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== 10.3#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.

Simplify: #math.equation(block: false, alt: "the fraction −20 minus 5 over 10")[$frac(−20 − 5, 10)$]. #linebreak() 
If you missed this problem, review .

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

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Simplify: #math.equation(block: false, alt: "4 plus the square root of 121")[$4 + sqrt(121)$]. #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "15")[$15$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Simplify: #math.equation(block: false, alt: "the square root of 128")[$sqrt(128)$]. #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "8 the square root of 2")[$8 sqrt(2)$]
]
]

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.’ 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 in general to solve a quadratic equation for #emph[x]. It may be helpful to look at one of the examples at the end of the last section where we solved an equation of the form #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$] as you read through the algebraic steps below, so you see them with numbers as well as ‘in general.’

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([We start with the standard form of a quadratic equation #linebreak() and solve it for #emph[x] by completing the square.], [#math.equation(block: false, alt: "a x squared plus b x plus c equals 0 a not equal to 0")[$#h(0.8em) a x^(2) + b x + c #h(0.5em) = #h(0.5em) 0 #h(2em) a ≠ 0$]]),
  [Isolate the variable terms on one side.], [#math.equation(block: false, alt: "a x squared plus b x, equals, − c")[$#h(2.3em) a x^(2) + b x & = & "−" c$]],
  [Make leading coefficient 1, by dividing by a.], [#math.equation(block: false, alt: "the fraction a x squared over a plus the fraction b over a x, equals, minus the fraction c over a")[$#h(2em) frac(a x^(2), a) + frac(b, a) x & = & − frac(c, a)$]],
  [Simplify.], [#math.equation(block: false, alt: "x squared plus the fraction b over a x, equals, minus the fraction c over a")[$#h(2.8em) x^(2) + frac(b, a) x & = & − frac(c, a)$]],
  [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 #linebreak() 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")[$attach(( frac(1, 2) #h(0.2em) frac(b, a) ), t: 2) = frac(b^(2), 4 a^(2))$]], [#math.equation(block: false, alt: "x squared plus the fraction b over a x plus the fraction b squared over 4 a squared, equals, minus the fraction c over a plus the fraction b squared over 4 a squared")[$x^(2) + frac(b, a) x + frac(b^(2), 4 a^(2)) & = & − frac(c, a) + frac(b^(2), 4 a^(2))$]],
  [The left side is a perfect square, factor it.], [#math.equation(block: false, alt: "open parenthesis x plus the fraction b over 2 a close parenthesis squared, equals, minus the fraction c over a plus the fraction b squared over 4 a squared")[$#h(2em) attach(( x + frac(b, 2 a) ), t: 2) & = & − frac(c, a) + frac(b^(2), 4 a^(2))$]],
  [Find the common denominator of the right side and write #linebreak() equivalent fractions with the common denominator.], [#math.equation(block: false, alt: "open parenthesis x plus the fraction b over 2 a close parenthesis squared, equals, the fraction b squared over 4 a squared minus the fraction c times 4 a over a times 4 a")[$#h(2em) attach(( x + frac(b, 2 a) ), t: 2) & = & frac(b^(2), 4 a^(2)) − frac(c · 4 a, a · 4 a)$]],
  [Simplify.], [#math.equation(block: false, alt: "open parenthesis x plus the fraction b over 2 a close parenthesis squared, equals, the fraction b squared over 4 a squared minus the fraction 4 a c over 4 a squared")[$#h(2em) attach(( x + frac(b, 2 a) ), t: 2) & = & frac(b^(2), 4 a^(2)) − frac(4 a c, 4 a^(2))$]],
  [Combine to one fraction.], [#math.equation(block: false, alt: "open parenthesis x plus the fraction b over 2 a close parenthesis squared, equals, the fraction b squared minus 4 a c over 4 a squared")[$#h(2em) attach(( x + frac(b, 2 a) ), t: 2) & = & frac(b^(2) − 4 a c, 4 a^(2))$]],
  [Use the square root property.], [#math.equation(block: false, alt: "x plus the fraction b over 2 a, equals, plus or minus the square root of the fraction b squared minus 4 a c over 4 a squared")[$#h(3em) x + frac(b, 2 a) & = & ± #h(0.2em) sqrt(frac(b^(2) − 4 a c, 4 a^(2)))$]],
  [Simplify.], [#math.equation(block: false, alt: "x plus the fraction b over 2 a, equals, plus or minus the fraction the square root of b squared minus 4 a c over 2 a")[$#h(3em) x + frac(b, 2 a) & = & ± #h(0.2em) frac(sqrt(b^(2) − 4 a c), 2 a)$]],
  [Add #math.equation(block: false, alt: "minus the fraction b over 2 a")[$− frac(b, 2 a)$] to both sides of the equation.], [#math.equation(block: false, alt: "x, equals, minus the fraction b over 2 a plus or minus the fraction the square root of b squared minus 4 a c over 2 a")[$#h(5.5em) x & = & − frac(b, 2 a) ± frac(sqrt(b^(2) − 4 a c), 2 a)$]],
  [Combine the terms on the right side.], [#math.equation(block: false, alt: "x, equals, the fraction − b plus or minus the square root of b squared minus 4 a c over 2 a")[$#h(5.5em) x & = & frac("−" b ± sqrt(b^(2) − 4 a c), 2 a)$]],
))

This last equation is the Quadratic Formula.

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

The solutions to a quadratic equation of the form #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$], #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 Quadratic Formula, we substitute the values of #math.equation(block: false, alt: "a , b , and c")[$a , b , #h(0.2em) "and" #h(0.2em) c$] into the expression on the right side of the formula. Then, we do all the math to simplify the expression. The result gives the solution(s) to the quadratic equation.

