, all divided by 2. This represents a solution to a quadratic equation.], alt: "A mathematical equation shows x equals 2 times the quantity of negative 3 plus or minus the square root of 5, all divided by 2. This represents a solution to a quadratic equation.", caption: none)],
  [Remove the common factors.], [#figure(figph[A mathematical equation is displayed, showing 'x = -3 ×1 ×5' in black font against a white background.], alt: "A mathematical equation is displayed, showing 'x = -3 ×1 ×5' in black font against a white background.", caption: none)],
  [Write as two solutions.], [#figure(figph[The image displays the solutions to a quadratic equation, showing x = -3 + sqrt(5) and x = -3 - sqrt(5).], alt: "The image displays the solutions to a quadratic equation, showing x = -3 + sqrt(5) and x = -3 - sqrt(5).", caption: none)],
  [Check: #linebreak() 
We leave the check for you!], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "x open parenthesis x plus 2 close parenthesis minus 5 equals 0 .")[$x ( x + 2 ) − 5 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals −1 plus the square root of 6 , x equals −1 minus the square root of 6")[$x = −1 + sqrt(6) , x = −1 − sqrt(6)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "3 y open parenthesis y minus 2 close parenthesis minus 3 equals 0 .")[$3 y ( y − 2 ) − 3 = 0 .$]

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

When we solved linear equations, if an equation had too many fractions we cleared the fractions by multiplying both sides of the equation by the LCD. This gave us an equivalent equation—without fractions— to solve. We can use the same strategy with quadratic equations.

#examplebox("Example 6")[][
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "the fraction 1 over 2 u squared plus the fraction 2 over 3 u equals the fraction 1 over 3 .")[$frac(1, 2) u^(2) + frac(2, 3) u = frac(1, 3) .$]

#solutionbox[
Our first step is to clear the fractions.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A quadratic equation is displayed on a white background, reading one-half u squared plus two-thirds u equals one-third.], alt: "A quadratic equation is displayed on a white background, reading one-half u squared plus two-thirds u equals one-third.", caption: none)]),
  [Multiply both sides by the LCD, 6, to clear the fractions.], [#figure(figph[A mathematical equation is displayed, showing 6 multiplied by the sum of one-half u squared and two-thirds u, equaling 6 multiplied by one-third.], alt: "A mathematical equation is displayed, showing 6 multiplied by the sum of one-half u squared and two-thirds u, equaling 6 multiplied by one-third.", caption: none)],
  [Multiply.], [#figure(figph[The image displays the mathematical quadratic equation 3u^2 + 4u = 2, presented in a clear, dark gray font against a plain white background.], alt: "The image displays the mathematical quadratic equation 3u^2 + 4u = 2, presented in a clear, dark gray font against a plain white background.", caption: none)],
  [Subtract 2 to get the equation in standard form.], [#figure(figph[Two lines of equations are displayed. The first line is 'ax^2 + bx + c = 0' in red. The second line, in black, is '3u^2 + 4u - 2 = 0'.], alt: "Two lines of equations are displayed. The first line is 'ax^2 + bx + c = 0' in red. The second line, in black, is '3u^2 + 4u - 2 = 0'.", caption: none)],
  [Identify the values of #emph[a], #emph[b], and #emph[c].], [#figure(figph[The image displays the equations a = 3 in light blue, b = 4 in red, and c = 2 in yellow-green, all arranged horizontally on a white background.], alt: "The image displays the equations a = 3 in light blue, b = 4 in red, and c = 2 in yellow-green, all arranged horizontally on a white background.", caption: none)],
  [Write the Quadratic Formula.], [#figure(figph[The image displays the quadratic formula, an algebraic formula used to solve quadratic equations of the form ax^2 + bx + c = 0, where 'u' represents the variable.], alt: "The image displays the quadratic formula, an algebraic formula used to solve quadratic equations of the form ax^2 + bx + c = 0, where 'u' represents the variable.", caption: none)],
  [Then substitute in the values of #emph[a], #emph[b], and #emph[c].], [#figure(figph[An example of the quadratic formula in use, with coefficients 'b=4', 'a=3', and 'c=-2' substituted to find the value of 'u'. The values are color-coded for clarity.], alt: "An example of the quadratic formula in use, with coefficients 'b=4', 'a=3', and 'c=-2' substituted to find the value of 'u'. The values are color-coded for clarity.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation, 'u = (-4 plus or minus sqrt(16 + 24)) / 6'], alt: "A mathematical equation, 'u = (-4 plus or minus sqrt(16 + 24)) / 6'", caption: none)],
  [], [#figure(figph[A mathematical equation shows 'u equals negative four plus or minus the square root of forty, all divided by six'.], alt: "A mathematical equation shows 'u equals negative four plus or minus the square root of forty, all divided by six'.", caption: none)],
  [Simplify the radical.], [#figure(figph[The image shows a mathematical equation where 'u' equals the fraction whose numerator is '-4 plus or minus 2 times the square root of 10' and whose denominator is '6'.], alt: "The image shows a mathematical equation where 'u' equals the fraction whose numerator is '-4 plus or minus 2 times the square root of 10' and whose denominator is '6'.", caption: none)],
  [Factor the common factor in the numerator.], [#figure(figph[A mathematical equation is displayed where 'u' is equal to 2 multiplied by the quantity of -2 plus or minus the square root of 10, all divided by 6, set against a plain white background.], alt: "A mathematical equation is displayed where 'u' is equal to 2 multiplied by the quantity of -2 plus or minus the square root of 10, all divided by 6, set against a plain white background.", caption: none)],
  [Remove the common factors.], [#figure(figph[A mathematical equation is displayed, showing u equals a fraction: negative 2 plus or minus the square root of 10, all divided by 3.], alt: "A mathematical equation is displayed, showing u equals a fraction: negative 2 plus or minus the square root of 10, all divided by 3.", caption: none)],
  [Rewrite to show two solutions.], [#figure(figph[Two mathematical expressions are shown for the variable u. The first is u equals negative two plus the square root of ten, all divided by three. The second is u equals negative two minus the square root of ten, all divided by three.], alt: "Two mathematical expressions are shown for the variable u. The first is u equals negative two plus the square root of ten, all divided by three. The second is u equals negative two minus the square root of ten, all divided by three.", caption: none)],
  [Check: #linebreak() 
We leave the check for you!], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "the fraction 1 over 4 c squared minus the fraction 1 over 3 c equals the fraction 1 over 12")[$frac(1, 4) c^(2) − frac(1, 3) c = frac(1, 12)$].

