# 11.1: The Square Root Property

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Let’s take a simple quadratic equation, $$x^2 = a$$ and solve:

$\begin{array}{rl}x^2=a &\text{Take the square root of both sides} \\ |x|=\pm\sqrt{a}&\text{Apply absolute value definition} \\ x=\pm\sqrt{a}&\text{Rewrite as two solutions} \\ x=\sqrt{a}\text{ or }x=-\sqrt{a}&\text{Solution}\end{array}\nonumber$

This is the square root property.

##### Square Root Property

$x^2=a\text{ if and only if }x=\pm\sqrt{a}\nonumber$

In other words,

$x^2=a\text{ if and only if }x=\sqrt{a}\text{ or }x=-\sqrt{a}\nonumber$

##### Example 11.1.1

Solve: $$x^2=81$$

Solution

We could rewrite the equation so that $$81$$ is on the left and then solve by factoring. However, for the sake of the property, we solve this equation by applying the square root property.

$\begin{array}{rl}x^2=81 &\text{The }x^2 \text{ is isolated and we apply the square root property} \\ x=\pm\sqrt{81} &\text{Simplify} \\ x=\pm 9&\text{Rewrite as two solutions} \\ x=9\text{ or }x=-9&\text{Solution}\end{array}\nonumber$

Notice, we could write the solution two ways: $$\pm 9$$, or, alternatively, $$9$$ or $$−9$$. As the problems become more challenging, it is common practice to write the solutions as two solutions.

##### Example 11.1.2

Solve: $$x^2=44$$

Solution

Notice, even if we moved $$44$$ to the left and tried to factor, we couldn’t because $$44$$ is not a perfect square. Hence, we need the square root property to solve.

$\begin{array}{rl}x^2=44 &\text{The }x^2\text{ is isolated and we apply the square root property} \\ x=\pm\sqrt{44}&\text{Simplify} \\ x=\pm\sqrt{4\cdot 11}&\text{Apply the product property} \\ x=\pm 2\sqrt{11}&\text{Rewrite as two solutions} \\ x=2\sqrt{11}\text{ or }x=-2\sqrt{11}&\text{Solution}\end{array}\nonumber$

##### Note

In 1545, French mathematician, Gerolamo Cardano, published his book The Great Art, or the Rules of Algebra. It included the solution of an equation with a fourth power, but it was considered absurd by many to take a quantity to the fourth power because there are only three dimensions.

##### Example 11.1.3

Solve: $$(x+4)^2=25$$

Solution

Even though the base has changed from $$x$$ to $$(x+ 4)$$, the method doesn’t change. Hence, we will apply the square root property to solve as long as the base is isolated.

$\begin{array}{rl}(x+4)^2=25&\text{The }(x+4)^2\text{ is isolated and we apply the square root property} \\ x+4=\pm\sqrt{25}&\text{Isolate }x \\ x=-4\pm\sqrt{25}&\text{Simplify }\sqrt{25} \\ x=-4\pm 5&\text{Rewrite as two solutions} \\ x=-4+5\text{ or }x=-4-5 &\text{Evaluate} \\ x=-1\text{ or }x=-9&\text{Solution}\end{array}\nonumber$

Here, we rewrote the solution as two different solutions in order to solve.

##### Example 11.1.4

Solve: $$(6x-9)^2=45$$

Solution

Even though the base has changed from $$x$$ to $$(6x − 9)$$, the method doesn’t change. Hence, we will apply the square root property to solve as long as the base is isolated.

$\begin{array}{rl}(6x-9)^2=45&\text{The }(6x-9)^2\text{ is isolated and we apply the square root property} \\ 6x-9=\pm\sqrt{45}&\text{Isolate the variable term} \\ 6x=9\pm\sqrt{45}&\text{Simplify }\sqrt{45} \\ 6x=9\pm\sqrt{9\cdot 5}&\text{Apply the product property} \\ 6x=9\pm 3\sqrt{5}&\text{Solve for }x \\ x=\dfrac{9\pm 3\sqrt{5}}{6}&\text{Factor a GCF} \\ x=\dfrac{\cancel{3}(3\pm\sqrt{5})}{\cancel{6}^2}&\text{Simplify} \\ x=\dfrac{3\pm\sqrt{5}}{2}&\text{Solution}\end{array}\nonumber$

Here, we leave the solution with the $$±$$ since the radicand was not a perfect square. Usually, when the radical is completely reduced out of the equation, we separate the solutions. Otherwise, we leave it as is.

## Isolate the Squared Term

Let’s take a look at examples where the leading term, or squared term, is not isolated. Recall, the squared term must be isolated in order to apply the square root property.

##### Example 11.1.5

Solve: $$5(3x-6)^2+7=27$$

Solution

We first need to isolate $$(3x − 6)^2$$ in order to apply the square root property. Then we can solve as usual.

