Solve Quadratic Equations Using the Square Root Property

Quadratic equations are equations of the form $a x^{2} + b x + c = 0$ , where $a \neq 0$ . They differ from linear equations by including a term with the variable raised to the second power. We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.

We have seen that some quadratic equations can be solved by factoring. In this chapter, we will use three other methods to solve quadratic equations.

Solve Quadratic Equations of the Form ax² = k Using the Square Root Property

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation $x^{2} = 9$ .

\begin{matrix} x^{2} & = & 9 \\ Put the equation in standard form. & x^{2} - 9 & = & 0 \\ Factor the left side. & (x - 3) (x + 3) & = & 0 \\ Use the Zero Product Property. & (x - 3) = 0, (x + 3) & = & 0 \\ Solve each equation. & x = 3, x & = & −3 \\ Combine the two solutions into \pm form. & x & = & \pm 3 \\ (The solution is read ‘ x is equal to positive or negative 3.’) \end{matrix}

We can easily use factoring to find the solutions of similar equations, like $x^{2} = 16$ and $x^{2} = 25$ , because 16 and 25 are perfect squares. But what happens when we have an equation like $x^{2} = 7$ ? Since 7 is not a perfect square, we cannot solve the equation by factoring.

These equations are all of the form $x^{2} = k$ .
We defined the square root of a number in this way:

If n^{2} = m, then n is a square root of m .

This leads to the Square Root Property.

Notice that the Square Root Property gives two solutions to an equation of the form $x^{2} = k$ : the principal square root of $k$ and its opposite. We could also write the solution as $x = \pm \sqrt{k}$ .

Now, we will solve the equation $x^{2} = 9$ again, this time using the Square Root Property.

\begin{matrix} x^{2} & = & 9 \\ Use the Square Root Property. & x & = & \pm \sqrt{9} \\ Simplify the radical. & x & = & \pm 3 \\ Rewrite to show the two solutions. & x = 3, x & = & −3 \end{matrix}

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation $x^{2} = 7$ .

\begin{matrix} \begin{matrix} Use the Square Root Property. \end{matrix} & \begin{matrix} x^{2} & = & 7 \\ x & = & \pm \sqrt{7} \end{matrix} \\ Rewrite to show two solutions. & x = \sqrt{7}, x = − \sqrt{7} \\ We cannot simplify \sqrt{7}, so we leave the answer as a radical. \end{matrix}

Solve: $x^{2} = 169$ .

Solution

$\begin{matrix} \begin{matrix} Use the Square Root Property. \\ Simplify the radical. \end{matrix} & \begin{matrix} x^{2} & = & 169 \\ x & = & \pm \sqrt{169} \\ x & = & \pm 13 \end{matrix} \\ Rewrite to show two solutions. & x = 13, x = −13 \end{matrix}$

How to Solve a Quadratic Equation of the Form ${a x}^{2} = k$ Using the Square Root Property

Solve: $x^{2} - 48 = 0$ .

Solution

The image shows the given equation, x squared minus 48 equals zero. Step one is to isolate the quadratic term and make its coefficient one so add 48 to both sides of the equation to get x squared by itself.

Step two is to use the Square Root Property to get x equals plus or minus the square root of 48.

Step three, simplify the square root of 48 by writing 48 as the product of 16 and three. The square root of 16 is four. The simplified solution is x equals plus or minus four square root of three.

Step four, check the solutions by substituting each solution into the original equation. When x equals four square root of three, replace x in the original equation with four square root of three to get four square root of three squared minus 48 equals zero. Simplify the left side to get 16 times three minus 48 equals zero which simplifies further to zero equals zero, a true statement. When x equals negative four square root of three, replace x in the original equation with negative four square root of three to get negative four square root of three squared minus 48 equals zero. Simplify the left side to get 16 times three minus 48 equals zero which simplifies further to zero equals zero, also a true statement.

To use the Square Root Property, the coefficient of the variable term must equal 1. In the next example, we must divide both sides of the equation by 5 before using the Square Root Property.

Solve: $5 m^{2} = 80$ .

Solution

The quadratic term is isolated.		$5 m^{2} = 80$
Divide by 5 to make its cofficient 1.		$\frac{5 m^{2}}{5} = \frac{80}{5}$
Simplify.		$m^{2} = 16$
Use the Square Root Property.		$m = \pm \sqrt{16}$
Simplify the radical.		$m = \pm 4$
Rewrite to show two solutions.		$m = 4, m = − 4$
Check the solutions.

The Square Root Property started by stating, ‘If $x^{2} = k$ , and $k \geq 0$ ’. What will happen if $k < 0$ ? This will be the case in the next example.

Solve: $q^{2} + 24 = 0$ .