#solutionbox[
#math.equation(block: true, alt: "c equals the fraction 2 plus the square root of 7 over 3 , c equals the fraction 2 minus the square root of 7 over 3")[$c = frac(2 + sqrt(7), 3) , #h(0.2em) #h(0.2em) c = frac(2 − sqrt(7), 3)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "the fraction 1 over 9 d squared minus the fraction 1 over 2 d equals minus the fraction 1 over 3")[$frac(1, 9) d^(2) − frac(1, 2) d = − frac(1, 3)$].

#solutionbox[
#math.equation(block: true, alt: "d equals the fraction 9 plus the square root of 33 over 4 , d equals the fraction 9 minus the square root of 33 over 4")[$d = frac(9 + sqrt(33), 4) , #h(0.2em) "d" = frac(9 − sqrt(33), 4)$]
]
]

Think about the equation (#emph[x] − 3)#super[2] = 0. We know from the #strong[Zero Product Property] that this equation has only one solution, #linebreak() #emph[x] = 3.

We will see in the next example how using the #strong[Quadratic Formula] to solve an equation whose standard form is a perfect square #strong[trinomial] equal to 0 gives just one solution. Notice that once the radicand is simplified it becomes 0 , which leads to only one solution.

#examplebox("Example 7")[][
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "4 x squared minus 20 x equals −25 .")[$4 x^(2) − 20 x = −25 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A quadratic equation, 4x^2 - 20x = -25, is displayed on a white background.], alt: "A quadratic equation, 4x^2 - 20x = -25, is displayed on a white background.", caption: none)]),
  [Add 25 to get the equation in standard form.], [#figure(figph[A quadratic equation in its general form (ax² + bx + c = 0) is shown, followed by a specific example (4x² - 20x + 25 = 0).], alt: "A quadratic equation in its general form (ax² + bx + c = 0) is shown, followed by a specific example (4x² - 20x + 25 = 0).", caption: none)],
  [Identify the values of #emph[a], #emph[b], and #emph[c].], [#figure(figph[The image displays mathematical variable assignments: a = 4, b = 20, c = 25, shown in blue, red, and yellow text, respectively, against a white background.], alt: "The image displays mathematical variable assignments: a = 4, b = 20, c = 25, shown in blue, red, and yellow text, respectively, against a white background.", caption: none)],
  [Write the quadratic formula.], [#figure(figph[The quadratic formula, x = (-b ×1 sqrt(b^2 - 4ac)) / 2a, is displayed in black text on a white background.], alt: "The quadratic formula, x = (-b ×1 sqrt(b^2 - 4ac)) / 2a, is displayed in black text on a white background.", caption: none)],
  [Then substitute in the values of #emph[a], #emph[b], and #emph[c].], [#figure(figph[An image showing the quadratic formula x = -b ± sqrt(b^2 - 4ac) / 2a with values b=-20, a=4, and c=25 substituted, displaying an intermediate step in solving a quadratic equation.], alt: "An image showing the quadratic formula x = -b ± sqrt(b^2 - 4ac) / 2a with values b=-20, a=4, and c=25 substituted, displaying an intermediate step in solving a quadratic equation.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation shows x equals the fraction with a numerator of 20 plus or minus the square root of 400 minus 400, and a denominator of 8.], alt: "A mathematical equation shows x equals the fraction with a numerator of 20 plus or minus the square root of 400 minus 400, and a denominator of 8.", caption: none)],
  [], [#figure(figph[A mathematical equation showing x equals twenty plus or minus the square root of zero, all divided by eight.], alt: "A mathematical equation showing x equals twenty plus or minus the square root of zero, all divided by eight.", caption: none)],
  [Simplify the radical.], [#figure(figph[A mathematical equation shows 'x = 20/8' written in black against a white background.], alt: "A mathematical equation shows 'x = 20/8' written in black against a white background.", caption: none)],
  [Simplify the fraction.], [#figure(figph[The image displays the mathematical equation 'x = 5/2' in black text against a plain white background, showing the value of the variable x as a fraction.], alt: "The image displays the mathematical equation 'x = 5/2' in black text against a plain white background, showing the value of the variable x as a fraction.", caption: none)],
  [Check: #linebreak() 
We leave the check for you!], [],
))