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

#solutionbox[
#figure(figph[The image shows the steps to solve the quadratic equation two x squared plus nine x minus five equals zero. Step one is to write the quadratic equation in standard form and identify the a, b, and c values. This equation is already in standard for. The value of a is two, the value of b is nine and the value of c is negative five.], alt: "The image shows the steps to solve the quadratic equation two x squared plus nine x minus five equals zero. Step one is to write the quadratic equation in standard form and identify the a, b, and c values. This equation is already in standard for. The value of a is two, the value of b is nine and the value of c is negative five.", caption: none)

#figure(figph[Step two is to write the quadratic formula. Then substitute in the values of a, b, and c. Substitute two for a, nine for b and negative five for c in the formula x equals the quantity negative b plus or minus the square root of b squared minus four times a times c divided by two times a. The formula becomes x equals negative nine plus or minus the square root of negative nine squared minus four time two times negative five all divided by two times two.], alt: "Step two is to write the quadratic formula. Then substitute in the values of a, b, and c. Substitute two for a, nine for b and negative five for c in the formula x equals the quantity negative b plus or minus the square root of b squared minus four times a times c divided by two times a. The formula becomes x equals negative nine plus or minus the square root of negative nine squared minus four time two times negative five all divided by two times two.", caption: none)

#figure(figph[Step three is to simplify the formula. Squaring negative nine and performing the multiplication to get negative nine plus or minus the square root of 81 minus negative 40 all divided by four. This simplifies further to negative nine plus or minus the square root of 121 all divided by four which reduces to negative nine plus or minus 11 all divided by four. Negative nine plus 11 divided by four is two fourths which reduces to one half. Negative nine minus 11 divided by four is negative 20 fourths which reduces to negative five.], alt: "Step three is to simplify the formula. Squaring negative nine and performing the multiplication to get negative nine plus or minus the square root of 81 minus negative 40 all divided by four. This simplifies further to negative nine plus or minus the square root of 121 all divided by four which reduces to negative nine plus or minus 11 all divided by four. Negative nine plus 11 divided by four is two fourths which reduces to one half. Negative nine minus 11 divided by four is negative 20 fourths which reduces to negative five.", caption: none)

#figure(figph[Step four is to check the solutions by putting each answer in the original equation to check. Replace x in two x squared plus nine x minus five equals zero with one half to get two times one half squared plus nine times one half minus five. Simplify to get one half plus nine halves minus five which is zero. Replace x in two x squared plus nine x minus five equals zero with negative five to get two times negative five squared plus nine times negative five minus five. Simplify to get 50 minus 45 minus five which is zero.], alt: "Step four is to check the solutions by putting each answer in the original equation to check. Replace x in two x squared plus nine x minus five equals zero with one half to get two times one half squared plus nine times one half minus five. Simplify to get one half plus nine halves minus five which is zero. Replace x in two x squared plus nine x minus five equals zero with negative five to get two times negative five squared plus nine times negative five minus five. Simplify to get 50 minus 45 minus five which is zero.", caption: none)
]
]

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

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

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

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

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

+ Write the Quadratic Formula in standard form. Identify the #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$] values.
+ Write the Quadratic Formula. Then substitute in the values of #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c .")[$c .$]
+ 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 ‘#math.equation(block: false, alt: "x equals")[$x =$]’.

#examplebox("Example 2")[][
Solve #math.equation(block: false, alt: "x squared minus 6 x plus 5 equals 0")[$x^(2) − 6 x + 5 = 0$] by using the Quadratic Formula.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A quadratic equation is displayed in black text against a white background: x squared minus 6x plus 5 equals 0.], alt: "A quadratic equation is displayed in black text against a white background: x squared minus 6x plus 5 equals 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, x^2 - 6x + 5 = 0.], alt: "The general form of a quadratic equation, ax^2 + bx + c = 0, is shown above a specific example, x^2 - 6x + 5 = 0.", caption: none)],
  [Identify the #emph[a, b, c] values.], [#figure(figph[The image displays mathematical variables with their assigned numerical values: 'a = 1' in light blue, 'b = -6' in red, and 'c = 5' in yellow, all against a plain white background.], alt: "The image displays mathematical variables with their assigned numerical values: 'a = 1' in light blue, 'b = -6' in red, and 'c = 5' in yellow, all against a plain white background.", caption: none)],
  [Write the Quadratic Formula.], [#figure(figph[The quadratic formula, x = (-b ×1×2(b×2 - 4ac)) / 2a, is displayed on a white background.], alt: "The quadratic formula, x = (-b ×1×2(b×2 - 4ac)) / 2a, is displayed on a white background.", caption: none)],
  [Then substitute in the values of #emph[a, b, c.]], [#figure(figph[The quadratic formula with specific values substituted: x = \[-(-6) plus or minus sqrt((-6)^2 - 4 \* 1 \* (5))\] / (2 \* 1), ready for calculation. -6 is colored red, 1 is teal, and 5 is lime green.], alt: "The quadratic formula with specific values substituted: x = [-(-6) plus or minus sqrt((-6)^2 - 4 * 1 * (5))] / (2 * 1), ready for calculation. -6 is colored red, 1 is teal, and 5 is lime green.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation is displayed on a white background: x equals a fraction where the numerator is 6 plus or minus the square root of 36 minus 20, and the denominator is 2.], alt: "A mathematical equation is displayed on a white background: x equals a fraction where the numerator is 6 plus or minus the square root of 36 minus 20, and the denominator is 2.", caption: none)

 #linebreak() #figure(figph[A mathematical equation shows 'x equals 6 plus or minus the square root of 16, all divided by 2.'], alt: "A mathematical equation shows 'x equals 6 plus or minus the square root of 16, all divided by 2.'", caption: none)

 #linebreak() #figure(figph[A mathematical equation shows X equals the fraction of (6 plus or minus 4) over 2, indicating two possible solutions for X.], alt: "A mathematical equation shows X equals the fraction of (6 plus or minus 4) over 2, indicating two possible solutions for X.", caption: none)],
  [Rewrite to show two solutions.], [#figure(figph[Two equations for 'x' are presented: x = (6 + 4) / 2 and x = (6 - 4) / 2, demonstrating two distinct solutions from a quadratic formula or similar calculation.], alt: "Two equations for 'x' are presented: x = (6 + 4) / 2 and x = (6 - 4) / 2, demonstrating two distinct solutions from a quadratic formula or similar calculation.", caption: none)],
  [Simplify.], [#figure(figph[The image displays a mathematical expression with two equations: 'x = 10/2' followed by a comma, and then 'x = 2/2' on a white background.], alt: "The image displays a mathematical expression with two equations: 'x = 10/2' followed by a comma, and then 'x = 2/2' on a white background.", caption: none)

 #linebreak() #figure(figph[The image displays mathematical equations showing two possible values for 'x': x equals 5, and x equals 1, set against a plain white background.], alt: "The image displays mathematical equations showing two possible values for 'x': x equals 5, and x equals 1, set against a plain white background.", caption: none)],
  [Check. #linebreak() #figure(figph[Solutions for x^2 - 6x + 5 = 0 are verified. Both x=5 and x=1 are shown to correctly satisfy the quadratic equation, confirming their validity as roots.], alt: "Solutions for x^2 - 6x + 5 = 0 are verified. Both x=5 and x=1 are shown to correctly satisfy the quadratic equation, confirming their validity as roots.", caption: none)], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "a squared minus 2 a minus 15 equals 0")[$a^(2) − 2 a − 15 = 0$] by using the Quadratic Formula.

