Solve Quadratic Equations Using the Quadratic Formula

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

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

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


Isolate the variable terms on one side.
Make the coefficient of $x^{2}$ equal to 1, by dividing by a.
Simplify.
To complete the square, find ${(\frac{1}{2} \cdot \frac{b}{a})}^{2}$ and add it to both sides of the equation.
${(\frac{1}{2} \frac{b}{a})}^{2} = \frac{b^{2}}{4 a^{2}}$
The left side is a perfect square, factor it.
Find the common denominator of the right side and write equivalent fractions with the common denominator.
Simplify.
Combine to one fraction.
Use the square root property.
Simplify the radical.
Add $- \frac{b}{2 a}$ to both sides of the equation.
Combine the terms on the right side.
	This equation is the Quadratic Formula.

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

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

How to Solve a Quadratic Equation Using the Quadratic Formula

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

Solution

Step 1 is to write the quadratic equation in standard form, a times x squared plus b x plus c equals zero, and identify the values a, b, and c. The equation 2 x squared plus 9 x minus 5 equals zero is in standard form. A equals 2, b equals 9, and c equals negative 5.

Step 2. Write the quadratic formula. Then substitute the values of a, b, and c. Substitute a equals 2, b equals 9, and c equals negative 5 into the equation x equals the quotient negative b plus or minus the square root of the difference b squared minus 4 a c divided by 2 a. So x equals the quotient negative 9 plus or minus the square root of the difference 9 squared minus the product 4 times 2 times negative 5 divided by the product 2 times 2.

$In step 3, simplify the fraction and solve for x. x equals the quotient negative 9 plus or minus the square root of the difference 81 minus negative 40 divided by 4. Simplify the radicand. x equals the quotient negative 9 plus or minus the square root of 121 divided by 4. Simplify the square root. x equals the quotient negative 9 plus or minus 11 divided by 4. Separate into two equations. The first equation is x equals the quotient negative 9 plus 11 divided by 4 which simplifies to 2 divided by 4. The first solution is x equals one half. The second equation is x equals the quotient negative 9 minus 11 divided by 4 which simplifies to negative 20 divided by 4. The second solution is x equals negative 5.$

The fourth, and final, step is to check the solution. Put each answer into the original equation to check. First, substitute x equals one half into the original equation, 2 x squared plus 9 x minus 5 equals 0. This yields 2 times the square of one half plus nine times one half minus 5. We need to show that this expression equals 0. Simplify the square. 2 times one fourth plus nine times one half minus 5 equals one half plus 9 halves minus 5, or 10 halves minus 5. 5 minus 5 equals 0, so x equals one half is indeed a solution. Next substitute x = negative 5 into the equation 2 x squared plus 9 x minus 5 equals 0. This yields 2 times the square of negative 5 plus 9 times negative 5 minus 5. We need to show that this expression equals 0. Simplify the square. 2 times 25 plus nine times negative 5 minus 5 equals 50 minus 45 minus 5, or 0. x equals negative 5 is a solution as well.

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 =”.

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

Solution


Write the equation in standard form by adding 5 to each side.
This equation is now in standard form.
Identify the values of $a, b, c .$
Write the Quadratic Formula.
Then substitute in the values of $a, b, c .$
Simplify.	$A mathematical equation shows x equals the fraction of (6 plus or minus the square root of 16) all over 2. This is a step in solving a quadratic equation.$
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.

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

Solution


This equation is in standard form.
Identify the values of a, b, and c.
Write the Quadratic Formula.
Then substitute in the values of a, b, and c.
Simplify.
	$A mathematical equation showing x equals the fraction with a numerator of -10 plus or minus the square root of 12, all divided by 4. This is a common step in the quadratic formula.$
Simplify the radical.	$The equation shows x equals the fraction with numerator -10 plus or minus 2 times the square root of 3, and denominator 4.$
Factor out the common factor in the numerator.	$The equation x equals the fraction 2 multiplied by the quantity -5 plus or minus the square root of 3, all divided by 4, is displayed on a white background.$
Remove the common factors.
Rewrite to show two solutions.
Check: We leave the check for you!

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

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

Solution


This equation is in standard form
Identify the values of $a, b, c .$
Write the Quadratic Formula.
Then substitute in the values of $a, b, c$ .
Simplify.	$A mathematical equation shows p equals a fraction where the numerator is -2 plus or minus the square root of (4 minus 108), and the denominator is 6.$
	$A mathematical equation showing the variable p is equal to a fraction where the numerator is -2 plus or minus the square root of -104, and the denominator is 6.$
Simplify the radical using complex numbers.
Simplify the radical.	$The image shows the mathematical equation for p, which is equal to a fraction with -2 plus or minus 2 times the square root of 26i in the numerator, and 6 in the denominator.$
Factor the common factor in the numerator.
Remove the common factors.	$A mathematical equation shows p equals the fraction with a numerator of -1 plus or minus the square root of 26 multiplied by i, and a denominator of 3.$
Rewrite in standard $a + b i$ form.
Write as two solutions.