$\begin{array}{rl} 5(3x-6)^2+7=27&\text{Isolate the variable term} \\ 5(3x-6)^2=20&\text{Isolate }(3x-6)^2 \\ (3x-6)^2=4&\text{Apply the square root property} \\ 3x-6=\pm\sqrt{4}&\text{Isolate the variable term} \\ 3x=6\pm\sqrt{4}&\text{Simplify }\sqrt{4} \\ 3x=6\pm 2&\text{Solve for }x \\ x=\dfrac{6\pm 2}{3}&\text{Rewrite as two solutions} \\ x=\dfrac{6+2}{3}\text{ or }x=\dfrac{6-2}{3}&\text{Evaluate} \\ x=\dfrac{8}{3}\text{ or }x=\dfrac{4}{3}&\text{Solution}\end{array}\nonumber$

Notice the radicand was a perfect square and so we were able to write the solutions as two separate numbers.

##### Example 11.1.6

Solve: $$5(r+4)^2+1=626$$

Solution

We first need to isolate $$(r + 4)^2$$ in order to apply the square root property. Then we can solve as usual.

$\begin{array}{rl} 5(r+4)^2+1=626&\text{Isolate the variable term} \\ 5(r+4)^2=625 &\text{Isolate }(r+4)^2 \\ (r+4)^2=125 &\text{Apply the square root property} \\ r+4=\pm\sqrt{125}&\text{Solve for }r \\ r=-4\pm\sqrt{125}&\text{Simplify }\sqrt{125} \\ r=-4\pm\sqrt{25\cdot 5}&\text{Apply the product property} \\ r=-4\pm 5\sqrt{5}&\text{Solution}\end{array}\nonumber$

##### Example 11.1.7

Solve: $$2n^2+5=4$$

Solution

We first need to isolate $$n^2$$ in order to apply the square root property. Then we can solve as usual.

$\begin{array}{rl} 2n^2+5=4&\text{Isolate the variable term} \\ 2n^2=-4&\text{Isolate }n^2 \\ n^2=-2&\text{Apply the square root property} \\ n=\pm\sqrt{-2}&\text{Reduce out an }i \\ n=\pm i\sqrt{2}&\text{Solution}\end{array}\nonumber$

Recall, a radicand of a square root that is less than zero is the imaginary part of a complex number. Now that we just discussed complex numbers in the previous chapter, we can solve any type of quadratic equation with real and complex solutions.

## Square Root Property Homework

Solve by applying the square root property.

##### Exercise 11.1.1

$$(x-3)^2=16$$

##### Exercise 11.1.2

$$(x-2)^2=49$$

##### Exercise 11.1.3

$$(x-7)^2=4$$

##### Exercise 11.1.4

$$(s-5)^2=16$$

##### Exercise 11.1.5

$$(p+5)^2=81$$

##### Exercise 11.1.6

$$(s+3)^2=4$$

##### Exercise 11.1.7

$$(t+9)^2=37$$

##### Exercise 11.1.8

$$(a+5)^2=87$$

##### Exercise 11.1.9

$$(v-2)^2=70$$

##### Exercise 11.1.10

$$(n-9)^2=63$$

##### Exercise 11.1.11

$$(v+4)^2=63$$

##### Exercise 11.1.12

$$(r+1)^2=125$$

##### Exercise 11.1.13

$$(9r+1)^2=9$$

##### Exercise 11.1.14

$$(7m-8)^2=36$$

##### Exercise 11.1.15

$$(3s-6)^2=25$$

##### Exercise 11.1.16

$$5(k-7)^2-6=369$$

##### Exercise 11.1.17

$$5(z+6)^2-10=365$$

##### Exercise 11.1.18

$$5(g-5)^2+13=103$$

##### Exercise 11.1.19

$$(2s+1)^2=0$$

##### Exercise 11.1.20

$$(z-4)^2=25$$

##### Exercise 11.1.21

$$(w+3)^2=49$$

##### Exercise 11.1.22

$$2n^2+7=5$$

##### Exercise 11.1.23

$$3n^2+2n=2n+24$$

##### Exercise 11.1.24

$$8n^2-29=25+2n^2$$

##### Exercise 11.1.25

$$2(r+9)^2-19=37$$

##### Exercise 11.1.26

$$3(n-3)^2+2=164$$

##### Exercise 11.1.27

$$3(y+8)^2+12=147$$

##### Exercise 11.1.28

$$6(4x-4)^2-5=145$$

##### Exercise 11.1.29

$$3(4x+6)^2-5=103$$

##### Exercise 11.1.30

$$7(2x+6)^2-5=170$$

##### Exercise 11.1.31

$$4(7x+6)^2-3=61$$

##### Exercise 11.1.32

$$3(7x+3)^2+7=55$$

##### Exercise 11.1.33

$$5(4x-5)^2-2=18$$

This page titled 11.1: The Square Root Property is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Darlene Diaz (ASCCC Open Educational Resources Initiative) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.