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11.1: The Square Root Property

  • Page ID
    45110
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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.