Solution

Demonstrates solving the quadratic equation q^2 + 24 = 0, illustrating steps that lead to the conclusion of no real solution.
	$q^{2} + 24 = 0$
Isolate the quadratic term.	$q^{2} = −24$
Use the Square Root Property.	$q = \pm \sqrt{−24}$
The $\sqrt{−24}$ is not a real number.	There is no real solution.

Remember, we first isolate the quadratic term and then make the coefficient equal to one.

Solve: $\frac{2}{3} u^{2} + 5 = 17$ .

Solution

		$\frac{2}{3} u^{2} + 5 = 17$
Isolate the quadratic term.		$\frac{2}{3} u^{2} = 12$
Multiply by $\frac{3}{2}$ to make the coefficient 1.		$\frac{3}{2} \cdot \frac{2}{3} u^{2} = \frac{3}{2} \cdot 12$
Simplify.		$u^{2} = 18$
Use the Square Root Property.		$u = \pm \sqrt{18}$
Simplify the radical.		$u = \pm \sqrt{9} \sqrt{2}$
Simplify.		$u = \pm 3 \sqrt{2}$
Rewrite to show two solutions.		$u = 3 \sqrt{2}, u = − 3 \sqrt{2}$
Check.

The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator.

Solve: $2 c^{2} - 4 = 45$ .

Solution

Step-by-step solution of a quadratic equation using the Square Root Property.
Isolate the quadratic term. Divide by 2 to make the coefficient 1. Simplify. Use the Square Root Property. Simplify the radical. Rationalize the denominator. Simplify.	$\begin{matrix} 2 c^{2} - 4 & = & 45 \\ 2 c^{2} & = & 49 \\ \frac{2 c^{2}}{2} & = & \frac{49}{2} \\ c^{2} & = & \frac{49}{2} \\ c & = & \pm \sqrt{\frac{49}{2}} \\ c & = & \pm \frac{\sqrt{49}}{\sqrt{2}} \\ c & = & \pm \frac{\sqrt{49} \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} \\ c & = & \pm \frac{7 \sqrt{2}}{2} \end{matrix}$
Rewrite to show two solutions.	$c = \frac{7 \sqrt{2}}{2}, c = - \frac{7 \sqrt{2}}{2}$
Check. We leave the check for you.

Solve Quadratic Equations of the Form a(x − h)² = k Using the Square Root Property

We can use the Square Root Property to solve an equation like ${(x - 3)}^{2} = 16$ , too. We will treat the whole binomial, $(x - 3)$ , as the quadratic term.

Solve: ${(x - 3)}^{2} = 16$ .

Solution

		${(x - 3)}^{2} = 16$
Use the Square Root Property.		$x - 3 = \pm \sqrt{16}$
Simplify.		$x - 3 = \pm 4$
Write as two equations.		$x - 3 = 4, x - 3 = − 4$
Solve.		$x = 7, x = − 1$
Check.

Solve: ${(y - 7)}^{2} = 12$ .

Solution

		${(y - 7)}^{2} = 12$
Use the Square Root Property.		$y - 7 = \pm \sqrt{12}$
Simplify the radical.		$y - 7 = \pm 2 \sqrt{3}$
Solve for y.		$y = 7 \pm 2 \sqrt{3}$
Rewrite to show two solutions.		$y = 7 + 2 \sqrt{3}, y = 7 - 2 \sqrt{3}$
Check.

Remember, when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.

Solve: ${(x - \frac{1}{2})}^{2} = \frac{5}{4} .$

Solution

This table illustrates the step-by-step solution of a quadratic equation using the Square Root Property.
Use the Square Root Property. Rewrite the radical as a fraction of square roots. Simplify the radical. Solve for x.	$\begin{matrix} {(x - \frac{1}{2})}^{2} & = & \frac{5}{4} \\ x - \frac{1}{2} & = & \pm \sqrt{\frac{5}{4}} \\ x - \frac{1}{2} & = & \pm \frac{\sqrt{5}}{\sqrt{4}} \\ x - \frac{1}{2} & = & \pm \frac{\sqrt{5}}{2} \\ x & = & \frac{1}{2} \pm \frac{\sqrt{5}}{2} \end{matrix}$
Rewrite to show two solutions.	$x = \frac{1}{2} + \frac{\sqrt{5}}{2}, x = \frac{1}{2} - \frac{\sqrt{5}}{2}$
Check. We leave the check for you.

We will start the solution to the next example by isolating the binomial.

Solve: ${(x - 2)}^{2} + 3 = 30$ .