Did you recognize that 4#emph[x]#super[2] − 20#emph[x] + 25 is a perfect square trinomial. It is equivalent to (2#emph[x] − 5)#super[2]? If you solve #linebreak() 4#emph[x]#super[2] − 20#emph[x] + 25 = 0 by factoring and then using the Square Root Property, do you get the same result?
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "r squared plus 10 r plus 25 equals 0 .")[$r^(2) + 10 r + 25 = 0 .$]

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

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve by using the Quadratic Formula: #math.equation(block: false, alt: "25 t squared minus 40 t equals −16 .")[$25 t^(2) − 40 t = −16 .$]

#solutionbox[
#math.equation(block: true, alt: "t equals the fraction 4 over 5")[$t = frac(4, 5)$]
]
]


=== Use the Discriminant to Predict the Number and Type of Solutions of a Quadratic Equation

When we solved the quadratic equations in the previous examples, sometimes we got two real solutions, one real solution, and sometimes two complex solutions. Is there a way to predict the number and type of solutions to a quadratic equation without actually solving the equation?

Yes, the expression under the radical of the Quadratic Formula makes it easy for us to determine the number and type of solutions. This expression is called the #strong[discriminant].

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

#figure(figph[In the Quadratic Formula, x equals the quotient of negative b plus or minus the square root of b squared minus 4 times a times c and 2 a, the value under the radical, b squared minus 4 times a times c, is called the discriminant.], alt: "In the Quadratic Formula, x equals the quotient of negative b plus or minus the square root of b squared minus 4 times a times c and 2 a, the value under the radical, b squared minus 4 times a times c, is called the discriminant.", caption: none)
]

Let’s look at the discriminant of the equations in some of the examples and the number and type of solutions to those quadratic equations.

#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([Quadratic Equation #linebreak() (in standard form)], [Discriminant #linebreak() #math.equation(block: false, alt: "b squared minus 4 a c")[$b^(2) − 4 a c$]], [Value of the Discriminant], [Number and Type of solutions]),
  [#math.equation(block: false, alt: "2 x squared plus 9 x minus 5 equals 0")[$2 x^(2) + 9 x − 5 = 0$]], [#math.equation(block: false, alt: "9 squared minus 4 times 2 open parenthesis −5 close parenthesis; 121")[$9^(2) − 4 · 2 ( −5 ) \ 121$]], [+], [2 real],
  [#math.equation(block: false, alt: "4 x squared minus 20 x plus 25 equals 0")[$4 x^(2) − 20 x + 25 = 0$]], [#math.equation(block: false, alt: "open parenthesis −20 close parenthesis squared minus 4 times 4 times 25; 0")[$attach(( −20 ), t: 2) − 4 · 4 · 25 \ 0$]], [0], [1 real],
  [#math.equation(block: false, alt: "3 p squared plus 2 p plus 9 equals 0")[$3 p^(2) + 2 p + 9 = 0$]], [#math.equation(block: false, alt: "2 squared minus 4 times 3 times 9; minus 104")[$2^(2) − 4 · 3 · 9 \  \ − 104$]], [−], [2 complex],
))