#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 #math.equation(block: false, alt: "b squared plus 10 b plus 24 equals 0")[$b^(2) + 10 b + 24 = 0$] by using the Quadratic Formula.

#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 Quadratic Formula. If we get a radical as a solution, the final answer must have the radical in its simplified form.

#examplebox("Example 3")[][
Solve #math.equation(block: false, alt: "4 y squared minus 5 y minus 3 equals 0")[$4 y^(2) − 5 y − 3 = 0$] by using the Quadratic Formula.

#solutionbox[
We can use the Quadratic Formula to solve for the variable in a quadratic equation, whether or not it is named ‘#emph[x]’.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical equation is displayed on a white background. The equation reads '4y^2 - 5y - 3 = 0'.], alt: "A mathematical equation is displayed on a white background. The equation reads '4y^2 - 5y - 3 = 0'.", caption: none)]),
  [This equation is in standard form.], [#figure(figph[Two quadratic equations are shown: the general form ax^2 + bx + c = 0 (in red) and a specific example 4y^2 - 5y - 3 = 0 (in black).], alt: "Two quadratic equations are shown: the general form ax^2 + bx + c = 0 (in red) and a specific example 4y^2 - 5y - 3 = 0 (in black).", caption: none)],
  [Identify the #emph[a, b, c] values.], [#figure(figph[The image displays the values of three variables: a=4 in light blue, b=-5 in red, and c=-3 in yellow-green, presented on a white background.], alt: "The image displays the values of three variables: a=4 in light blue, b=-5 in red, and c=-3 in yellow-green, presented on a white background.", caption: none)],
  [Write the Quadratic Formula.], [#figure(figph[The image displays the quadratic formula, y = (-b ×1 sqrt(b×2 - 4ac)) / 2a, a fundamental equation in algebra for solving quadratic equations.], alt: "The image displays the quadratic formula, y = (-b ×1 sqrt(b×2 - 4ac)) / 2a, a fundamental equation in algebra for solving quadratic equations.", caption: none)],
  [Then substitute in the values of #emph[a, b, c.]], [#figure(figph[The quadratic formula is displayed with specific values substituted: y = (-(-5) +/- sqrt((-5)^2 - 4 \* 4 \* (-3))) / (2 \* 4).], alt: "The quadratic formula is displayed with specific values substituted: y = (-(-5) +/- sqrt((-5)^2 - 4 * 4 * (-3))) / (2 * 4).", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation showing y equals 5 plus or minus the square root of 25 plus 48, all divided by 8.], alt: "A mathematical equation showing y equals 5 plus or minus the square root of 25 plus 48, all divided by 8.", caption: none)

 #linebreak() #figure(figph[A mathematical equation for y is shown on a white background, which states y equals a fraction with 5 plus or minus the square root of 73 in the numerator, and 8 in the denominator.], alt: "A mathematical equation for y is shown on a white background, which states y equals a fraction with 5 plus or minus the square root of 73 in the numerator, and 8 in the denominator.", caption: none)],
  [Rewrite to show two solutions.], [#figure(figph[Two solutions for y are displayed: y = (5 + 'square root of 73') / 8 and y = (5 - 'square root of 73') / 8.], alt: "Two solutions for y are displayed: y = (5 + 'square root of 73') / 8 and y = (5 - 'square root of 73') / 8.", caption: none)],
  [Check. We leave the check to you.], [],
))
]
]

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

#solutionbox[
#math.equation(block: true, alt: "p equals the fraction −4 plus or minus the square root of 6 over 2")[$p = frac(−4 ± sqrt(6), 2)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "5 q squared minus 11 q plus 3 equals 0")[$5 q^(2) − 11 q + 3 = 0$] by using the Quadratic Formula.

#solutionbox[
#math.equation(block: true, alt: "q equals the fraction 11 plus or minus the square root of 61 over 10")[$q = frac(11 ± sqrt(61), 10)$]
]
]

#examplebox("Example 4")[][
Solve #math.equation(block: false, alt: "2 x squared plus 10 x plus 11 equals 0")[$2 x^(2) + 10 x + 11 = 0$] by using the Quadratic Formula.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A quadratic equation displayed on a white background: 2x^2 + 10x + 11 = 0.], alt: "A quadratic equation displayed on a white background: 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 (in red), is shown above a specific example, 2x^2 + 10x + 11 = 0 (in black).], alt: "The general form of a quadratic equation, ax^2 + bx + c = 0 (in red), is shown above a specific example, 2x^2 + 10x + 11 = 0 (in black).", caption: none)],
  [Identify the #emph[a, b, c] values.], [#figure(figph[The image displays the variables and their assigned numerical values: a = 2, b = 10, and c = 11, with the numbers presented in a colorful, stylized font against a white background.], alt: "The image displays the variables and their assigned numerical values: a = 2, b = 10, and c = 11, with the numbers presented in a colorful, stylized font against a white background.", caption: none)],
  [Write the Quadratic Formula.], [#figure(figph[The quadratic formula, used to find the solutions for x in a quadratic equation, is displayed as x = (-b ± sqrt(b^2 - 4ac)) / 2a.], alt: "The quadratic formula, used to find the solutions for x in a quadratic equation, is displayed as x = (-b ± sqrt(b^2 - 4ac)) / 2a.", caption: none)],
  [Then substitute in the values of #emph[a, b, c.]], [#figure(figph[The quadratic formula with specific values of a=2, b=10, and c=11 substituted to solve for x.], alt: "The quadratic formula with specific values of a=2, b=10, and c=11 substituted to solve for x.", caption: none)],
  [Simplify.], [#figure(figph[An intermediate calculation solving a quadratic equation, showing x equals negative ten plus or minus the square root of one hundred minus eighty-eight, all divided by four.], alt: "An intermediate calculation solving a quadratic equation, showing x equals negative ten plus or minus the square root of one hundred minus eighty-eight, all divided by four.", caption: none)