Remember, to use the Quadratic Formula, the equation must be written in standard form, ax² + bx + c = 0. Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

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

Solution

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


Distribute to get the equation in standard form.
This equation is now in standard form
Identify the values of $a, b, c .$
Write the Quadratic Formula.
Then substitute in the values of $a, b, c$ .	$Mathematical equation showing the quadratic formula with 'b=6' (red), 'a=1' (light blue), and 'c=4' (yellow) color-coded and substituted into the expression x = -(6) ×plusminus ×sqrt{(6)^2 - 4 * 1 * 4} / (2 * 1).$
Simplify.	$The image shows a step in solving a quadratic equation, with x equal to a fraction where the numerator is -6 plus or minus the square root of (36 minus 16), and the denominator is 2.$
	$A mathematical equation is displayed: x equals the fraction with a numerator of -6 plus or minus the square root of 20, and a denominator of 2.$
Simplify the radical.	$A mathematical equation shows x equals the fraction with a numerator of -6 plus or minus 2 times the square root of 5, all divided by 2.$
Factor the common factor in the numerator.
Remove the common factors.
Write as two solutions.
Check: We leave the check for 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.

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

Solution

Our first step is to clear the fractions.


Multiply both sides by the LCD, 6, to clear the fractions.
Multiply.
Subtract 2 to get the equation in standard form.
Identify the values of a, b, and c.
Write the Quadratic Formula.
Then substitute in the values of a, b, and c.
Simplify.

Simplify the radical.	$The image shows a mathematical equation where 'u' equals the fraction whose numerator is '-4 plus or minus 2 times the square root of 10' and whose denominator is '6'.$
Factor the common factor in the numerator.
Remove the common factors.	$A mathematical equation is displayed, showing u equals a fraction: negative 2 plus or minus the square root of 10, all divided by 3.$
Rewrite to show two solutions.
Check: We leave the check for you!

Think about the equation (x − 3)² = 0. We know from the Zero Product Property that this equation has only one solution,
x = 3.

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

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

Solution


Add 25 to get the equation in standard form.
Identify the values of a, b, and c.
Write the quadratic formula.
Then substitute in the values of a, b, and c.
Simplify.	$A mathematical equation shows x equals the fraction with a numerator of 20 plus or minus the square root of 400 minus 400, and a denominator of 8.$

Simplify the radical.
Simplify the fraction.	$The image displays the mathematical equation 'x = 5/2' in black text against a plain white background, showing the value of the variable x as a fraction.$
Check: We leave the check for you!

Did you recognize that 4x² − 20x + 25 is a perfect square trinomial. It is equivalent to (2x − 5)²? If you solve
4x² − 20x + 25 = 0 by factoring and then using the Square Root Property, do you get the same result?

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

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

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

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

Quadratic Equation (in standard form)	Discriminant $b^{2} - 4 a c$	Value of the Discriminant	Number and Type of solutions
$2 x^{2} + 9 x - 5 = 0$	$\begin{matrix} 9^{2} - 4 \cdot 2 (−5) \\ 121 \end{matrix}$	+	2 real
$4 x^{2} - 20 x + 25 = 0$	$\begin{matrix} {(−20)}^{2} - 4 \cdot 4 \cdot 25 \\ 0 \end{matrix}$	0	1 real
$3 p^{2} + 2 p + 9 = 0$	$\begin{matrix} 2^{2} - 4 \cdot 3 \cdot 9 \\ - 104 \end{matrix}$	−	2 complex

When the value under the radical in the Quadratic Formula, the discriminant, is positive, the equation has two real solutions. When the value under the radical in the Quadratic Formula, the discriminant, is zero, the equation has one real solution. When the value under the radical in the Quadratic Formula, the discriminant, is negative, the equation has two complex solutions.

Determine the number of solutions to each quadratic equation.