#figure(figph[When the value under the radical in the Quadratic Formula, the discriminant, is positive, the equation has two real solutions. When the value under the radical in the Quadratic Formula, the discriminant, is zero, the equation has one real solution. When the value under the radical in the Quadratic Formula, the discriminant, is negative, the equation has two complex solutions.], alt: "When the value under the radical in the Quadratic Formula, the discriminant, is positive, the equation has two real solutions. When the value under the radical in the Quadratic Formula, the discriminant, is zero, the equation has one real solution. When the value under the radical in the Quadratic Formula, the discriminant, is negative, the equation has two complex solutions.", caption: none)

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Using the Discriminant, #emph[b]#super[2] − 4#emph[ac], to Determine the Number and Type of Solutions of a Quadratic Equation]

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 ,")[$a ≠ 0 ,$]

- If #emph[b]#super[2] − 4#emph[ac] \> 0, the equation has 2 real solutions.
- if #emph[b]#super[2] − 4#emph[ac] = 0, the equation has 1 real solution.
- if #emph[b]#super[2] − 4#emph[ac] \< 0, the equation has 2 complex solutions.
]

#examplebox("Example 8")[][
Determine the number of solutions to each quadratic equation.

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

#solutionbox[
To determine the number of solutions of each quadratic equation, we will look at its discriminant.

ⓐ

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "3 x squared plus 7 x minus 9 equals 0")[$3 x^(2) + 7 x − 9 = 0$]]),
  [The equation is in standard form, identify #emph[a], #emph[b], and #emph[c].], [#math.equation(block: false, alt: "a equals 3 , b equals 7 , c equals −9")[$a = 3 , #h(0.5em) b = 7 , #h(0.2em) c = −9$]],
  [Write the discriminant.], [#math.equation(block: false, alt: "b squared minus 4 a c")[$b^(2) − 4 a c$]],
  [Substitute in the values of #emph[a], #emph[b], and #emph[c].], [#math.equation(block: false, alt: "open parenthesis 7 close parenthesis squared minus 4 times 3 times open parenthesis −9 close parenthesis")[$attach(( 7 ), t: 2) − 4 · 3 · ( −9 )$]],
  [Simplify.], [#math.equation(block: false, alt: "49 plus 108")[$49 + 108$]],
  [], [#math.equation(block: false, alt: "157")[$157$]],
))

Since the discriminant is positive, there are 2 real solutions to the equation.

ⓑ

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "5 n squared plus n plus 4 equals 0")[$5 n^(2) + n + 4 = 0$]]),
  [The equation is in standard form, identify #emph[a], #emph[b], and #emph[c].], [#math.equation(block: false, alt: "a equals 5 , b equals 1 , c equals 4")[$a = 5 , #h(0.5em) b = 1 , #h(0.5em) c = 4$]],
  [Write the discriminant.], [#math.equation(block: false, alt: "b squared minus 4 a c")[$b^(2) − 4 a c$]],
  [Substitute in the values of #emph[a], #emph[b], and #emph[c].], [#math.equation(block: false, alt: "open parenthesis 1 close parenthesis squared minus 4 times 5 times 4")[$attach(( 1 ), t: 2) − 4 · 5 · 4$]],
  [Simplify.], [#math.equation(block: false, alt: "1 minus 80")[$1 − 80$]],
  [], [#math.equation(block: false, alt: "−79")[$−79$]],
))

Since the discriminant is negative, there are 2 complex solutions to the equation.

ⓒ

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "9 y squared minus 6 y plus 1 equals 0")[$9 y^(2) − 6 y + 1 = 0$]]),
  [The equation is in standard form, identify #emph[a], #emph[b], and #emph[c].], [#math.equation(block: false, alt: "a equals 9 , b equals −6 , c equals 1")[$a = 9 , b = −6 , c = 1$]],
  [Write the discriminant.], [#math.equation(block: false, alt: "b squared minus 4 a c")[$b^(2) − 4 a c$]],
  [Substitute in the values of #emph[a], #emph[b], and #emph[c].], [#math.equation(block: false, alt: "open parenthesis −6 close parenthesis squared minus 4 times 9 times 1")[$attach(( −6 ), t: 2) − 4 · 9 · 1$]],
  [Simplify.], [#math.equation(block: false, alt: "36 minus 36")[$36 − 36$]],
  [], [#math.equation(block: false, alt: "0")[$0$]],
))

Since the discriminant is 0, there is 1 real solution to the equation.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Determine the numberand type of solutions to each quadratic equation.