 #linebreak() #figure(figph[A mathematical equation showing x equals negative 10 plus or minus the square root of 12, all divided by 4.], alt: "A mathematical equation showing x equals negative 10 plus or minus the square root of 12, all divided by 4.", caption: none)],
  [Simplify the radical.], [#figure(figph[A mathematical equation displays the value of x as negative 10 plus or minus 2 multiplied by the square root of 3, all divided by 4.], alt: "A mathematical equation displays the value of x as negative 10 plus or minus 2 multiplied by the square root of 3, all divided by 4.", caption: none)],
  [Factor out the common factor in the numerator.], [#figure(figph[A mathematical equation shows x = 2(-5 ± square root of 3)/4, illustrating a step in solving a quadratic equation with a plus-minus sign indicating two possible solutions.], alt: "A mathematical equation shows x = 2(-5 ± square root of 3)/4, illustrating a step in solving a quadratic equation with a plus-minus sign indicating two possible solutions.", caption: none)],
  [Remove the common factors.], [#figure(figph[The equation shows x equals a fraction where the numerator is -5 plus or minus the square root of 3, and the denominator is 2, representing solutions from a quadratic formula.], alt: "The equation shows x equals a fraction where the numerator is -5 plus or minus the square root of 3, and the denominator is 2, representing solutions from a quadratic formula.", caption: none)],
  [Rewrite to show two solutions.], [#figure(figph[The image displays two solutions for the variable x: x = (-5 + sqrt(3))/2 and x = (-5 - sqrt(3))/2, likely derived from the quadratic formula.], alt: "The image displays two solutions for the variable x: x = (-5 + sqrt(3))/2 and x = (-5 - sqrt(3))/2, likely derived from the quadratic formula.", caption: none)],
  [Check. We leave the check to you.], [],
))
]
]

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

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

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

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

We cannot take the square root of a negative number. So, when we substitute #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$] into the Quadratic Formula, if the quantity inside the radical is negative, the quadratic equation has no real solution. We will see this in the next example.

#examplebox("Example 5")[][
Solve #math.equation(block: false, alt: "3 p squared plus 2 p plus 9 equals 0")[$3 p^(2) + 2 p + 9 = 0$] by using the Quadratic Formula.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([This equation is in standard form.], [#figure(figph[Two quadratic equations are shown, with the general form ax^2 + bx + c = 0 in red, and a specific example 3p^2 + 2p + 9 = 0 in black, both set equal to zero.], alt: "Two quadratic equations are shown, with the general form ax^2 + bx + c = 0 in red, and a specific example 3p^2 + 2p + 9 = 0 in black, both set equal to zero.", caption: none)]),
  [Identify the #emph[a, b, c] values.], [#figure(figph[The equation a=3, b=2, c=9 is shown in a colorful font, with 'a' in light blue, 'b' in red, and 'c' in yellow, all against a white background.], alt: "The equation a=3, b=2, c=9 is shown in a colorful font, with 'a' in light blue, 'b' in red, and 'c' in yellow, all against a white background.", caption: none)],
  [Write the Quadratic Formula.], [#figure(figph[The image displays the quadratic formula, p = \[-b ± sqrt(b^2 - 4ac)\] / 2a, a fundamental mathematical equation used to find the roots of a quadratic equation. It is shown in standard algebraic notation.], alt: "The image displays the quadratic formula, p = [-b ± sqrt(b^2 - 4ac)] / 2a, a fundamental mathematical equation used to find the roots of a quadratic equation. It is shown in standard algebraic notation.", caption: none)],
  [Then substitute in the values of #emph[a, b, c.]], [#figure(figph[A mathematical formula for 'p' is shown, representing a step in the quadratic equation. It displays specific numerical values, including 2, 3, and 9, substituted into the numerator and denominator, with some values color-coded.], alt: "A mathematical formula for 'p' is shown, representing a step in the quadratic equation. It displays specific numerical values, including 2, 3, and 9, substituted into the numerator and denominator, with some values color-coded.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation shown is p = (-2 ×1 sqrt(4 - 108))/6, which is a step in applying the quadratic formula, featuring a negative discriminant leading to complex solutions.], alt: "A mathematical equation shown is p = (-2 ×1 sqrt(4 - 108))/6, which is a step in applying the quadratic formula, featuring a negative discriminant leading to complex solutions.", caption: none)],
  [Simplify the radical.], [#figure(figph[A mathematical equation shows p equals a fraction. The numerator is -2 plus or minus the square root of -104, and the denominator is 6.], alt: "A mathematical equation shows p equals a fraction. The numerator is -2 plus or minus the square root of -104, and the denominator is 6.", caption: none)],
  [We cannot take the square root of a negative number.], [There is no real solution.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "4 a squared minus 3 a plus 8 equals 0")[$4 a^(2) − 3 a + 8 = 0$] by using the Quadratic Formula.

#solutionbox[
no real solution
]
]

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

#solutionbox[
no real solution
]
]

The quadratic equations we have solved so far in this section were all written in standard form, #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$]. Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