Solution

Step-by-step solution to a quadratic equation using the Square Root Property.
	$\begin{matrix} {(x - 2)}^{2} + 3 & = & 30 \end{matrix}$
Isolate the binomial term.	$\begin{matrix} {(x - 2)}^{2} & = & 27 \end{matrix}$
Use the Square Root Property.	$\begin{matrix} x - 2 & = & \pm \sqrt{27} \end{matrix}$
Simplify the radical.	$\begin{matrix} x - 2 & = & \pm 3 \sqrt{3} \end{matrix}$
Solve for x.	$\begin{matrix} x & = & 2 \pm 3 \sqrt{3} \end{matrix}$
Rewrite to show two solutions.	$x = 2 + 3 \sqrt{3}, x = 2 - 3 \sqrt{3}$
Check. We leave the check for you.

Solve: ${(3 v - 7)}^{2} = −12$ .

Solution

Demonstration of applying the square root property to solve (3v-7)^2 = -12, resulting in no real solution.
Use the Square Root Property.	$\begin{matrix} {(3 v - 7)}^{2} & = & −12 \\ 3 v - 7 & = & \pm \sqrt{−12} \end{matrix}$
The $\sqrt{−12}$ is not a real number.	There is no real solution.

The left sides of the equations in the next two examples do not seem to be of the form $a {(x - h)}^{2}$ . But they are perfect square trinomials, so we will factor to put them in the form we need.

Solve: $p^{2} - 10 p + 25 = 18$ .

Solution

The left side of the equation is a perfect square trinomial. We will factor it first.

Demonstrates solving a perfect square trinomial using the Square Root Property, showing steps like factoring, simplifying radicals, and presenting two solutions for p.
Factor the perfect square trinomial. Use the Square Root Property. Simplify the radical. Solve for p.	$\begin{matrix} p^{2} - 10 p + 25 & = & 18 \\ {(p - 5)}^{2} & = & 18 \\ p - 5 & = & \pm \sqrt{18} \\ p - 5 & = & \pm 3 \sqrt{2} \\ p & = & 5 \pm 3 \sqrt{2} \end{matrix}$
Rewrite to show two solutions.	$p = 5 + 3 \sqrt{2,} p = 5 - 3 \sqrt{2}$
Check. We leave the check for you.

Solve: $4 n^{2} + 4 n + 1 = 16$ .

Solution

Again, we notice the left side of the equation is a perfect square trinomial. We will factor it first.

		$4 n^{2} + 4 n + 1 = 16$
Factor the perfect square trinomial.		${(2 n + 1)}^{2} = 16$
Use the Square Root Property.		$2 n + 1 = \pm \sqrt{16}$
Simplify the radical.		$2 n + 1 = \pm 4$
Solve for n.		$2 n = − 1 \pm 4$
Divide each side by 2.		$\begin{matrix} \frac{2 n}{2} & = & \frac{- 1 \pm 4}{2} \\ n & = & \frac{- 1 \pm 4}{2} \end{matrix}$
Rewrite to show two solutions.		$n = \frac{- 1 + 4}{2}, n = \frac{- 1 - 4}{2}$
Simplify each equation.		$n = \frac{3}{2}, n = - \frac{5}{2}$
Check.

Key Concepts

Square Root Property
If $x^{2} = k$ , and $k \geq 0$ , then $x = \sqrt{k} or x = − \sqrt{k}$ .

Practice Makes Perfect

Solve Quadratic Equations of the form $a x^{2} = k$ Using the Square Root Property

In the following exercises, solve the following quadratic equations.

$a^{2} = 49$

Solution

$a = \pm 7$

$b^{2} = 144$

$r^{2} - 24 = 0$

Solution

$r = \pm 2 \sqrt{6}$

$t^{2} - 75 = 0$

$u^{2} - 300 = 0$

Solution

$u = \pm 10 \sqrt{3}$

$v^{2} - 80 = 0$

$4 m^{2} = 36$

Solution

$m = \pm 3$

$3 n^{2} = 48$

$x^{2} + 20 = 0$

Solution

no real solution

$y^{2} + 64 = 0$

$\frac{2}{5} a^{2} + 3 = 11$

Solution

$a = \pm 2 \sqrt{5}$

$\frac{3}{2} b^{2} - 7 = 41$

$7 p^{2} + 10 = 26$

Solution

$p = \pm \frac{4 \sqrt{7}}{7}$

$2 q^{2} + 5 = 30$

Solve Quadratic Equations of the Form $a {(x - h)}^{2} = k$ Using the Square Root Property

In the following exercises, solve the following quadratic equations.