ⓐ #math.equation(block: false, alt: "8 m squared minus 3 m plus 6 equals 0")[$8 m^(2) − 3 m + 6 = 0$] ⓑ #math.equation(block: false, alt: "5 z squared plus 6 z minus 2 equals 0")[$5 z^(2) + 6 z − 2 = 0$] ⓒ #math.equation(block: false, alt: "9 w squared plus 24 w plus 16 equals 0 .")[$9 w^(2) + 24 w + 16 = 0 .$]

#solutionbox[
ⓐ 2 complex solutions; ⓑ 2 real solutions; ⓒ 1 real solution
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Determine the number and type of solutions to each quadratic equation.

ⓐ #math.equation(block: false, alt: "b squared plus 7 b minus 13 equals 0")[$b^(2) + 7 b − 13 = 0$] ⓑ #math.equation(block: false, alt: "5 a squared minus 6 a plus 10 equals 0")[$5 a^(2) − 6 a + 10 = 0$] ⓒ #math.equation(block: false, alt: "4 r squared minus 20 r plus 25 equals 0 .")[$4 r^(2) − 20 r + 25 = 0 .$]

#solutionbox[
ⓐ 2 real solutions; ⓑ 2 complex solutions; ⓒ 1 real solution
]
]


=== Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

We summarize the four methods that we have used to solve quadratic equations below.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Methods for Solving Quadratic Equations]

+ Factoring
+ Square Root Property
+ Completing the Square
+ Quadratic Formula
]

Given that we have four methods to use to solve a quadratic equation, how do you decide which one to use? Factoring is often the quickest method and so we try it first. If the equation is #math.equation(block: false, alt: "a x squared equals k")[$a x^(2) = k$] or #math.equation(block: false, alt: "a open parenthesis x minus h close parenthesis squared equals k")[$a attach(( x − h ), t: 2) = k$] we use the Square Root Property. For any other equation, it is probably best to use the Quadratic Formula.  Remember, you can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method.

What about the method of Completing the Square? Most people find that method cumbersome and prefer not to use it. We needed to include it in the list of methods because we completed the square in general to derive the Quadratic Formula. You will also use the process of Completing the Square in other areas of algebra.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Identify the most appropriate method to solve a quadratic equation.]

+ Try #strong[Factoring] first. If the quadratic factors easily, this method is very quick.
+ Try the #strong[Square Root Property] next. If the equation fits the form #math.equation(block: false, alt: "a x squared equals k")[$a x^(2) = k$] or #math.equation(block: false, alt: "a open parenthesis x minus h close parenthesis squared equals k ,")[$a attach(( x − h ), t: 2) = k ,$] it can easily be solved by using the Square Root Property.
+ Use the #strong[Quadratic Formula]. Any other quadratic equation is best solved by using the Quadratic Formula.
]

The next example uses this strategy to decide how to solve each quadratic equation.

#examplebox("Example 9")[][
Identify the most appropriate method to use to solve each quadratic equation.

ⓐ #math.equation(block: false, alt: "5 z squared equals 17")[$5 z^(2) = 17$] ⓑ #math.equation(block: false, alt: "4 x squared minus 12 x plus 9 equals 0")[$4 x^(2) − 12 x + 9 = 0$] ⓒ #math.equation(block: false, alt: "8 u squared plus 6 u equals 11 .")[$8 u^(2) + 6 u = 11 .$]

#solutionbox[
ⓐ  #linebreak() 
#math.equation(block: false, alt: "5 z squared equals 17")[$#h(2em) 5 z^(2) = 17$]

Since the equation is in the #math.equation(block: false, alt: "a x squared equals k ,")[$a x^(2) = k ,$] the most appropriate method is to use the Square Root Property.

ⓑ  #linebreak() 
#math.equation(block: false, alt: "4 x squared minus 12 x plus 9 equals 0")[$#h(3.5em) 4 x^(2) − 12 x + 9 = 0$]

We recognize that the left side of the equation is a perfect square trinomial, and so factoring will be the most appropriate method.

ⓒ

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "8 u squared plus 6 u equals 11")[$8 u^(2) + 6 u = 11$]]),
  [Put the equation in standard form.], [#math.equation(block: false, alt: "8 u squared plus 6 u minus 11 equals 0")[$8 u^(2) + 6 u − 11 = 0$]],
))

While our first thought may be to try factoring, thinking about all the possibilities for trial and error method leads us to choose the Quadratic Formula as the most appropriate method.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Identify the most appropriate method to use to solve each quadratic equation.

ⓐ #math.equation(block: false, alt: "x squared plus 6 x plus 8 equals 0")[$x^(2) + 6 x + 8 = 0$] ⓑ #math.equation(block: false, alt: "open parenthesis n minus 3 close parenthesis squared equals 16")[$attach(( n − 3 ), t: 2) = 16$] ⓒ #math.equation(block: false, alt: "5 p squared minus 6 p equals 9 .")[$5 p^(2) − 6 p = 9 .$]

#solutionbox[
ⓐ factoring; ⓑ Square Root Property; ⓒ Quadratic Formula
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Identify the most appropriate method to use to solve each quadratic equation.

