Solve Quadratic Equations Using the Square Root Property

A quadratic equation is an equation of the form ax² + bx + c = 0, where

a \neq 0

. Quadratic equations differ from linear equations by including a quadratic term with the variable raised to the second power of the form ax². 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 learn three other methods to use in case a quadratic equation cannot be factored.

Solve Quadratic Equations of the form

a x^{2} = k

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation x² = 9.

We can easily use factoring to find the solutions of similar equations, like x² = 16 and x² = 25, because 16 and 25 are perfect squares. In each case, we would get two solutions,

x = 4, x = −4

But what happens when we have an equation like x² = 7? Since 7 is not a perfect square, we cannot solve the equation by factoring.

Previously we learned that since 169 is the square of 13, we can also say that 13 is a square root of 169. Also, (−13)² = 169, so −13 is also a square root of 169. Therefore, both 13 and −13 are square roots of 169. So, every positive number has two square roots—one positive and one negative. We earlier defined the square root of a number in this way:

Since these equations are all of the form x² = k, the square root definition tells us the solutions are the two square roots of k. This leads to the Square Root Property.

Notice that the Square Root Property gives two solutions to an equation of the form x² = k, the principal square root of

k

Now we will solve the equation x² = 9 again, this time using the Square Root Property.

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation x² = 7.

How to solve a Quadratic Equation of the form *ax*² = *k* Using the Square Root Property

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

![Step one is to isolate the quadratic term and make its coefficient one. For the equation x squared minus fifty equals zero, first add fifty to both sides to get x squared by itself. The new equation is x squared equals fifty.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_09_01_001a_img.jpg) ![Step two is to use Square Root Property. Remember to write the plus or minus symbol. The equation created is x equals the positive or negative square root of 50.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_09_01_001b_img.jpg) ![Step three is to simplify the radical if possible. Continue to write equivalent equations. X equals the positive or negative square root of twenty five times the square root of five. X equals positive or negative five times the square root of five. Rewrite to show two solutions: x equals five times the square root of five or x equals negative five times the square root of five.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_09_01_001c_img.jpg) ![Step four is to check the solutions. Substitute x equals five times the square root of five and or x equals negative five times the square root of five into the original equation. Start with x equals five times the square root of five. x squared minus fifty equals zero. Does the square of five times the square root of five minus fifty equal zero? Does twenty five times two minus fifty equal zero? Yes, because zero equals zero. Next check x equals negative five times the square root of five. x squared minus fifty equals zero. Does the square of negative five times the square root of five minus fifty equal zero? Does twenty five times two minus fifty equal zero? Yes, because zero equals zero.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_09_01_001d_img.jpg)

The steps to take to use the Square Root Property to solve a quadratic equation are listed here.

In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient 3 before using the Square Root Property.

Our method also works when fractions occur in the equation, we solve as any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.

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

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 of the form a(x − h)² = k as well. Notice that the quadratic term, x, in the original form ax² = k is replaced with (x − h).

On the left is the equation a times x square equals k. Replacing x in this equation with the expression x minus h changes the equation. It is now a times the square of x minus h equals k.

The first step, like before, is to isolate the term that has the variable squared. In this case, a binomial is being squared. Once the binomial is isolated, by dividing each side by the coefficient of a, then the Square Root Property can be used on (x − h)².

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

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.

Key Concepts

Practice Makes Perfect

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

In the following exercises, solve each equation.

a^{2} = 49

a = \pm 7

b^{2} = 144

r^{2} - 24 = 0

r = \pm 2 \sqrt{6}

t^{2} - 75 = 0

u^{2} - 300 = 0

u = \pm 10 \sqrt{3}

v^{2} - 80 = 0

4 m^{2} = 36

m = \pm 3

3 n^{2} = 48

\frac{4}{3} x^{2} = 48

x = \pm 6

\frac{5}{3} y^{2} = 60

x^{2} + 25 = 0

x = \pm 5 i

y^{2} + 64 = 0

x^{2} + 63 = 0

x = \pm 3 \sqrt{7} i

y^{2} + 45 = 0

\frac{4}{3} x^{2} + 2 = 110

x = \pm 9

\frac{2}{3} y^{2} - 8 = −2

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

a = \pm 2 \sqrt{5}

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

7 p^{2} + 10 = 26

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

2 q^{2} + 5 = 30

5 y^{2} - 7 = 25

y = \pm \frac{4 \sqrt{10}}{5}

3 x^{2} - 8 = 46

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

In the following exercises, solve each equation.