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

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical equation is displayed on a white background: x(x + 6) + 4 = 0.], alt: "A mathematical equation is displayed on a white background: x(x + 6) + 4 = 0.", caption: none)]),
  [Distribute to get the equation in standard form.], [#figure(figph[A quadratic equation is displayed: x squared plus 6x plus 4 equals 0.], alt: "A quadratic equation is displayed: x squared plus 6x plus 4 equals 0.", caption: none)],
  [This equation is now in standard form.], [#figure(figph[Two quadratic equations are displayed on a white background: the general form ax^2 + bx + c = 0 in red, and a specific example x^2 + 6x + 4 = 0 in black.], alt: "Two quadratic equations are displayed on a white background: the general form ax^2 + bx + c = 0 in red, and a specific example x^2 + 6x + 4 = 0 in black.", caption: none)],
  [Identify the #emph[a, b, c] values.], [#figure(figph[The image displays text showing variable assignments: a = 1 (in light blue), b = 6 (in red), and c = 4 (in yellow) on a white background.], alt: "The image displays text showing variable assignments: a = 1 (in light blue), b = 6 (in red), and c = 4 (in yellow) on a white background.", 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 four a c, all divided by two a.], alt: "The quadratic formula, x equals negative b, plus or minus the square root of b squared minus four a c, all divided by two a.", caption: none)],
  [Then substitute in the values of #emph[a, b, c.]], [#figure(figph[The quadratic formula showing the substitution of coefficients a=1, b=6, and c=4 (highlighted in colors) to solve for x.], alt: "The quadratic formula showing the substitution of coefficients a=1, b=6, and c=4 (highlighted in colors) to solve for x.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation shows X equals a fraction with a numerator of -6 plus or minus the square root of 36 minus 16, all divided by 2. This is a step in solving a quadratic equation.], alt: "A mathematical equation shows X equals a fraction with a numerator of -6 plus or minus the square root of 36 minus 16, all divided by 2. This is a step in solving a quadratic equation.", caption: none)],
  [Simplify inside the radical.], [#figure(figph[A mathematical equation shows x equals negative 6 plus or minus the square root of 20, all divided by 2.], alt: "A mathematical equation shows x equals negative 6 plus or minus the square root of 20, all divided by 2.", caption: none)],
  [Simplify the radical.], [#figure(figph[A mathematical equation for x is displayed as a fraction: x equals negative six plus or minus two times the square root of five, all divided by two.], alt: "A mathematical equation for x is displayed as a fraction: x equals negative six plus or minus two times the square root of five, all divided by two.", caption: none)],
  [Factor out the common factor in the numerator.], [#figure(figph[An algebraic equation showing x equals 2 times the quantity of -3 plus or minus the square root of 5, all divided by 2. This is a step in solving a quadratic equation.], alt: "An algebraic equation showing x equals 2 times the quantity of -3 plus or minus the square root of 5, all divided by 2. This is a step in solving a quadratic equation.", caption: none)],
  [Remove the common factors.], [#figure(figph[The image displays the mathematical equation x = -3 ± √5, presented in a clear, centered format on a white background.], alt: "The image displays the mathematical equation x = -3 ± √5, presented in a clear, centered format on a white background.", caption: none)],
  [Rewrite to show two solutions.], [#figure(figph[Two mathematical expressions show the values of x: x equals -3 + sqrt(5) and x equals -3 - sqrt(5).], alt: "Two mathematical expressions show the values of x: x equals -3 + sqrt(5) and x equals -3 - sqrt(5).", caption: none)],
  [Check. We leave the check to you.], [],
))
]
]

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

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

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y open parenthesis 3 y minus 1 close parenthesis minus 2 equals 0")[$y ( 3 y − 1 ) − 2 = 0$] by using the Quadratic Formula.

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

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 7")[][
Solve #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")[$frac(1, 2) u^(2) + frac(2, 3) u = frac(1, 3)$] by using the Quadratic Formula.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical equation is displayed on a white background: 1/2 u^2 + 2/3 u = 1/3.], alt: "A mathematical equation is displayed on a white background: 1/2 u^2 + 2/3 u = 1/3.", 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 quadratic equation 3u^2 + 4u = 2, presented clearly against a white background.], alt: "The image displays the quadratic equation 3u^2 + 4u = 2, presented clearly against a white background.", caption: none)],
  [Subtract 2 to get the equation in standard form.], [#figure(figph[Two quadratic equations are displayed: the general form ax^2 + bx + c = 0 in red, and a specific example 3u^2 + 4u - 2 = 0 in black, illustrating the mathematical structure.], alt: "Two quadratic equations are displayed: the general form ax^2 + bx + c = 0 in red, and a specific example 3u^2 + 4u - 2 = 0 in black, illustrating the mathematical structure.", caption: none)],
  [Identify the #emph[a, b, c] values.], [#figure(figph[The image displays mathematical variable assignments on a white background, with 'a = 3' in light blue, 'b = 4' in red, and 'c = -2' in yellow, representing numerical values for each variable.], alt: "The image displays mathematical variable assignments on a white background, with 'a = 3' in light blue, 'b = 4' in red, and 'c = -2' in yellow, representing numerical values for each variable.", caption: none)],
  [Write the Quadratic Formula.], [#figure(figph[The quadratic formula, an algebraic expression used to find the roots of a quadratic equation. It is written as u = (-b ×1×2 sqrt(b×0 - 4ac)) / (2a).], alt: "The quadratic formula, an algebraic expression used to find the roots of a quadratic equation. It is written as u = (-b ×1×2 sqrt(b×0 - 4ac)) / (2a).", caption: none)],
  [Then substitute in the values of #emph[a, b, c.]], [#figure(figph[The quadratic formula is shown with specific values substituted for a=3, b=4, and c=-2 to solve for the variable u.], alt: "The quadratic formula is shown with specific values substituted for a=3, b=4, and c=-2 to solve for the variable u.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation for 'u' is shown, calculated as the fraction of '-4 plus or minus the square root of (16 + 24)' all divided by '6'.], alt: "A mathematical equation for 'u' is shown, calculated as the fraction of '-4 plus or minus the square root of (16 + 24)' all divided by '6'.", caption: none)

 #linebreak() #figure(figph[A mathematical equation shows 'u = -4 plus or minus the square root of 40, all divided by 6' on a white background.], alt: "A mathematical equation shows 'u = -4 plus or minus the square root of 40, all divided by 6' on a white background.", caption: none)],
  [Simplify the radical.], [#figure(figph[A mathematical equation shows 'u equals negative 4 plus or minus 2 times the square root of 10, all divided by 6' on a white background.], alt: "A mathematical equation shows 'u equals negative 4 plus or minus 2 times the square root of 10, all divided by 6' on a white background.", caption: none)],
  [Factor out the common factor in the numerator.], [#figure(figph[A mathematical equation shows u equals a fraction: numerator is 2 multiplied by the quantity -2 plus or minus the square root of 10, and the denominator is 6.], alt: "A mathematical equation shows u equals a fraction: numerator is 2 multiplied by the quantity -2 plus or minus the square root of 10, and the denominator is 6.", caption: none)],
  [Remove the common factors.], [#figure(figph[The image displays a mathematical equation, expressing the variable 'u' as a fraction: the numerator is -2 plus or minus the square root of 10, and the denominator is 3.], alt: "The image displays a mathematical equation, expressing the variable 'u' as a fraction: the numerator is -2 plus or minus the square root of 10, and the denominator is 3.", caption: none)],
  [Rewrite to show two solutions.], [#figure(figph[Two mathematical expressions show the solutions for 'u' as u = (-2 +  sqrt(10))/3 and u = (-2 - sqrt(10))/3.], alt: "Two mathematical expressions show the solutions for 'u' as u = (-2 + sqrt(10))/3 and u = (-2 - sqrt(10))/3.", caption: none)],
  [Check. We leave the check to you.], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #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")[$frac(1, 4) c^(2) − frac(1, 3) c = frac(1, 12)$] by using the Quadratic Formula.

