Sumário

Solve Quadratic Equations Using the Quadratic Formula

Learning Objectives

By the end of this section, you will be able to:

Solve quadratic equations using the quadratic formula
Use the discriminant to predict the number of solutions of a quadratic equation
Identify the most appropriate method to use to solve a quadratic equation

Be Prepared 10.3

Before you get started, take this readiness quiz.

Simplify: $\frac{−20 - 5}{10} \frac{−20 - 5}{10}$ .
If you missed this problem, review Example 1.74.
Simplify: $4 + \sqrt{121} 4 + \sqrt{121}$ .
If you missed this problem, review Example 9.29.
Simplify: $\sqrt{128} \sqrt{128}$ .
If you missed this problem, review Example 9.12.

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 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 $a x^{2} + b x + c = 0 a x^{2} + b x + c = 0$ as you read through the algebraic steps below, so you see them with numbers as well as ‘in general.’

$\begin{matrix} We start with the standard form of a quadratic equation & a x^{2} + b x + c & = & 0 a \neq 0 \\ and solve it for x by completing the square. \\ Isolate the variable terms on one side. & a x^{2} + b x & = & − c \\ Make leading coefficient 1, by dividing by a. & \frac{a x^{2}}{a} + \frac{b}{a} x & = & - \frac{c}{a} \\ Simplify. & x^{2} + \frac{b}{a} x & = & - \frac{c}{a} \\ To complete the square, find {(\frac{1}{2} \cdot \frac{b}{a})}^{2} and add it to both \\ sides of the equation. {(\frac{1}{2} \frac{b}{a})}^{2} = \frac{b^{2}}{4 a^{2}} & 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. & {(x + \frac{b}{2 a})}^{2} & = & - \frac{c}{a} + \frac{b^{2}}{4 a^{2}} \\ \begin{matrix} Find the common denominator of the right side and write \\ equivalent fractions with the common denominator. \end{matrix} & {(x + \frac{b}{2 a})}^{2} & = & \frac{b^{2}}{4 a^{2}} - \frac{c \cdot 4 a}{a \cdot 4 a} \\ Simplify. & {(x + \frac{b}{2 a})}^{2} & = & \frac{b^{2}}{4 a^{2}} - \frac{4 a c}{4 a^{2}} \\ Combine to one fraction. & {(x + \frac{b}{2 a})}^{2} & = & \frac{b^{2} - 4 a c}{4 a^{2}} \\ Use the square root property. & x + \frac{b}{2 a} & = & \pm \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}} \\ Simplify. & x + \frac{b}{2 a} & = & \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} \\ Add - \frac{b}{2 a} to both sides of the equation. & x & = & - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} \\ Combine the terms on the right side. & x & = & \frac{− b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{matrix} \begin{matrix} We start with the standard form of a quadratic equation & a x^{2} + b x + c = 0 \\ and solve it for \\ Isolate the variable terms on one side. & a x^{2} + b x = − c \\ Make leading coefficient 1, by dividing by a. & \frac{a x^{2}}{a} + \frac{b}{a} x = - \frac{c}{a} & Simplify. & x^{2} + \frac{b}{a} x = - \frac{c}{a} & To complete the square, find & sides of the equation. & 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. & {(x + \frac{b}{2 a})}^{2} = - \frac{c}{a} + \frac{b^{2}}{4 a^{2}} & \begin{matrix} Find the common denominator of the right side and write \\ equivalent fractions with the common denominator. \end{matrix} & {(x + \frac{b}{2 a})}^{2} = \frac{b^{2}}{4 a^{2}} - \frac{c \cdot 4 a}{a \cdot 4 a} & Simplify. & {(x + \frac{b}{2 a})}^{2} = \frac{b^{2}}{4 a^{2}} - \frac{4 a c}{4 a^{2}} & Combine to one fraction. & {(x + \frac{b}{2 a})}^{2} = \frac{b^{2} - 4 a c}{4 a^{2}} & Use the square root property. & x + \frac{b}{2 a} = \pm & Simplify. & x + \frac{b}{2 a} = \pm & Add & x = - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} & Combine the terms on the right side. & x = \frac{− b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{matrix}$

This last equation is the Quadratic Formula.