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

Solution

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

ⓐ

Steps demonstrating the calculation of the discriminant for the quadratic equation 3x^2 + 7x - 9 = 0.
	$3 x^{2} + 7 x - 9 = 0$
The equation is in standard form, identify a, b, and c.	$a = 3, b = 7, c = −9$
Write the discriminant.	$b^{2} - 4 a c$
Substitute in the values of a, b, and c.	${(7)}^{2} - 4 \cdot 3 \cdot (−9)$
Simplify.	$49 + 108$
	$157$

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

ⓑ

Step-by-step calculation of the discriminant for the quadratic equation 5n^2 + n + 4 = 0.
	$5 n^{2} + n + 4 = 0$
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, and c.	${(1)}^{2} - 4 \cdot 5 \cdot 4$
Simplify.	$1 - 80$
	$−79$

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

ⓒ

Step-by-step calculation of the discriminant for the quadratic equation 9y^2 - 6y + 1 = 0.
	$9 y^{2} - 6 y + 1 = 0$
The equation is in standard form, identify a, b, and c.	$a = 9, b = −6, c = 1$
Write the discriminant.	$b^{2} - 4 a c$
Substitute in the values of a, b, and c.	${(−6)}^{2} - 4 \cdot 9 \cdot 1$
Simplify.	$36 - 36$
	$0$

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

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

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

Given that we have four methods to use to solve a quadratic equation, how do you decide which one to use? Factoring is often the quickest method and so we try it first. If the equation is $a x^{2} = k$ or $a {(x - h)}^{2} = k$ we use the Square Root Property. For any other equation, it is probably best to use the Quadratic Formula. Remember, you can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method.

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

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

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

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

Solution

ⓐ
$\begin{matrix} 5 z^{2} = 17 \end{matrix}$

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

ⓑ
$\begin{matrix} 4 x^{2} - 12 x + 9 = 0 \end{matrix}$

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

ⓒ

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

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

Key Concepts

Quadratic Formula
- The solutions to a quadratic equation of the form ax² + bx + c = 0, $a \neq 0$ are given by the formula:
  
  $x = \frac{− b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
How to solve a quadratic equation using the Quadratic Formula.
1. Write the quadratic equation in standard form, ax² + bx + c = 0. Identify the values of a, b, c.
2. Write the Quadratic Formula. Then substitute in the values of a, b, c.
3. Simplify.
4. Check the solutions.
Using the Discriminant, b² − 4ac, to Determine the Number and Type of Solutions of a Quadratic Equation
- For a quadratic equation of the form ax² + bx + c = 0, a≠0,
  - If b² − 4ac > 0, the equation has 2 real solutions.
  - if b² − 4ac = 0, the equation has 1 real solution.
  - if b² − 4ac < 0, the equation has 2 complex solutions.
Methods to Solve Quadratic Equations:
- Factoring
- Square Root Property
- Completing the Square
- Quadratic Formula
How to identify the most appropriate method to solve a quadratic equation.
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 ax² = 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.

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

Solution

$m = −1, m = \frac{3}{4}$

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

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

Solution

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

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

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

Solution

$p = −4, p = −3$

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

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

Solution

$r = −3, r = 11$

$t^{2} + 13 t = −40$

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

Solution

$u = \frac{−7 \pm \sqrt{73}}{6}$

$2 p^{2} + 8 p + 5 = 0$

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

Solution

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

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

$x^{2} + 8 x - 4 = 0$

Solution

$x = −4 \pm 2 \sqrt{5}$

$y^{2} + 4 y - 4 = 0$

$3 y^{2} + 5 y - 2 = 0$

Solution

$y = −2, y = \frac{1}{3}$

$6 x^{2} + 2 x - 20 = 0$

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

Solution

$x = - \frac{3}{4} \pm \frac{\sqrt{15}}{4} i$

$2 x^{2} - x + 1 = 0$

$8 x^{2} - 6 x + 2 = 0$

Solution

$x = \frac{3}{8} \pm \frac{\sqrt{7}}{8} i$

$8 x^{2} - 4 x + 1 = 0$

$(v + 1) (v - 5) - 4 = 0$

Solution

$v = 2 \pm \sqrt{13}$

$(x + 1) (x - 3) = 2$

$(y + 4) (y - 7) = 18$

Solution

$y = \frac{3 \pm \sqrt{193}}{2}$

$(x + 2) (x + 6) = 21$

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

Solution

$m = −1, m = \frac{3}{4}$

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

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

Solution

$b = \frac{−2 \pm \sqrt{22}}{6}$

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

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

Solution

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

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

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

Solution

$q = - \frac{3}{5}$

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

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

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

Solution

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.

Solution

Writing Exercises

Solve the equation $x^{2} + 10 x = 120$

ⓐ by completing the square

ⓑ using the Quadratic Formula

Solution

Answers will vary.

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

ⓐ by completing the square

ⓑ using the Quadratic Formula

Self Check

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

This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can solve quadratic equations using the quadratic formula.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can use the discriminant to predict the number of solutions of a quadratic equation.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can identify the most appropriate method to use to solve a quadratic equation.” “Confidently,” “with some help,” or “No, I don’t get it.”

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

discriminant

In the Quadratic Formula,

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

the quantity b² − 4ac is called the discriminant.