${(x + 2)}^{2} = 9$

Solution

$x = 1, x = −5$

${(y - 5)}^{2} = 36$

${(u - 6)}^{2} = 64$

Solution

$u = 14, u = −2$

${(v + 10)}^{2} = 121$

${(m - 6)}^{2} = 20$

Solution

$m = 6 \pm 2 \sqrt{5}$

${(n + 5)}^{2} = 32$

${(r - \frac{1}{2})}^{2} = \frac{3}{4}$

Solution

$r = \frac{1}{2} \pm \frac{\sqrt{3}}{2}$

${(t - \frac{5}{6})}^{2} = \frac{11}{25}$

${(a - 7)}^{2} + 5 = 55$

Solution

$a = 7 \pm 5 \sqrt{2}$

${(b - 1)}^{2} - 9 = 39$

${(5 c + 1)}^{2} = −27$

Solution

no real solution

${(8 d - 6)}^{2} = −24$

$m^{2} - 4 m + 4 = 8$

Solution

$m = 2 \pm 2 \sqrt{2}$

$n^{2} + 8 n + 16 = 27$

$25 x^{2} - 30 x + 9 = 36$

Solution

$x = - \frac{3}{5}, x = \frac{9}{5}$

$9 y^{2} + 12 y + 4 = 9$

Mixed Practice

In the following exercises, solve using the Square Root Property.

$2 r^{2} = 32$

Solution

$r = \pm 4$

$4 t^{2} = 16$

${(a - 4)}^{2} = 28$

Solution

$a = 4 \pm 2 \sqrt{7}$

${(b + 7)}^{2} = 8$

$9 w^{2} - 24 w + 16 = 1$

Solution

$w = 1, w = \frac{5}{3}$

$4 z^{2} + 4 z + 1 = 49$

$a^{2} - 18 = 0$

Solution

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

$b^{2} - 108 = 0$

${(p - \frac{1}{3})}^{2} = \frac{7}{9}$

Solution

$p = \frac{1}{3} \pm \frac{\sqrt{7}}{3}$

${(q - \frac{3}{5})}^{2} = \frac{3}{4}$

$m^{2} + 12 = 0$

Solution

no real solution

$n^{2} + 48 = 0$

$u^{2} - 14 u + 49 = 72$

Solution

$u = 7 \pm 6 \sqrt{2}$

$v^{2} + 18 v + 81 = 50$

${(m - 4)}^{2} + 3 = 15$

Solution

$m = 4 \pm 2 \sqrt{3}$

${(n - 7)}^{2} - 8 = 64$

${(x + 5)}^{2} = 4$

Solution

$x = −3, x = −7$

${(y - 4)}^{2} = 64$

$6 c^{2} + 4 = 29$

Solution

$c = \pm \frac{5 \sqrt{6}}{6}$

$2 d^{2} - 4 = 77$

${(x - 6)}^{2} + 7 = 3$

Solution

no real solution

${(y - 4)}^{2} + 10 = 9$

Everyday Math

Paola has enough mulch to cover 48 square feet. She wants to use it to make three square vegetable gardens of equal sizes. Solve the equation $3 s^{2} = 48$ to find $s$ , the length of each garden side.

Solution

4 feet

Kathy is drawing up the blueprints for a house she is designing. She wants to have four square windows of equal size in the living room, with a total area of 64 square feet. Solve the equation $4 s^{2} = 64$ to find $s$ , the length of the sides of the windows.

Writing Exercises

Explain why the equation $x^{2} + 12 = 8$ has no solution.

Solution

Answers will vary.

Explain why the equation $y^{2} + 8 = 12$ has two solutions.

Self Check

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

This table has three 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 of the form a x squared equals k using the square root property.” and the last row reads “solve quadratic equations of the form a times the quantity x minus h squared equals k using the square root property.” The remaining columns are blank.

ⓑ If most of your checks were:

…confidently: Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help: This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no-I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

quadratic equation

A quadratic equation is an equation of the form

a x^{2} + b x + c = 0

, where

a \neq 0 .

Square Root Property

The Square Root Property states that, if

x^{2} = k

and

k \geq 0

, then

x = \sqrt{k} or x = − \sqrt{k .}

Solve Quadratic Equations Using the Square Root Property

Solve Quadratic Equations of the Form ax2 = k Using the Square Root Property

Solution

How to Solve a Quadratic Equation of the Form ax2=k Using the Square Root Property

Solution

Solution

Solution

Solution

Solution

Solve Quadratic Equations of the Form a(x − h)2 = k Using the Square Root Property

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Key Concepts

Practice Makes Perfect

Everyday Math

Writing Exercises

Self Check

Solve Quadratic Equations of the Form ax² = k Using the Square Root Property

How to Solve a Quadratic Equation of the Form ${a x}^{2} = k$ Using the Square Root Property

Solve Quadratic Equations of the Form a(x − h)² = k Using the Square Root Property