Quadratic Formula

The solutions to a quadratic equation of the form $a x^{2} + b x + c = 0 a x^{2} + b x + c = 0$ , $a \neq 0 a \neq 0$ are given by the formula:

x = \frac{− b \pm \sqrt{b^{2} - 4 a c}}{2 a} x = \frac{− b \pm \sqrt{b^{2} - 4 a c}}{2 a}

To use the Quadratic Formula, we substitute the values of $a, b, and c a, b,$ 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.

Example 10.28

How to Solve a Quadratic Equation Using the Quadratic Formula

Solve $2 x^{2} + 9 x - 5 = 0 2 x^{2} + 9 x - 5 = 0$ by using the Quadratic Formula.

Mostrar solução

Solution

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.

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.

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.

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.

Try It 10.55

Solve $3 y^{2} - 5 y + 2 = 0 3 y^{2} - 5 y + 2 = 0$ by using the Quadratic Formula.

Try It 10.56

Solve $4 z^{2} + 2 z - 6 = 0 4 z^{2} + 2 z - 6 = 0$ by using the Quadratic Formula.

How To

Solve a quadratic equation using the Quadratic Formula.

Step 1. Write the Quadratic Formula in standard form. Identify the $a a$ , $b b$ , and $c c$ values.
Step 2. Write the Quadratic Formula. Then substitute in the values of $a a$ , $b b$ , and $c . c .$
Step 3. Simplify.
Step 4. 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 ‘ $x = x =$ ’.

Example 10.29

Solve $x^{2} - 6 x + 5 = 0 x^{2} - 6 x + 5 = 0$ by using the Quadratic Formula.

Mostrar solução

Solution


This equation is in standard form.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.
Rewrite to show two solutions.
Simplify.
Check.

Try It 10.57

Solve $a^{2} - 2 a - 15 = 0 a^{2} - 2 a - 15 = 0$ by using the Quadratic Formula.

Try It 10.58

Solve $b^{2} + 10 b + 24 = 0 b^{2} + 10 b + 24 = 0$ by using the Quadratic Formula.

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.

Example 10.30

Solve $4 y^{2} - 5 y - 3 = 0 4 y^{2} - 5 y - 3 = 0$ by using the Quadratic Formula.

Mostrar solução

Solution

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


This equation is in standard form.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.
Rewrite to show two solutions.
Check. We leave the check to you.

Try It 10.59

Solve $2 p^{2} + 8 p + 5 = 0 2 p^{2} + 8 p + 5 = 0$ by using the Quadratic Formula.

Try It 10.60

Solve $5 q^{2} - 11 q + 3 = 0 5 q^{2} - 11 q + 3 = 0$ by using the Quadratic Formula.

Example 10.31

Solve $2 x^{2} + 10 x + 11 = 0 2 x^{2} + 10 x + 11 = 0$ by using the Quadratic Formula.

Mostrar solução

Solution


This equation is in standard form.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.
Simplify the radical.
Factor out the common factor in the numerator.
Remove the common factors.
Rewrite to show two solutions.
Check. We leave the check to you.

Try It 10.61

Solve $3 m^{2} + 12 m + 7 = 0 3 m^{2} + 12 m + 7 = 0$ by using the Quadratic Formula.

Try It 10.62

Solve $5 n^{2} + 4 n - 4 = 0 5 n^{2} + 4 n - 4 = 0$ by using the Quadratic Formula.

We cannot take the square root of a negative number. So, when we substitute $a a$ , $b b$ , and $c c$ 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.

Example 10.32

Solve $3 p^{2} + 2 p + 9 = 0 3 p^{2} + 2 p + 9 = 0$ by using the Quadratic Formula.

Mostrar solução

Solution

This equation is in standard form.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.
Simplify the radical.
We cannot take the square root of a negative number.		There is no real solution.

Try It 10.63

Solve $4 a^{2} - 3 a + 8 = 0 4 a^{2} - 3 a + 8 = 0$ by using the Quadratic Formula.

Try It 10.64

Solve $5 b^{2} + 2 b + 4 = 0 5 b^{2} + 2 b + 4 = 0$ by using the Quadratic Formula.

The quadratic equations we have solved so far in this section were all written in standard form, $a x^{2} + b x + c = 0 a x^{2} + b x + c = 0$ . Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

Example 10.33

Solve $x (x + 6) + 4 = 0 x (x + 6) + 4 = 0$ by using the Quadratic Formula.