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

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #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 2")[$frac(1, 9) d^(2) − frac(1, 2) d = − frac(1, 2)$] by using the Quadratic Formula.

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

Think about the equation #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis squared equals 0")[$attach(( x − 3 ), t: 2) = 0$]. We know from the Zero Products Principle that this equation has only one solution: #math.equation(block: false, alt: "x equals 3")[$x = 3$].

We will see in the next example how using the Quadratic Formula to solve an equation with a perfect square also gives just one solution.

#examplebox("Example 8")[][
Solve #math.equation(block: false, alt: "4 x squared minus 20 x equals −25")[$4 x^(2) − 20 x = −25$] by using the Quadratic Formula.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A quadratic equation is displayed on a white background, which reads 4x^2 - 20x = -25.], alt: "A quadratic equation is displayed on a white background, which reads 4x^2 - 20x = -25.", caption: none)]),
  [Add 25 to get the equation in standard form.], [#figure(figph[Two quadratic equations are shown: the general form ax^2 + bx + c = 0 (red) and a specific example 4x^2 - 20x + 25 = 0 (black).], alt: "Two quadratic equations are shown: the general form ax^2 + bx + c = 0 (red) and a specific example 4x^2 - 20x + 25 = 0 (black).", caption: none)],
  [Identify the #emph[a, b, c] values.], [#figure(figph[The image displays the variables a, b, and c assigned numerical values. Specifically, a = 4, b = -20, and c = 25, presented in a horizontal line on a white background with varying colors for each assignment.], alt: "The image displays the variables a, b, and c assigned numerical values. Specifically, a = 4, b = -20, and c = 25, presented in a horizontal line on a white background with varying colors for each assignment.", caption: none)],
  [Write the Quadratic Formula.], [#figure(figph[The quadratic formula is shown, giving the solutions for x: x = (-b ± sqrt(b^2 - 4ac)) / (2a). This fundamental algebraic equation is used to find the roots of a quadratic polynomial.], alt: "The quadratic formula is shown, giving the solutions for x: x = (-b ± sqrt(b^2 - 4ac)) / (2a). This fundamental algebraic equation is used to find the roots of a quadratic polynomial.", caption: none)],
  [Then substitute in the values of #emph[a, b, c.]], [#figure(figph[The quadratic formula with specific values substituted: x = (-(-20) plus or minus sqrt((-20) squared minus 4 times four times 25) divded by (2 times 4). -20 is colored red, 4 is teal, and 25 is lime green], alt: "The quadratic formula with specific values substituted: x = (-(-20) plus or minus sqrt((-20) squared minus 4 times four times 25) divded by (2 times 4). -20 is colored red, 4 is teal, and 25 is lime green", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation shows x equals a fraction where the numerator is 20 plus or minus the square root of 400 minus 400, and the denominator is 8.], alt: "A mathematical equation shows x equals a fraction where the numerator is 20 plus or minus the square root of 400 minus 400, and the denominator is 8.", caption: none)

 #linebreak() #figure(figph[A mathematical equation shows x equals a fraction where the numerator is 20 plus or minus the square root of 0, and the denominator is 8.], alt: "A mathematical equation shows x equals a fraction where the numerator is 20 plus or minus the square root of 0, and the denominator is 8.", caption: none)],
  [Simplify the radical.], [#figure(figph[A mathematical equation is displayed on a white background, showing 'x = 20/8'.], alt: "A mathematical equation is displayed on a white background, showing 'x = 20/8'.", caption: none)],
  [Simplify the fraction.], [#figure(figph[The image displays a mathematical equation written in black text on a white background. The equation is 'x = 5/2'.], alt: "The image displays a mathematical equation written in black text on a white background. The equation is 'x = 5/2'.", caption: none)],
  [Check. We leave the check to you.], [],
))

Did you recognize that #math.equation(block: false, alt: "4 x squared minus 20 x plus 25")[$4 x^(2) − 20 x + 25$] is a perfect square?
]
]

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

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

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

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


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

When we solved the quadratic equations in the previous examples, sometimes we got two solutions, sometimes one solution, sometimes no real solutions. Is there a way to predict the number of solutions to a quadratic equation without actually solving the equation?

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

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

In the Quadratic Formula #math.equation(block: false, alt: "x equals the fraction − b plus or minus the square root of b squared minus 4 a c over 2 a")[$x = frac("−" b ± sqrt(b^(2) − 4 a c), 2 a)$], the quantity #math.equation(block: false, alt: "b squared minus 4 a c")[$b^(2) − 4 a c$] is called the #strong[discriminant].
]

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

#figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([], [#strong[Quadratic Equation (in standard form)]], [#strong[Discriminant #math.equation(block: false, alt: "b squared minus 4 a c")[$b^(2) − 4 a c$]]], [#strong[Sign of the Discriminant]], [#strong[Number of real 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 equals 121")[$9^(2) − 4 · 2 ( −5 ) = 121$]], [+], [2],
  [], [#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 equals 0")[$attach(( −20 ), t: 2) − 4 · 4 · 25 = 0$]], [0], [1],
  [], [#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 equals −104")[$2^(2) − 4 · 3 · 9 = −104$]], [−], [0],
))

When the discriminant is #strong[positive] #math.equation(block: false, alt: "open parenthesis x equals the fraction − b plus or minus the square root of plus over 2 a close parenthesis")[$( x = frac("−" b ± sqrt(+), 2 a) )$] the quadratic equation has #strong[two solutions].

When the discriminant is #strong[zero] #math.equation(block: false, alt: "open parenthesis x equals the fraction − b plus or minus the square root of 0 over 2 a close parenthesis")[$( x = frac("−" b ± sqrt(0), 2 a) )$] the quadratic equation has #strong[one solution].

When the discriminant is #strong[negative] #math.equation(block: false, alt: "open parenthesis x equals the fraction − b plus or minus the square root of minus over 2 a close parenthesis")[$( x = frac("−" b ± sqrt(−), 2 a) )$] the quadratic equation has #strong[no real solutions].

