Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis

10.1 Solve Quadratic Equations Using the Square Root Property

Learning Objectives

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

Solve quadratic equations 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

Be Prepared 10.1

Before you get started, take this readiness quiz.

Simplify: $\sqrt{75}$ .
If you missed this problem, review Example 9.12.

Be Prepared 10.2

Simplify: $\sqrt{\frac{64}{3}}$ .
If you missed this problem, review Example 9.67.

Be Prepared 10.3

Factor: $4 x^{2} - 12 x + 9$ .
If you missed this problem, review Example 7.43.

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.

Square Root Property

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

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}

Example 10.1

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

Solution

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

Try It 10.1

Solve: $x^{2} = 81$ .

Try It 10.2

Solve: $y^{2} = 121$ .

Example 10.2

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

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

Solution

Try It 10.3

Solve: $x^{2} - 50 = 0$ .

Try It 10.4

Solve: $y^{2} - 27 = 0$ .

How To

Solve a quadratic equation using the Square Root Property.

Step 1. Isolate the quadratic term and make its coefficient one.
Step 2. Use Square Root Property.
Step 3. Simplify the radical.
Step 4. Check the solutions.

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.

Example 10.3

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.

Try It 10.5

Solve: $2 x^{2} = 98$ .

Try It 10.6

Solve: $3 z^{2} = 108$ .

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.

Example 10.4

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

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.

Try It 10.7

Solve: $c^{2} + 12 = 0$ .

Try It 10.8

Solve: $d^{2} + 81 = 0$ .

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

Example 10.5

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.

Try It 10.9

Solve: $\frac{1}{2} x^{2} + 4 = 24$ .

Try It 10.10

Solve: $\frac{3}{4} y^{2} - 3 = 18$ .

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

Example 10.6

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

Solution

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.

Try It 10.11

Solve: $5 r^{2} - 2 = 34$ .

Try It 10.12

Solve: $3 t^{2} + 6 = 70$ .

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.

Example 10.7

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.

Try It 10.13

Solve: ${(q + 5)}^{2} = 1$ .

Try It 10.14

Solve: ${(r - 3)}^{2} = 25$ .

Example 10.8

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.

Try It 10.15

Solve: ${(a - 3)}^{2} = 18$ .

Try It 10.16

Solve: ${(b + 2)}^{2} = 40$ .

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

Example 10.9

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

Solution

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.

Try It 10.17

Solve: ${(x - \frac{1}{3})}^{2} = \frac{5}{9} .$

Try It 10.18

Solve: ${(y - \frac{3}{4})}^{2} = \frac{7}{16} .$

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

Example 10.10

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

Solution

\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.

Try It 10.19

Solve: ${(a - 5)}^{2} + 4 = 24$ .

Try It 10.20

Solve: ${(b - 3)}^{2} - 8 = 24$ .

Example 10.11

Solve: ${(3 v - 7)}^{2} = −12$ .

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.

Try It 10.21

Solve: ${(3 r + 4)}^{2} = −8$ .

Try It 10.22

Solve: ${(2 t - 8)}^{2} = −10$ .

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.

Example 10.12

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

Solution

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

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.

Try It 10.23

Solve: $x^{2} - 6 x + 9 = 12$ .

Try It 10.24

Solve: $y^{2} + 12 y + 36 = 32$ .

Example 10.13

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.

Try It 10.25

Solve: $9 m^{2} - 12 m + 4 = 25$ .

Try It 10.26

Solve: $16 n^{2} + 40 n + 25 = 4$ .

Media

Access these online resources for additional instruction and practice with solving quadratic equations:

Section 10.1 Exercises

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.

1.

$a^{2} = 49$

2.

$b^{2} = 144$

3.

$r^{2} - 24 = 0$

4.

$t^{2} - 75 = 0$

5.

$u^{2} - 300 = 0$

6.

$v^{2} - 80 = 0$

7.

$4 m^{2} = 36$

8.

$3 n^{2} = 48$

9.

$x^{2} + 20 = 0$

10.

$y^{2} + 64 = 0$

11.

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

12.

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

13.

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

14.

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

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.

15.

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

16.

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

17.

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

18.

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

19.

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

20.

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

21.

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

22.

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

23.

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

24.

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

25.

${(5 c + 1)}^{2} = −27$

26.

${(8 d - 6)}^{2} = −24$

27.

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

28.

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

29.

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

30.

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

Mixed Practice

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

31.

$2 r^{2} = 32$

32.

$4 t^{2} = 16$

33.

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

34.

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

35.

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

36.

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

37.

$a^{2} - 18 = 0$

38.

$b^{2} - 108 = 0$

39.

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

40.

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

41.

$m^{2} + 12 = 0$

42.

$n^{2} + 48 = 0$

43.

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

44.

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

45.

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

46.

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

47.

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

48.

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

49.

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

50.

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

51.

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

52.

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

Everyday Math

53.

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.

54.

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

55.

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

56.

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.

ⓑ 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.

10.1 Solve Quadratic Equations Using the Square Root Property