Mostrar solução

Solution


Distribute to get the equation in standard form.
This equation is now in standard form.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.
Simplify inside the radical.
Simplify the radical.
Factor out the common factor in the numerator.
Remove the common factors.
Rewrite to show two solutions.
Check. We leave the check to you.

Try It 10.65

Solve $x (x + 2) - 5 = 0 x (x + 2) - 5 = 0$ by using the Quadratic Formula.

Try It 10.66

Solve $y (3 y - 1) - 2 = 0 y (3 y - 1) - 2 = 0$ by using the Quadratic Formula.

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.

Example 10.34

Solve $\frac{1}{2} u^{2} + \frac{2}{3} u = \frac{1}{3} \frac{1}{2} u^{2} + \frac{2}{3} u = \frac{1}{3}$ by using the Quadratic Formula.

Mostrar solução

Solution


Multiply both sides by the LCD, 6, to clear the fractions.
Multiply.
Subtract 2 to get the equation in standard form.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.
Simplify the radical.
Factor out the common factor in the numerator.
Remove the common factors.
Rewrite to show two solutions.
Check. We leave the check to you.

Try It 10.67

Solve $\frac{1}{4} c^{2} - \frac{1}{3} c = \frac{1}{12} \frac{1}{4} c^{2} - \frac{1}{3} c = \frac{1}{12}$ by using the Quadratic Formula.

Try It 10.68

Solve $\frac{1}{9} d^{2} - \frac{1}{2} d = - \frac{1}{2} \frac{1}{9} d^{2} - \frac{1}{2} d = - \frac{1}{2}$ by using the Quadratic Formula.

Think about the equation ${(x - 3)}^{2} = 0 {(x - 3)}^{2} = 0$ . We know from the Zero Products Principle that this equation has only one solution: $x = 3 x = 3$ .

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.

Example 10.35

Solve $4 x^{2} - 20 x = −25 4 x^{2} - 20 x = −25$ by using the Quadratic Formula.

Mostrar solução

Solution


Add 25 to get the equation in standard form.
Identify the a, b, c values.
Write the Quadratic Formula.
Then substitute in the values of a, b, c.
Simplify.
Simplify the radical.
Simplify the fraction.
Check. We leave the check to you.

Did you recognize that $4 x^{2} - 20 x + 25 4 x^{2} - 20 x + 25$ is a perfect square?

Try It 10.69

Solve $r^{2} + 10 r + 25 = 0 r^{2} + 10 r + 25 = 0$ by using the Quadratic Formula.

Try It 10.70

Solve $25 t^{2} - 40 t = −16 25 t^{2} - 40 t = −16$ by using the Quadratic Formula.

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 discriminant.

Discriminant

In the Quadratic Formula $x = \frac{− b \pm \sqrt{b^{2} - 4 a c}}{2 a} x = \frac{− b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ , the quantity $b^{2} - 4 a c b^{2} - 4 a c$ is called the discriminant.

Let’s look at the discriminant of the equations in Example 10.28, Example 10.32, and Example 10.35, and the number of solutions to those quadratic equations.

	Quadratic Equation (in standard form)	Discriminant $b^{2} - 4 a c b^{2} - 4 a c$	Sign of the Discriminant	Number of real solutions
Example 10.28	$2 x^{2} + 9 x - 5 = 0 2 x^{2} + 9 x - 5 = 0$	$9^{2} - 4 \cdot 2 (−5) = 121 9^{2} - 4 \cdot 2 (−5) = 121$	+	2
Example 10.35	$4 x^{2} - 20 x + 25 = 0 4 x^{2} - 20 x + 25 = 0$	${(−20)}^{2} - 4 \cdot 4 \cdot 25 = 0 {(−20)}^{2} - 4 \cdot 4 \cdot 25 = 0$	0	1
Example 10.32	$3 p^{2} + 2 p + 9 = 0 3 p^{2} + 2 p + 9 = 0$	$2^{2} - 4 \cdot 3 \cdot 9 = −104 2^{2} - 4 \cdot 3 \cdot 9 = −104$	−	0

When the discriminant is positive $(x = \frac{− b \pm \sqrt{+}}{2 a}) (x = \frac{− b \pm \sqrt{+}}{2 a})$ the quadratic equation has two solutions.