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

u = 14, u = −2

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

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

m = 6 \pm 2 \sqrt{5}

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

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

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

{(x + \frac{1}{5})}^{2} = \frac{7}{25}

{(y + \frac{2}{3})}^{2} = \frac{8}{81}

y = - \frac{2}{3} \pm \frac{2 \sqrt{2}}{9}

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

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

a = 7 \pm 5 \sqrt{2}

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

4 {(x + 3)}^{2} - 5 = 27

x = −3 \pm 2 \sqrt{2}

5 {(x + 3)}^{2} - 7 = 68

{(5 c + 1)}^{2} = −27

c = - \frac{1}{5} \pm \frac{3 \sqrt{3}}{5} i

{(8 d - 6)}^{2} = −24

{(4 x - 3)}^{2} + 11 = −17

x = \frac{3}{4} \pm \frac{\sqrt{7}}{2} i

{(2 y + 1)}^{2} - 5 = −23

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

m = 2 \pm 2 \sqrt{2}

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

x^{2} - 6 x + 9 = 12

x = 3 + 2 \sqrt{3}, x = 3 - 2 \sqrt{3}

y^{2} + 12 y + 36 = 32

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

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

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

36 x^{2} - 24 x + 4 = 81

x = - \frac{7}{6}, x = \frac{11}{6}

64 x^{2} + 144 x + 81 = 25

Mixed Practice

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

2 r^{2} = 32

r = \pm 4

4 t^{2} = 16

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

a = 4 \pm 2 \sqrt{7}

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

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

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

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

a^{2} - 18 = 0

a = \pm 3 \sqrt{2}

b^{2} - 108 = 0

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

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

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

m^{2} + 12 = 0

no real solution

n^{2} + 48 = 0 .

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

u = 7 \pm 6 \sqrt{2}

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

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

m = 4 \pm 2 \sqrt{3}

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

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

x = −3, x = −7

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

6 c^{2} + 4 = 29

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

2 d^{2} - 4 = 77

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

no real solution

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

Writing Exercises

In your own words, explain the Square Root Property.

Answers will vary.

In your own words, explain how to use the Square Root Property to solve the quadratic equation ${(x + 2)}^{2} = 16$

Self Check

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

Choose how would you respond to the statement “I can solve quadratic equations of the form a times the square of x minus h equals k using the Square Root Property.” “Confidently,” “with some help,” or “No, I don’t get it.”

ⓑ 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. Who 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.

	$3 z^{2} = 108$
The quadratic term is isolated. Divide by 3 to make its coefficient 1.	$\frac{3 z^{2}}{3} = \frac{108}{3}$
Simplify.	$z^{2} = 36$
Use the Square Root Property.	$z = \pm \sqrt{36}$
Simplify the radical.	$z = \pm 6$
Rewrite to show two solutions.	$z = 6, z = −6$
Check the solutions:

	$x^{2} + 72 = 0$
Isolate the quadratic term.	$x^{2} = −72$
Use the Square Root Property.	$x = \pm \sqrt{−72}$
Simplify using complex numbers.	$x = \pm \sqrt{72} i$
Simplify the radical.	$x = \pm 6 \sqrt{2} i$
Rewrite to show two solutions.	$x = 6 \sqrt{2} i, x = −6 \sqrt{2} i$
Check the solutions:

	$\frac{2}{3} u^{2} + 5 = 17$
Isolate the quadratic term.
Multiply by $\frac{3}{2}$ to make the coefficient 1.
Simplify.
Use the Square Root Property.
Simplify the radical.
Simplify.
Rewrite to show two solutions.
Check:


Isolate the quadratic term.
Divide by $2$ to make the coefficient 1.
Simplify.
Use the Square Root Property.
Rewrite the radical as a fraction of square roots.
Rationalize the denominator.
Simplify.
Rewrite to show two solutions.
Check: We leave the check for you.

	$4 {(y - 7)}^{2} = 48$
Divide both sides by the coefficient 4.	${(y - 7)}^{2} = 12$
Use the Square Root Property on the binomial	$y - 7 = \pm \sqrt{12}$
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:

	$4 n^{2} + 4 n + 1 = 16$
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.	$\frac{2 n}{2} = \frac{−1 \pm 4}{2}$ $n = \frac{−1 \pm 4}{2}$
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: