10.3 Solve Quadratic Equations Using the Quadratic Formula

When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes.’ 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 $a x^{2} + b x + c = 0$ as you read through the algebraic steps below, so you see them with numbers as well as ‘in general.’

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

This last equation is the Quadratic Formula.

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

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 ‘ $x =$ ’.

Example 2

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


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.	$A mathematical equation is displayed on a white background: x equals a fraction where the numerator is 6 plus or minus the square root of 36 minus 20, and the denominator is 2.$ $A mathematical equation shows X equals the fraction of (6 plus or minus 4) over 2, indicating two possible solutions for X.$
Rewrite to show two solutions.
Simplify.
Check.

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 3

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

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.	$A mathematical equation for y is shown on a white background, which states y equals a fraction with 5 plus or minus the square root of 73 in the numerator, and 8 in the denominator.$
Rewrite to show two solutions.
Check. We leave the check to you.

Example 4

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


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.	$The equation shows x equals a fraction where the numerator is -5 plus or minus the square root of 3, and the denominator is 2, representing solutions from a quadratic formula.$
Rewrite to show two solutions.
Check. We leave the check to you.

We cannot take the square root of a negative number. So, when we substitute $a$ , $b$ , and $c$ 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 5

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

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.	$A mathematical equation shows p equals a fraction. The numerator is -2 plus or minus the square root of -104, and the denominator is 6.$
We cannot take the square root of a negative number.	There is no real solution.

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

Example 6

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


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.	$A mathematical equation shows X equals a fraction with a numerator of -6 plus or minus the square root of 36 minus 16, all divided by 2. This is a step in solving a quadratic equation.$
Simplify inside the radical.
Simplify the radical.	$A mathematical equation for x is displayed as a fraction: x equals negative six plus or minus two times the square root of five, all divided by two.$
Factor out the common factor in the numerator.
Remove the common factors.
Rewrite to show two solutions.
Check. We leave the check to you.

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 7

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


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.	$A mathematical equation for 'u' is shown, calculated as the fraction of '-4 plus or minus the square root of (16 + 24)' all divided by '6'.$
Simplify the radical.
Factor out the common factor in the numerator.	$A mathematical equation shows u equals a fraction: numerator is 2 multiplied by the quantity -2 plus or minus the square root of 10, and the denominator is 6.$
Remove the common factors.	$The image displays a mathematical equation, expressing the variable 'u' as a fraction: the numerator is -2 plus or minus the square root of 10, and the denominator is 3.$
Rewrite to show two solutions.
Check. We leave the check to you.

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

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 8

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


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.	$A mathematical equation shows x equals a fraction where the numerator is 20 plus or minus the square root of 400 minus 400, and the denominator is 8.$ $A mathematical equation shows x equals a fraction where the numerator is 20 plus or minus the square root of 0, and the denominator is 8.$
Simplify the radical.
Simplify the fraction.
Check. We leave the check to you.

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

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.

Let’s look at the discriminant of the equations in Example 1, Example 5, and Example 8, and the number of solutions to those quadratic equations.

	Quadratic Equation (in standard form)	Discriminant $b squared minus 4 a c$	Sign of the Discriminant	Number of real solutions
Example 1	$2 x^{2} + 9 x - 5 = 0$	$9^{2} - 4 \cdot 2 (−5) = 121$	+	2
Example 8	$4 x^{2} - 20 x + 25 = 0$	${(−20)}^{2} - 4 \cdot 4 \cdot 25 = 0$	0	1
Example 5	$3 p^{2} + 2 p + 9 = 0$	$2^{2} - 4 \cdot 3 \cdot 9 = −104$	−	0

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

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

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

Example 9

Determine the number of solutions to each quadratic equation:

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

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

ⓐ

	$\begin{matrix} 2 v^{2} - 3 v + 6 & = & 0 \end{matrix}$
The equation is in standard form, identify a, b, c.	$\begin{matrix} a = 2, b = −3, c & = & 6 \end{matrix}$
Write the discriminant.	$b squared minus 4 a c$
Substitute in the values of a, b, c.	${(3)}^{2} - 4 \cdot 2 \cdot 6$
Simplify.	$9 minus 48; −39$
Because the discriminant is negative, there are no real solutions to the equation.