When the discriminant is zero $(x = \frac{− b \pm \sqrt{0}}{2 a}) (x = \frac{− b \pm \sqrt{0}}{2 a})$ the quadratic equation has one solution.

When the discriminant is negative $(x = \frac{− b \pm \sqrt{-}}{2 a}) (x = \frac{− b \pm \sqrt{-}}{2 a})$ the quadratic equation has no real solutions.

How To

Use the discriminant, $b^{2} - 4 a c b^{2} - 4 a c$ , to determine the number of solutions of a Quadratic Equation.

For a quadratic equation of the form $a x^{2} + b x + c = 0 a x^{2} + b x + c = 0$ , $a \neq 0 a \neq 0$ ,

if $b^{2} - 4 a c > 0 b^{2} - 4 a c > 0$ , the equation has two solutions.
if $b^{2} - 4 a c = 0 b^{2} - 4 a c = 0$ , the equation has one solution.
if $b^{2} - 4 a c < 0 b^{2} - 4 a c < 0$ , the equation has no real solutions.

Example 10.36

Determine the number of solutions to each quadratic equation:

ⓐ $2 v^{2} - 3 v + 6 = 0 2 v^{2} - 3 v + 6 = 0$ ⓑ $3 x^{2} + 7 x - 9 = 0 3 x^{2} + 7 x - 9 = 0$ ⓒ $5 n^{2} + n + 4 = 0 5 n^{2} + n + 4 = 0$ ⓓ $9 y^{2} - 6 y + 1 = 0 9 y^{2} - 6 y + 1 = 0$

Mostrar solução

Solution

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

ⓐ
$\begin{matrix} 2 v^{2} - 3 v + 6 = 0 \\ The equation is in standard form, identify a, b, c . & a = 2, b = −3, c = 6 \\ Write the discriminant. & b^{2} - 4 a c \\ Substitute in the values of a, b, c . & {(3)}^{2} - 4 \cdot 2 \cdot 6 \\ Simplify. & \begin{matrix} 9 - 48 \\ −39 \end{matrix} \\ Because the discriminant is negative, there are no real \\ solutions to the equation. \end{matrix} \begin{matrix} The equation is in standard form, identify \\ Write the discriminant. \\ Substitute in the values of \\ Simplify. \\ Because the discriminant is negative, there are no real \\ solutions to the equation. \end{matrix}$
ⓑ
$\begin{matrix} 3 x^{2} + 7 x - 9 = 0 \\ The equation is in standard form, identify a, b, c . & a = 3, b = 7, c = −9 \\ Write the discriminant. & b^{2} - 4 a c \\ Substitute in the values of a, b, c . & {(7)}^{2} - 4 \cdot 3 \cdot (−9) \\ Simplify. & \begin{matrix} 49 + 108 \\ 157 \end{matrix} \\ Because the discriminant is positive, there are two \\ solutions to the equation. \end{matrix} \begin{matrix} The equation is in standard form, identify \\ Write the discriminant. \\ Substitute in the values of \\ Simplify. \\ Because the discriminant is positive, there are two \\ solutions to the equation. \end{matrix}$
ⓒ
$\begin{matrix} 5 n^{2} + n + 4 = 0 \\ The equation is in standard form, identify a, b, and c . & a = 5, b = 1, c = 4 \\ Write the discriminant. & b^{2} - 4 a c \\ Substitute in the values of a, b, c . & {(1)}^{2} - 4 \cdot 5 \cdot 4 \\ Simplify. & \begin{matrix} 1 - 80 \\ −79 \end{matrix} \\ Because the discriminant is negative, there are no real \\ solutions to the equation. \end{matrix} \begin{matrix} The equation is in standard form, identify \\ Write the discriminant. \\ Substitute in the values of \\ Simplify. \\ Because the discriminant is negative, there are no real \\ solutions to the equation. \end{matrix}$
ⓓ
$\begin{matrix} 9 y^{2} - 6 y + 1 = 0 \\ The equation is in standard form, identify a, b, c . & a = 9, b = −6, c = 1 \\ Write the discriminant. & b^{2} - 4 a c \\ Substitute in the values of a, b, c . & {(−6)}^{2} - 4 \cdot 9 \cdot 1 \\ Simplify. & \begin{matrix} 36 - 36 \\ 0 \end{matrix} \\ Because the discriminant is 0, there is one solution to the equation. \end{matrix} \begin{matrix} 9 y^{2} - 6 y + 1 = 0 \\ The equation is in standard form, identify & a = 9, b = −6, c = 1 \\ Write the discriminant. & b^{2} - 4 a c \\ Substitute in the values of & {(−6)}^{2} - 4 \cdot 9 \cdot 1 \\ Simplify. & \begin{matrix} 36 - 36 \\ 0 \end{matrix} \\ Because the discriminant is 0, there is one solution to the equation. \end{matrix}$