ⓐ #math.equation(block: false, alt: "8 a squared plus 3 a minus 9 equals 0")[$8 a^(2) + 3 a − 9 = 0$] ⓑ #math.equation(block: false, alt: "4 b squared plus 4 b plus 1 equals 0")[$4 b^(2) + 4 b + 1 = 0$] ⓒ #math.equation(block: false, alt: "5 c squared equals 125 .")[$5 c^(2) = 125 .$]

#solutionbox[
ⓐ Quadratic Forumula; #linebreak() ⓑ Factoring or Square Root Property ⓒ Square Root Property
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Access these online resources for additional instruction and practice with using the Quadratic Formula.

- #link("https://openstax.org/l/37QuadForm1")[Using the Quadratic Formula]
- #link("https://openstax.org/l/37QuadForm2")[Solve a Quadratic Equation Using the Quadratic Formula with Complex Solutions]
- #link("https://openstax.org/l/37QuadForm3")[Discriminant in Quadratic Formula]
]


=== Key Concepts

- Quadratic Formula
- The solutions to 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")[$a ≠ 0$] are given by the formula: #linebreak() 
#math.equation(block: true, alt: "x equals the fraction − b plus or minus the square root of b squared minus 4 a c over 2 a")[$x = frac("−" b ± sqrt(b^(2) − 4 a c), 2 a)$]
- How to solve a quadratic equation using the Quadratic Formula.
+ Write the quadratic equation in standard form, #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0. Identify the values of #emph[a], #emph[b], #emph[c].
+ Write the Quadratic Formula. Then substitute in the values of #emph[a], #emph[b], #emph[c].
+ Simplify.
+ Check the solutions.
- Using the Discriminant, #emph[b]#super[2] − 4#emph[ac], to Determine the Number and Type of Solutions of a Quadratic Equation
- 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 ,")[$a ≠ 0 ,$]
- If #emph[b]#super[2] − 4#emph[ac] \> 0, the equation has 2 real solutions.
- if #emph[b]#super[2] − 4#emph[ac] = 0, the equation has 1 real solution.
- if #emph[b]#super[2] − 4#emph[ac] \< 0, the equation has 2 complex solutions.
- Methods to Solve Quadratic Equations:
- Factoring
- Square Root Property
- Completing the Square
- Quadratic Formula
- How to identify the most appropriate method to solve a quadratic equation.
+ Try Factoring first. If the quadratic factors easily, this method is very quick.
+ Try the #strong[Square Root Property] next. If the equation fits the form #emph[ax]#super[2] = #emph[k] or #emph[a] (#emph[x] − #emph[h])#super[2] = #emph[k], it can easily be solved by using the Square Root Property.
+ Use the #strong[Quadratic Formula.] Any other quadratic equation is best solved by using the Quadratic Formula.


==== Practice Makes Perfect

#strong[Solve Quadratic Equations Using the Quadratic Formula]

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

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

#solutionbox[
#math.equation(block: true, alt: "m equals −1 , m equals the fraction 3 over 4")[$m = −1 , m = frac(3, 4)$]
]

#math.equation(block: true, alt: "4 n squared minus 9 n plus 5 equals 0")[$4 n^(2) − 9 n + 5 = 0$]

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

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

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

#math.equation(block: true, alt: "p squared plus 7 p plus 12 equals 0")[$p^(2) + 7 p + 12 = 0$]

#solutionbox[
#math.equation(block: true, alt: "p equals −4 , p equals −3")[$p = −4 , p = −3$]
]

#math.equation(block: true, alt: "q squared plus 3 q minus 18 equals 0")[$q^(2) + 3 q − 18 = 0$]

#math.equation(block: true, alt: "r squared minus 8 r equals 33")[$r^(2) − 8 r = 33$]

#solutionbox[
#math.equation(block: true, alt: "r equals −3 , r equals 11")[$r = −3 , r = 11$]
]

#math.equation(block: true, alt: "t squared plus 13 t equals −40")[$t^(2) + 13 t = −40$]

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

#solutionbox[
#math.equation(block: true, alt: "u equals the fraction −7 plus or minus the square root of 73 over 6")[$u = frac(−7 ± sqrt(73), 6)$]
]

#math.equation(block: true, alt: "2 p squared plus 8 p plus 5 equals 0")[$2 p^(2) + 8 p + 5 = 0$]

#math.equation(block: true, alt: "2 a squared minus 6 a plus 3 equals 0")[$2 a^(2) − 6 a + 3 = 0$]