ⓑ

	$3 x^{2} + 7 x - 9 = 0$
The equation is in standard form, identify a, b, c.	$a = 3, b = 7, c = −9$
Write the discriminant.	$b squared minus 4 a c$
Substitute in the values of a, b, c.	${(7)}^{2} - 4 \cdot 3 \cdot (−9)$
Simplify.	$49 plus 108; 157$
Because the discriminant is positive, there are two solutions to the equation.

ⓒ

	$\begin{matrix} 5 n^{2} + n + 4 & = & 0 \end{matrix}$
The equation is in standard form, identify a, b, c.	$\begin{matrix} a = 5, b = 1, c & = & 4 \end{matrix}$
Write the discriminant.	$b squared minus 4 a c$
Substitute in the values of a, b, c.	${(1)}^{2} - 4 \cdot 5 \cdot 4$
Simplify.	$1 minus 80; −79$
Because the discriminant is negative, there are no real solutions to the equation.

ⓓ

	$\begin{matrix} 9 y^{2} - 6 y + 1 & = & 0 \end{matrix}$
The equation is in standard form, identify a, b, c.	$\begin{matrix} a = 9, b = −6, c & = & 1 \end{matrix}$
Write the discriminant.	$b squared minus 4 a c$
Substitute in the values of a, b, c.	${(−6)}^{2} - 4 \cdot 9 \cdot 1$
Simplify.	$36 minus 36; 0$
Because the discriminant is 0, there is one solution to the equation.

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.

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.

Key Concepts

Quadratic Formula The solutions to a quadratic equation of the form $a x^{2} + b x + c = 0,$ $a not equal to 0$ are given by the formula:

$x = \frac{− b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Solve a Quadratic Equation Using the Quadratic Formula
To solve a quadratic equation using the Quadratic Formula.
1. Write the quadratic formula in standard form. Identify the $a, b, c$ values.
2. Write the quadratic formula. Then substitute in the values of $a, b, c .$
3. Simplify.
4. Check the solutions.
Using the Discriminant, $b squared minus 4 a c$ , to Determine the Number of Solutions of a Quadratic Equation
For a quadratic equation of the form ax2+bx+c=0, a≠0,
- if $b squared minus 4 a c greater than 0$ , the equation has 2 solutions.
- if $b^{2} - 4 a c = 0$ , the equation has 1 solution.
- if $b squared minus 4 a c less than 0$ , the equation has no real solutions.
To identify the most appropriate method to solve a quadratic equation:
1. Try Factoring first. If the quadratic factors easily this method is very quick.
2. Try the Square Root Property next. If the equation fits the form $a x^{2} = k$ or $a {(x - h)}^{2} = k$ , it can easily be solved by using the Square Root Property.
3. Use the Quadratic Formula. Any other quadratic equation is best solved by using the Quadratic Formula.

Practice Makes Perfect

Solve Quadratic Equations Using the Quadratic Formula

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

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

$m = −1, m = \frac{3}{4}$

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

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

$p = \frac{1}{2}, p = 3$

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

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

$p = −4, p = −3$

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

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

$r = −3, r = 11$

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

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

$u = \frac{−7 \pm \sqrt{73}}{6}$

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

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

$a = \frac{3 \pm \sqrt{3}}{2}$

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

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

no real solution

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

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

$v = \frac{−5 \pm \sqrt{65}}{2}$

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

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

$m = −1, m = \frac{3}{4}$

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

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

$c = - \frac{3}{4}$

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

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

$m = - \frac{7}{5}, m = 1$

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

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

$p = −3, p = 9$

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

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

$r = \frac{−3 \pm \sqrt{89}}{8}$

$3 t (t - 2) = 2$

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

$a = \frac{−6 \pm \sqrt{26}}{2}$

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

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

$b = \frac{−2 \pm \sqrt{22}}{6}$

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

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

$x = \frac{−6 \pm \sqrt{42}}{4}$

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

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.

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

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

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

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

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.

Everyday Math

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

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

Writing Exercises

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

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