Try It 10.71

Determine the number of solutions to each quadratic equation:

ⓐ $8 m^{2} - 3 m + 6 = 0 8 m^{2} - 3 m + 6 = 0$ ⓑ $5 z^{2} + 6 z - 2 = 0 5 z^{2} + 6 z - 2 = 0$ ⓒ $9 w^{2} + 24 w + 16 = 0 9 w^{2} + 24 w + 16 = 0$ ⓓ $9 u^{2} - 2 u + 4 = 0 9 u^{2} - 2 u + 4 = 0$

Try It 10.72

Determine the number of solutions to each quadratic equation:

ⓐ $b^{2} + 7 b - 13 = 0 b^{2} + 7 b - 13 = 0$ ⓑ $5 a^{2} - 6 a + 10 = 0 5 a^{2} - 6 a + 10 = 0$ ⓒ $4 r^{2} - 20 r + 25 = 0 4 r^{2} - 20 r + 25 = 0$ ⓓ $7 t^{2} - 11 t + 3 = 0 7 t^{2} - 11 t + 3 = 0$

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.

How To

Identify the most appropriate method to solve a Quadratic Equation.

Step 1. Try Factoring first. If the quadratic factors easily, this method is very quick.
Step 2. Try the Square Root Property next. If the equation fits the form $a x^{2} = k a x^{2} = k$ or $a {(x - h)}^{2} = k a {(x - h)}^{2} = k$ , it can easily be solved by using the Square Root Property.
Step 3. Use the 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.

Example 10.37

Identify the most appropriate method to use to solve each quadratic equation:

ⓐ $5 z^{2} = 17 5 z^{2} = 17$ ⓑ $4 x^{2} - 12 x + 9 = 0 4 x^{2} - 12 x + 9 = 0$ ⓒ $8 u^{2} + 6 u = 11 8 u^{2} + 6 u = 11$

Mostrar solução

Solution

ⓐ $5 z^{2} = 17 5 z^{2} = 17$

Since the equation is in the $a x^{2} = k a x^{2} = k$ , the most appropriate method is to use the Square Root Property.

ⓑ $4 x^{2} - 12 x + 9 = 0 4 x^{2} - 12 x + 9 = 0$

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

Put the equation in standard form. $8 u^{2} + 6 u - 11 = 0 8 u^{2} + 6 u - 11 = 0$

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

Try It 10.73

Identify the most appropriate method to use to solve each quadratic equation:

ⓐ $x^{2} + 6 x + 8 = 0 x^{2} + 6 x + 8 = 0$ ⓑ ${(n - 3)}^{2} = 16 {(n - 3)}^{2} = 16$ ⓒ $5 p^{2} - 6 p = 9 5 p^{2} - 6 p = 9$

Try It 10.74

Identify the most appropriate method to use to solve each quadratic equation:

ⓐ $8 a^{2} + 3 a - 9 = 0 8 a^{2} + 3 a - 9 = 0$ ⓑ $4 b^{2} + 4 b + 1 = 0 4 b^{2} + 4 b + 1 = 0$ ⓒ $5 c^{2} = 125 5 c^{2} = 125$

Media

Access these online resources for additional instruction and practice with using the Quadratic Formula:

Practice Makes Perfect

Solve Quadratic Equations Using the Quadratic Formula

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

99.

$4 m^{2} + m - 3 = 0 4 m^{2} + m - 3 = 0$

100.

$4 n^{2} - 9 n + 5 = 0 4 n^{2} - 9 n + 5 = 0$

101.

$2 p^{2} - 7 p + 3 = 0 2 p^{2} - 7 p + 3 = 0$

102.

$3 q^{2} + 8 q - 3 = 0 3 q^{2} + 8 q - 3 = 0$

103.

$p^{2} + 7 p + 12 = 0 p^{2} + 7 p + 12 = 0$

104.