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

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

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

#solutionbox[
#math.equation(block: true, alt: "x equals −4 plus or minus 2 the square root of 5")[$x = −4 ± 2 sqrt(5)$]
]

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

#math.equation(block: true, alt: "3 y squared plus 5 y minus 2 equals 0")[$3 y^(2) + 5 y − 2 = 0$]

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

#math.equation(block: true, alt: "6 x squared plus 2 x minus 20 equals 0")[$6 x^(2) + 2 x − 20 = 0$]

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

#solutionbox[
#math.equation(block: true, alt: "x equals minus the fraction 3 over 4 plus or minus the fraction the square root of 15 over 4 i")[$x = − frac(3, 4) ± frac(sqrt(15), 4) i$]
]

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

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

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 3 over 8 plus or minus the fraction the square root of 7 over 8 i")[$x = frac(3, 8) ± frac(sqrt(7), 8) i$]
]

#math.equation(block: true, alt: "8 x squared minus 4 x plus 1 equals 0")[$8 x^(2) − 4 x + 1 = 0$]

#math.equation(block: true, alt: "open parenthesis v plus 1 close parenthesis open parenthesis v minus 5 close parenthesis minus 4 equals 0")[$( v + 1 ) ( v − 5 ) − 4 = 0$]

#solutionbox[
#math.equation(block: true, alt: "v equals 2 plus or minus the square root of 13")[$v = 2 ± sqrt(13)$]
]

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

#math.equation(block: true, alt: "open parenthesis y plus 4 close parenthesis open parenthesis y minus 7 close parenthesis equals 18")[$( y + 4 ) ( y − 7 ) = 18$]

#solutionbox[
#math.equation(block: true, alt: "y equals the fraction 3 plus or minus the square root of 193 over 2")[$y = frac(3 ± sqrt(193), 2)$]
]

#math.equation(block: true, alt: "open parenthesis x plus 2 close parenthesis open parenthesis x plus 6 close parenthesis equals 21")[$( x + 2 ) ( x + 6 ) = 21$]

#math.equation(block: true, alt: "the fraction 1 over 3 m squared plus the fraction 1 over 12 m equals the fraction 1 over 4")[$frac(1, 3) m^(2) + frac(1, 12) m = frac(1, 4)$]

#solutionbox[
#math.equation(block: true, alt: "m equals −1 , m equals the fraction 3 over 4")[$m = −1 , m = frac(3, 4)$]
]

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

#math.equation(block: true, alt: "the fraction 3 over 4 b squared plus the fraction 1 over 2 b equals the fraction 3 over 8")[$frac(3, 4) b^(2) + frac(1, 2) b = frac(3, 8)$]

#solutionbox[
#math.equation(block: true, alt: "b equals the fraction −2 plus or minus the square root of 22 over 6")[$b = frac(−2 ± sqrt(22), 6)$]
]

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

#math.equation(block: true, alt: "16 c squared plus 24 c plus 9 equals 0")[$16 c^(2) + 24 c + 9 = 0$]

#solutionbox[
#math.equation(block: true, alt: "c equals minus the fraction 3 over 4")[$c = − frac(3, 4)$]
]

#math.equation(block: true, alt: "25 d squared minus 60 d plus 36 equals 0")[$25 d^(2) − 60 d + 36 = 0$]

#math.equation(block: true, alt: "25 q squared plus 30 q plus 9 equals 0")[$25 q^(2) + 30 q + 9 = 0$]

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

#math.equation(block: true, alt: "16 y squared plus 8 y plus 1 equals 0")[$16 y^(2) + 8 y + 1 = 0$]

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

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

ⓐ #math.equation(block: false, alt: "4 x squared minus 5 x plus 16 equals 0")[$4 x^(2) − 5 x + 16 = 0$] ⓑ #math.equation(block: false, alt: "36 y squared plus 36 y plus 9 equals 0")[$36 y^(2) + 36 y + 9 = 0$] ⓒ #math.equation(block: false, alt: "6 m squared plus 3 m minus 5 equals 0")[$6 m^(2) + 3 m − 5 = 0$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "no real solutions")[$"no real solutions"$] ⓑ #math.equation(block: false, alt: "1")[$1$] #linebreak() ⓒ #math.equation(block: false, alt: "2")[$2$]
]

ⓐ #math.equation(block: false, alt: "9 v squared minus 15 v plus 25 equals 0")[$9 v^(2) − 15 v + 25 = 0$] ⓑ #math.equation(block: false, alt: "100 w squared plus 60 w plus 9 equals 0")[$100 w^(2) + 60 w + 9 = 0$] ⓒ #math.equation(block: false, alt: "5 c squared plus 7 c minus 10 equals 0")[$5 c^(2) + 7 c − 10 = 0$]