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Use the discriminant, #math.equation(block: false, alt: "b squared minus 4 a c")[$b^(2) − 4 a c$], to determine the number of solutions of a Quadratic Equation.]

For a quadratic equation of the form #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$], #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$],

- if #math.equation(block: false, alt: "b squared minus 4 a c greater than 0")[$b^(2) − 4 a c > 0$], the equation has two solutions.
- if #math.equation(block: false, alt: "b squared minus 4 a c equals 0")[$b^(2) − 4 a c = 0$], the equation has one solution.
- if #math.equation(block: false, alt: "b squared minus 4 a c less than 0")[$b^(2) − 4 a c < 0$], the equation has no real solutions.
]

#examplebox("Example 9")[][
Determine the number of solutions to each quadratic equation:

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

ⓐ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "2 v squared minus 3 v plus 6, equals, 0")[$#h(1.2em) 2 v^(2) − 3 v + 6 & = & 0$]]),
  [The equation is in standard form, identify #emph[a], #emph[b], #emph[c].], [#math.equation(block: false, alt: "a equals 2 , b equals −3 , c, equals, 6")[$a = 2 , b = −3 , c & = & 6$]],
  [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], #emph[c].], [#math.equation(block: false, alt: "open parenthesis 3 close parenthesis squared minus 4 times 2 times 6")[$attach(( 3 ), t: 2) − 4 · 2 · 6$]],
  [Simplify.], [#math.equation(block: false, alt: "9 minus 48; −39")[$9 − 48 \ −39$]],
  [Because the discriminant is negative, there are no real #linebreak() solutions to the equation.], [],
))

ⓑ  #linebreak()

#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")[$#h(1em) 3 x^(2) + 7 x − 9 = 0$]]),
  [The equation is in standard form, identify #emph[a], #emph[b], #emph[c].], [#math.equation(block: false, alt: "a equals 3 , b equals 7 , c equals −9")[$a = 3 , b = 7 , c = −9$]],
  [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], #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; 157")[$49 + 108 \ 157$]],
  [Because the discriminant is positive, there are two #linebreak() solutions to the equation.], [],
))

ⓒ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "5 n squared plus n plus 4, equals, 0")[$#h(1.1em) 5 n^(2) + n + 4 & = & 0$]]),
  [#linebreak() The equation is in standard form, identify #emph[a], #emph[b], #emph[c].], [#math.equation(block: false, alt: "a equals 5 , b equals 1 , c, equals, 4")[$a = 5 , b = 1 , c & = & 4$]],
  [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], #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; −79")[$1 − 80 \ −79$]],
  [Because the discriminant is negative, there are no real #linebreak() solutions to the equation.], [],
))

ⓓ  #linebreak()

#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")[$#h(1.1em) 9 y^(2) − 6 y + 1 & = & 0$]]),
  [#linebreak() The equation is in standard form, identify #emph[a], #emph[b], #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], #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; 0")[$36 − 36 \ 0$]],
  [Because the discriminant is 0, there is one solution to the equation.], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Determine the number 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")[$9 w^(2) + 24 w + 16 = 0$] ⓓ #math.equation(block: false, alt: "9 u squared minus 2 u plus 4 equals 0")[$9 u^(2) − 2 u + 4 = 0$]

#solutionbox[
ⓐ no real solutions ⓑ 2 ⓒ 1 ⓓ no real solutions
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Determine the number 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")[$4 r^(2) − 20 r + 25 = 0$] ⓓ #math.equation(block: false, alt: "7 t squared minus 11 t plus 3 equals 0")[$7 t^(2) − 11 t + 3 = 0$]

#solutionbox[
ⓐ 2 ⓑ no real solutions ⓒ 1 ⓓ 2
]
]


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

We have used four methods to solve quadratic equations:

- Factoring
- Square Root Property
- Completing the Square
- Quadratic Formula

You can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method to use.

#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 quadratic equation can be solved by using the Quadratic Formula.
]

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 this chapter 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.

#examplebox("Example 10")[][
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[
ⓐ #math.equation(block: false, alt: "5 z squared equals 17")[$5 z^(2) = 17$]

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.

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

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

ⓒ #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 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[
ⓐ factor ⓑ 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 Formula ⓑ factoring ⓒ 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/25Quadformula")[Solving Quadratic Equations: Solving with the Quadratic Formula]
- #link("https://openstax.org/l/25Usequadform")[How to solve a quadratic equation in standard form using the Quadratic Formula (example)]
- #link("https://openstax.org/l/25Byquadformula")[Solving Quadratic Equations using the Quadratic Formula—Example 3]
- #link("https://openstax.org/l/25Solvequadform")[Solve Quadratic Equations using Quadratic Formula]
]


=== Key Concepts

- #strong[Quadratic Formula] The solutions to a quadratic equation of the form #math.equation(block: false, alt: "a x squared plus b x plus c equals 0 ,")[$a x^(2) + b x + c = 0 ,$] #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")[$x = frac("−" b ± sqrt(b^(2) − 4 a c), 2 a)$]
- #strong[Solve a Quadratic Equation Using the Quadratic Formula] #linebreak() 
To solve a quadratic equation using the Quadratic Formula.
+ Write the quadratic formula in standard form. Identify the #math.equation(block: false, alt: "a , b , c")[$a , b , c$] values.
+ Write the quadratic formula. Then substitute in the values of #math.equation(block: false, alt: "a , b , c .")[$a , b , c .$]
+ Simplify.
+ Check the solutions.
- #strong[Using the Discriminant, #math.equation(block: false, alt: "b squared minus 4 a c")[$b^(2) − 4 a c$], to Determine the Number of Solutions of a Quadratic Equation] #linebreak() 
For a quadratic equation of the form #math.equation(block: false, alt: "a x squared plus b x plus c equals 0 ,")[$a x^(2) + b x + c = 0 ,$] #math.equation(block: false, alt: "a not equal to 0 ,")[$a ≠ 0 ,$]
- if #math.equation(block: false, alt: "b squared minus 4 a c greater than 0")[$b^(2) − 4 a c > 0$], the equation has 2 solutions.
- if #math.equation(block: false, alt: "b squared minus 4 a c equals 0")[$b^(2) − 4 a c = 0$], the equation has 1 solution.
- if #math.equation(block: false, alt: "b squared minus 4 a c less than 0")[$b^(2) − 4 a c < 0$], the equation has no real solutions.
- #strong[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 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 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 minus 33 equals 0")[$r^(2) − 8 r − 33 = 0$]