$q^{2} + 3 q - 18 = 0 q^{2} + 3 q - 18 = 0$

105.

$r^{2} - 8 r - 33 = 0 r^{2} - 8 r - 33 = 0$

106.

$t^{2} + 13 t + 40 = 0 t^{2} + 13 t + 40 = 0$

107.

$3 u^{2} + 7 u - 2 = 0 3 u^{2} + 7 u - 2 = 0$

108.

$6 z^{2} - 9 z + 1 = 0 6 z^{2} - 9 z + 1 = 0$

109.

$2 a^{2} - 6 a + 3 = 0 2 a^{2} - 6 a + 3 = 0$

110.

$5 b^{2} + 2 b - 4 = 0 5 b^{2} + 2 b - 4 = 0$

111.

$2 x^{2} + 3 x + 9 = 0 2 x^{2} + 3 x + 9 = 0$

112.

$6 y^{2} - 5 y + 2 = 0 6 y^{2} - 5 y + 2 = 0$

113.

$v (v + 5) - 10 = 0 v (v + 5) - 10 = 0$

114.

$3 w (w - 2) - 8 = 0 3 w (w - 2) - 8 = 0$

115.

$\frac{1}{3} m^{2} + \frac{1}{12} m = \frac{1}{4} \frac{1}{3} m^{2} + \frac{1}{12} m = \frac{1}{4}$

116.

$\frac{1}{3} n^{2} + n = - \frac{1}{2} \frac{1}{3} n^{2} + n = - \frac{1}{2}$

117.

$16 c^{2} + 24 c + 9 = 0 16 c^{2} + 24 c + 9 = 0$

118.

$25 d^{2} - 60 d + 36 = 0 25 d^{2} - 60 d + 36 = 0$

119.

$5 m^{2} + 2 m - 7 = 0 5 m^{2} + 2 m - 7 = 0$

120.

$8 n^{2} - 3 n + 3 = 0 8 n^{2} - 3 n + 3 = 0$

121.

$p^{2} - 6 p - 27 = 0 p^{2} - 6 p - 27 = 0$

122.

$25 q^{2} + 30 q + 9 = 0 25 q^{2} + 30 q + 9 = 0$

123.

$4 r^{2} + 3 r - 5 = 0 4 r^{2} + 3 r - 5 = 0$

124.

$3 t (t - 2) = 2 3 t (t - 2) = 2$

125.

$2 a^{2} + 12 a + 5 = 0 2 a^{2} + 12 a + 5 = 0$

126.

$4 d^{2} - 7 d + 2 = 0 4 d^{2} - 7 d + 2 = 0$

127.

$\frac{3}{4} b^{2} + \frac{1}{2} b = \frac{3}{8} \frac{3}{4} b^{2} + \frac{1}{2} b = \frac{3}{8}$

128.

$\frac{1}{9} c^{2} + \frac{2}{3} c = 3 \frac{1}{9} c^{2} + \frac{2}{3} c = 3$

129.

$2 x^{2} + 12 x - 3 = 0 2 x^{2} + 12 x - 3 = 0$

130.

$16 y^{2} + 8 y + 1 = 0 16 y^{2} + 8 y + 1 = 0$

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.

131.

ⓐ $4 x^{2} - 5 x + 16 = 0 4 x^{2} - 5 x + 16 = 0$
ⓑ $36 y^{2} + 36 y + 9 = 0 36 y^{2} + 36 y + 9 = 0$
ⓒ $6 m^{2} + 3 m - 5 = 0 6 m^{2} + 3 m - 5 = 0$
ⓓ $18 n^{2} - 7 n + 3 = 0 18 n^{2} - 7 n + 3 = 0$

132.

ⓐ $9 v^{2} - 15 v + 25 = 0 9 v^{2} - 15 v + 25 = 0$
ⓑ $100 w^{2} + 60 w + 9 = 0 100 w^{2} + 60 w + 9 = 0$
ⓒ $5 c^{2} + 7 c - 10 = 0 5 c^{2} + 7 c - 10 = 0$
ⓓ $15 d^{2} - 4 d + 8 = 0 15 d^{2} - 4 d + 8 = 0$

133.