ⓐ #math.equation(block: false, alt: "r squared plus 12 r plus 36 equals 0")[$r^(2) + 12 r + 36 = 0$] ⓑ #math.equation(block: false, alt: "8 t squared minus 11 t plus 5 equals 0")[$8 t^(2) − 11 t + 5 = 0$] ⓒ #math.equation(block: false, alt: "3 v squared minus 5 v minus 1 equals 0")[$3 v^(2) − 5 v − 1 = 0$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "1")[$1$] ⓑ #math.equation(block: false, alt: "no real solutions")[$"no real solutions"$] #linebreak() ⓒ #math.equation(block: false, alt: "2")[$2$]
]

ⓐ #math.equation(block: false, alt: "25 p squared plus 10 p plus 1 equals 0")[$25 p^(2) + 10 p + 1 = 0$] ⓑ #math.equation(block: false, alt: "7 q squared minus 3 q minus 6 equals 0")[$7 q^(2) − 3 q − 6 = 0$] ⓒ #math.equation(block: false, alt: "7 y squared plus 2 y plus 8 equals 0")[$7 y^(2) + 2 y + 8 = 0$]

#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: "x squared minus 5 x minus 24 equals 0")[$x^(2) − 5 x − 24 = 0$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis y plus 5 close parenthesis squared equals 12")[$attach(( y + 5 ), t: 2) = 12$] #linebreak() ⓒ #math.equation(block: false, alt: "14 m squared plus 3 m equals 11")[$14 m^(2) + 3 m = 11$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "factor")[$"factor"$] #linebreak() ⓑ #math.equation(block: false, alt: "square root")[$"square root"$] #linebreak() ⓒ #math.equation(block: false, alt: "Quadratic Formula")[$"Quadratic Formula"$]
]

ⓐ #math.equation(block: false, alt: "open parenthesis 8 v plus 3 close parenthesis squared equals 81")[$attach(( 8 v + 3 ), t: 2) = 81$] #linebreak() ⓑ #math.equation(block: false, alt: "w squared minus 9 w minus 22 equals 0")[$w^(2) − 9 w − 22 = 0$] #linebreak() ⓒ #math.equation(block: false, alt: "4 n squared minus 10 n equals 6")[$4 n^(2) − 10 n = 6$]

ⓐ #math.equation(block: false, alt: "6 a squared plus 14 a equals 20")[$6 a^(2) + 14 a = 20$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis x minus the fraction 1 over 4 close parenthesis squared equals the fraction 5 over 16")[$attach(( x − frac(1, 4) ), t: 2) = frac(5, 16)$] #linebreak() ⓒ #math.equation(block: false, alt: "y squared minus 2 y equals 8")[$y^(2) − 2 y = 8$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "Quadratic Formula")[$"Quadratic Formula"$] #linebreak() ⓑ #math.equation(block: false, alt: "square root")[$"square root"$] #linebreak() ⓒ #math.equation(block: false, alt: "factor")[$"factor"$]
]

ⓐ #math.equation(block: false, alt: "8 b squared plus 15 b equals 4")[$8 b^(2) + 15 b = 4$] #linebreak() ⓑ #math.equation(block: false, alt: "the fraction 5 over 9 v squared minus the fraction 2 over 3 v equals 1")[$frac(5, 9) v^(2) − frac(2, 3) v = 1$] #linebreak() ⓒ #math.equation(block: false, alt: "open parenthesis w plus the fraction 4 over 3 close parenthesis squared equals the fraction 2 over 9")[$attach(( w + frac(4, 3) ), t: 2) = frac(2, 9)$]


==== Writing Exercises

Solve the equation #math.equation(block: false, alt: "x squared plus 10 x equals 120")[$x^(2) + 10 x = 120$]

ⓐ by completing the square

ⓑ using the Quadratic Formula

ⓒ Which method do you prefer? Why?

#solutionbox[
Answers will vary.
]

Solve the equation #math.equation(block: false, alt: "12 y squared plus 23 y equals 24")[$12 y^(2) + 23 y = 24$]

ⓐ by completing the square

ⓑ using the Quadratic Formula

ⓒ Which method do you prefer? Why?


==== Self Check

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

#figure(figph[This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can solve quadratic equations using the quadratic formula.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can use the discriminant to predict the number of solutions of a quadratic equation.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can identify the most appropriate method to use to solve a quadratic equation.” “Confidently,” “with some help,” or “No, I don’t get it.”], alt: "This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can solve quadratic equations using the quadratic formula.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can use the discriminant to predict the number of solutions of a quadratic equation.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can identify the most appropriate method to use to solve a quadratic equation.” “Confidently,” “with some help,” or “No, I don’t get it.”", caption: none)

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?