#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 plus 40 equals 0")[$t^(2) + 13 t + 40 = 0$]

#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: "6 z squared minus 9 z plus 1 equals 0")[$6 z^(2) − 9 z + 1 = 0$]

#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: "2 x squared plus 3 x plus 9 equals 0")[$2 x^(2) + 3 x + 9 = 0$]

#solutionbox[
no real solution
]

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

#math.equation(block: true, alt: "v open parenthesis v plus 5 close parenthesis minus 10 equals 0")[$v ( v + 5 ) − 10 = 0$]

#solutionbox[
#math.equation(block: true, alt: "v equals the fraction −5 plus or minus the square root of 65 over 2")[$v = frac(−5 ± sqrt(65), 2)$]
]

#math.equation(block: true, alt: "3 w open parenthesis w minus 2 close parenthesis minus 8 equals 0")[$3 w ( w − 2 ) − 8 = 0$]

#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: "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: "5 m squared plus 2 m minus 7 equals 0")[$5 m^(2) + 2 m − 7 = 0$]

#solutionbox[
#math.equation(block: true, alt: "m equals minus the fraction 7 over 5 , m equals 1")[$m = − frac(7, 5) , m = 1$]
]

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

#math.equation(block: true, alt: "p squared minus 6 p minus 27 equals 0")[$p^(2) − 6 p − 27 = 0$]

#solutionbox[
#math.equation(block: true, alt: "p equals −3 , p equals 9")[$p = −3 , p = 9$]
]

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

#math.equation(block: true, alt: "4 r squared plus 3 r minus 5 equals 0")[$4 r^(2) + 3 r − 5 = 0$]

#solutionbox[
#math.equation(block: true, alt: "r equals the fraction −3 plus or minus the square root of 89 over 8")[$r = frac(−3 ± sqrt(89), 8)$]
]

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

#math.equation(block: true, alt: "2 a squared plus 12 a plus 5 equals 0")[$2 a^(2) + 12 a + 5 = 0$]

#solutionbox[
#math.equation(block: true, alt: "a equals the fraction −6 plus or minus the square root of 26 over 2")[$a = frac(−6 ± sqrt(26), 2)$]
]

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

#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: "2 x squared plus 12 x minus 3 equals 0")[$2 x^(2) + 12 x − 3 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction −6 plus or minus the square root of 42 over 4")[$x = frac(−6 ± sqrt(42), 4)$]
]

#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 Solutions of a Quadratic Equation]

In the following exercises, determine the number of solutions to 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")[$6 m^(2) + 3 m − 5 = 0$]
+ ⓓ #math.equation(block: false, alt: "18 n squared minus 7 n plus 3 equals 0")[$18 n^(2) − 7 n + 3 = 0$]

#solutionbox[
ⓐ no real solutions ⓑ 1 #linebreak() ⓒ 2 ⓓ no real solutions
]

+ ⓐ #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: "15 d squared minus 4 d plus 8 equals 0")[$15 d^(2) − 4 d + 8 = 0$]

+ ⓐ #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: "4 u squared minus 12 u plus 9 equals 0")[$4 u^(2) − 12 u + 9 = 0$]
+ ⓓ #math.equation(block: false, alt: "3 v squared minus 5 v minus 1 equals 0")[$3 v^(2) − 5 v − 1 = 0$]

#solutionbox[
ⓐ 1 ⓑ no real solutions #linebreak() ⓒ 1 ⓓ 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")[$7 y^(2) + 2 y + 8 = 0$]
+ ⓓ #math.equation(block: false, alt: "25 z squared minus 60 z plus 36 equals 0")[$25 z^(2) − 60 z + 36 = 0$]

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

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

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

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

#solutionbox[
ⓐ square root ⓑ square root #linebreak() ⓒ factor
]

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


==== Everyday Math

A flare is fired straight up from a ship at sea. Solve the equation #math.equation(block: false, alt: "16 open parenthesis t squared minus 13 t plus 40 close parenthesis equals 0")[$16 ( t^(2) − 13 t + 40 ) = 0$] for #math.equation(block: false, alt: "t")[$t$], the number of seconds it will take for the flare to be at an altitude of 640 feet.

#solutionbox[
5 seconds, 8 seconds
]

An architect is designing a hotel lobby. She wants to have a triangular window looking out to an atrium, with the width of the window 6 feet more than the height. Due to energy restrictions, the area of the window must be 140 square feet. Solve the equation #math.equation(block: false, alt: "the fraction 1 over 2 h squared plus 3 h equals 140")[$frac(1, 2) h^(2) + 3 h = 140$] for #math.equation(block: false, alt: "h")[$h$], the height of the window.


==== Writing Exercises

Solve the equation #math.equation(block: false, alt: "x squared plus 10 x equals 200")[$x^(2) + 10 x = 200$] #linebreak() ⓐ by completing the square #linebreak() ⓑ using the Quadratic Formula #linebreak() ⓒ Which method do you prefer? Why?

#solutionbox[
ⓐ #math.equation(block: false, alt: "−20 , 10")[$−20 , 10$] ⓑ #math.equation(block: false, alt: "−20 , 10")[$−20 , 10$] #linebreak() ⓒ answers will vary
]

Solve the equation #math.equation(block: false, alt: "12 y squared plus 23 y equals 24")[$12 y^(2) + 23 y = 24$] #linebreak() ⓐ by completing the square #linebreak() ⓑ using the Quadratic Formula #linebreak() ⓒ 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 has four rows and four columns. The first row is a header row and it labels each column. The first column is labeled "I can …", the second "Confidently", the third “With some help” and the last "No–I don’t get it". In the “I can…” column the next row reads “solve quadratic equations using the quadratic formula.” The next row reads “use the discriminant to predict the number of solutions of a quadratic equation.” and the last row reads “identify the most appropriate method to use to solve a quadratic equation.” The remaining columns are blank.], alt: "This table has four rows and four columns. The first row is a header row and it labels each column. The first column is labeled \"I can …\", the second \"Confidently\", the third “With some help” and the last \"No–I don’t get it\". In the “I can…” column the next row reads “solve quadratic equations using the quadratic formula.” The next row reads “use the discriminant to predict the number of solutions of a quadratic equation.” and the last row reads “identify the most appropriate method to use to solve a quadratic equation.” The remaining columns are blank.", caption: none)

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