ⓐ $r^{2} + 12 r + 36 = 0 r^{2} + 12 r + 36 = 0$
ⓑ $8 t^{2} - 11 t + 5 = 0 8 t^{2} - 11 t + 5 = 0$
ⓒ $4 u^{2} - 12 u + 9 = 0 4 u^{2} - 12 u + 9 = 0$
ⓓ $3 v^{2} - 5 v - 1 = 0 3 v^{2} - 5 v - 1 = 0$

134.

ⓐ $25 p^{2} + 10 p + 1 = 0 25 p^{2} + 10 p + 1 = 0$
ⓑ $7 q^{2} - 3 q - 6 = 0 7 q^{2} - 3 q - 6 = 0$
ⓒ $7 y^{2} + 2 y + 8 = 0 7 y^{2} + 2 y + 8 = 0$
ⓓ $25 z^{2} - 60 z + 36 = 0 25 z^{2} - 60 z + 36 = 0$

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.

135.

ⓐ $x^{2} - 5 x - 24 = 0 x^{2} - 5 x - 24 = 0$
ⓑ ${(y + 5)}^{2} = 12 {(y + 5)}^{2} = 12$
ⓒ $14 m^{2} + 3 m = 11 14 m^{2} + 3 m = 11$

136.

ⓐ ${(8 v + 3)}^{2} = 81 {(8 v + 3)}^{2} = 81$
ⓑ $w^{2} - 9 w - 22 = 0 w^{2} - 9 w - 22 = 0$
ⓒ $4 n^{2} - 10 = 6 4 n^{2} - 10 = 6$

137.

ⓐ $6 a^{2} + 14 = 20 6 a^{2} + 14 = 20$
ⓑ ${(x - \frac{1}{4})}^{2} = \frac{5}{16} {(x - \frac{1}{4})}^{2} = \frac{5}{16}$
ⓒ $y^{2} - 2 y = 8 y^{2} - 2 y = 8$

138.

ⓐ $8 b^{2} + 15 b = 4 8 b^{2} + 15 b = 4$
ⓑ $\frac{5}{9} v^{2} - \frac{2}{3} v = 1 \frac{5}{9} v^{2} - \frac{2}{3} v = 1$
ⓒ ${(w + \frac{4}{3})}^{2} = \frac{2}{9} {(w + \frac{4}{3})}^{2} = \frac{2}{9}$

Everyday Math

139.

A flare is fired straight up from a ship at sea. Solve the equation $16 (t^{2} - 13 t + 40) = 0 16 (t^{2} - 13 t + 40) = 0$ for $t t$ , the number of seconds it will take for the flare to be at an altitude of 640 feet.

140.

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 $\frac{1}{2} h^{2} + 3 h = 140 \frac{1}{2} h^{2} + 3 h = 140$ for $h h$ , the height of the window.

Writing Exercises

141.

Solve the equation $x^{2} + 10 x = 200 x^{2} + 10 x = 200$
ⓐ by completing the square
ⓑ using the Quadratic Formula
ⓒ Which method do you prefer? Why?

142.

Solve the equation $12 y^{2} + 23 y = 24 12 y^{2} + 23 y = 24$
ⓐ 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.

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.

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

Learning Objectives

Be Prepared 10.3

Quadratic Formula

Example 10.28

How to Solve a Quadratic Equation Using the Quadratic Formula

Solution

Try It 10.55

Try It 10.56

How To

Solve a quadratic equation using the Quadratic Formula.

Example 10.29

Solution

Try It 10.57

Try It 10.58

Example 10.30

Solution

Try It 10.59

Try It 10.60

Example 10.31

Solution

Try It 10.61

Try It 10.62

Example 10.32

Solution

Try It 10.63

Try It 10.64

Example 10.33

Solution

Try It 10.65

Try It 10.66

Example 10.34

Solution

Try It 10.67

Try It 10.68

Example 10.35

Solution

Try It 10.69

Try It 10.70

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

Discriminant

How To

Use the discriminant, b2−4acb2−4ac, to determine the number of solutions of a Quadratic Equation.

Example 10.36

Solution

Try It 10.71

Try It 10.72

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

How To

Identify the most appropriate method to solve a Quadratic Equation.

Example 10.37

Solution

Try It 10.73

Try It 10.74

Media

Practice Makes Perfect

Everyday Math

Writing Exercises

Self Check

Use the discriminant, $b^{2} - 4 a c b^{2} - 4 a c$ , to determine the number of solutions of a Quadratic Equation.