ORCCA Solving Radical Equations

Section 6.4 Solving Radical Equations

Objectives: PCC Course Content and Outcome Guide

permalinkIn this section, we will learn how to solve equations involving radicals. The basic strategy to solve radical equations is to isolate the radical on one side of the equation and then raise to a power on both sides to cancel the radical.

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Figure 6.4.1. Alternative Video Lessons

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Subsection 6.4.1 Solving Radical Equations

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Definition 6.4.2. Radical Equation.

A radical equation is an equation in which there is a variable inside at least one radical.

Examples include the equations $\sqrt{x-2}=3+x$ and $1+\sqrt[3]{2-x}=x\text{.}$

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Example 6.4.3.

The formula $T=2\pi\sqrt{\frac{L}{g}}$ is used to calculate the period of a pendulum and is attributed to the scientist Christiaan Huygens ¹. In the formula, $T$ stands for the pendulum's period (how long one back-and-forth oscillation takes) in seconds, $L$ stands for the pendulum's length in meters, and $g$ is approximately 9.8 ^m⁄_s² which is the gravitational acceleration constant on Earth.

An engineer is designing a pendulum. Its period must be $10$ seconds. How long should the pendulum's length be?

We will substitute $\substitute{10}$ into the formula for $T$ and also the value of $g\text{,}$ and then solve for $L\text{:}$

\begin{matrix} 10 & = 2 π \sqrt{\frac{L}{9.8}} \frac{1}{2 π} \cdot 10 & = \frac{1}{2 π} \cdot 2 π \sqrt{\frac{L}{9.8}} \frac{5}{π} & = \sqrt{\frac{L}{9.8}} {(\frac{5}{π})}^{2} & = {(\sqrt{\frac{L}{9.8}})}^{2} & canceling square root by squaring both sides \frac{25}{π^{2}} & = \frac{L}{9.8} 9.8 \cdot \frac{25}{π^{2}} & = 9.8 \cdot \frac{L}{9.8} 24.82 & \approx L \end{matrix}

$\begin{align*} \substitute{10}\amp=2\pi\sqrt{\frac{L}{\substitute{9.8}}}\\ \multiplyleft{\frac{1}{2\pi}}10\amp=\multiplyleft{\frac{1}{2\pi}}2\pi\sqrt{\frac{L}{9.8}}\\ \frac{5}{\pi}\amp=\sqrt{\frac{L}{9.8}}\\ \left(\frac{5}{\pi}\right)^{\highlight{2}}\amp=\left(\sqrt{\frac{L}{9.8}}\right)^{\highlight{2}}\amp\text{canceling square root by squaring both sides}\\ \frac{25}{\pi^2}\amp=\frac{L}{9.8}\\ \multiplyleft{9.8}\frac{25}{\pi^2}\amp=\multiplyleft{9.8}\frac{L}{9.8}\\ 24.82\amp\approx L \end{align*}$

To build a pendulum with a period of $10$ seconds, its length should be approximately $24.82$ meters.

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Remark 6.4.4.

Squaring both sides of an equation is “dangerous,” as it could create extraneous solutions, which will not make the equation true. For example, if we square both sides of $1=-1\text{,}$ we have:

$\begin{align*} 1\amp=-1\amp\amp\text{false}\\ (1)^2\amp=(-1)^2\amp\amp\text{square both sides}\ldots\\ 1\amp=1\amp\amp\text{true} \end{align*}$

By squaring both sides of an equation, we can sometimes turn a false equation into a true one. This is why we must check solutions when we square both sides of an equation.

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Example 6.4.5.

Solve the equation $1+\sqrt{y-1}=4$ for $y\text{.}$

Explanation

We will isolate the radical first, and then square both sides.

\begin{align*} 1+\sqrt{y-1}\amp=4\\ \sqrt{y-1}\amp=3\\ \left(\sqrt{y-1}\right)^{\highlight{2}}\amp=3^{\highlight{2}}\\ y-1\amp=9\\ y\amp=10 \end{align*}

Because we squared both sides of an equation, we must check the solution.

\begin{align*} 1+\sqrt{\substitute{10}-1}\amp\stackrel{?}{=}4\\ 1+\sqrt{9}\amp\stackrel{?}{=}4\\ 1+3\amp\stackrel{\checkmark}{=}4 \end{align*}

So, $10$ is the solution to the equation $1+\sqrt{y-1}=4\text{.}$

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Example 6.4.6.

Solve the equation $5+\sqrt{q}=3$ for $q\text{.}$

Explanation

First, isolate the radical and square both sides.

\begin{align*} 5+\sqrt{q}\amp=3\\ \sqrt{q}\amp=-2\\ \left(\sqrt{q}\right)^{\highlight{2}}\amp=(-2)^{\highlight{2}}\\ q\amp=4 \end{align*}

Because we squared both sides of an equation, we must check the solution.

\begin{align*} 5+\sqrt{\substitute{4}}\amp\stackrel{?}{=}3\\ 5+2\amp\stackrel{?}{=}3\\ 7\amp\stackrel{\text{no}}{=} 3 \end{align*}

Thus, the potential solution $-2$ is actually extraneous and we have no real solutions to the equation $5+\sqrt{q}=3\text{.}$ The solution set is the empty set, $\emptyset\text{.}$

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Remark 6.4.7.

In the previous example, it would be legitimate to observe that there are no solutions at earlier stages. From the very beginning, how could $5$ plus a positive quantity result in $3\text{?}$ Or at the second step, since square roots are non-negative, how could a square root equal $-2\text{?}$

You do not have to be able to make these observations. If you follow the general steps for solving radical equations and you remember to check the possible solutions you find, then that will be enough.

permalinkSometimes, we need to square both sides of an equation twice before finding the solutions, like in the next example.

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Example 6.4.8.

Solve the equation $\sqrt{p-5}=5-\sqrt{p}$ for $p\text{.}$

Explanation

We cannot isolate two radicals, so we will simply square both sides, and later try to isolate the remaining radical.

\begin{align*} \sqrt{p-5}\amp=5-\sqrt{p}\\ \left(\sqrt{p-5}\right)^{\highlight{2}}\amp=\left(5-\sqrt{p}\right)^{\highlight{2}}\\ p-5\amp=25-10\sqrt{p}+p \amp\text{ after expanding the binomial squared}\\ -5\amp=25-10\sqrt{p}\\ -30\amp=-10\sqrt{p}\\ 3\amp=\sqrt{p}\\ 3^{\highlight{2}}\amp=\left(\sqrt{p}\right)^{\highlight{2}}\\ 9\amp=p \end{align*}

Because we squared both sides of an equation, we must check the solution.

\begin{align*} \sqrt{\substitute{9}-5}\amp\stackrel{?}{=}5-\sqrt{9}\\ \sqrt{4}\amp\stackrel{?}{=}5-3\\ 2\amp\stackrel{\checkmark}{=}2 \end{align*}

So $9$ is the solution. The solution set is $\{9\}\text{.}$

permalinkLet's look at an example of solving an equation with a cube root. There is very little difference between solving an equation with one cube root and solving an equation with one square root. Instead of squaring both sides, you cube both sides.

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Example 6.4.9.

Solve for $q$ in $\sqrt[3]{2-q}+2=5\text{.}$

Explanation

\begin{align*} \sqrt[3]{2-q}+2\amp=5\\ \sqrt[3]{2-q}\amp=3\\ \left(\sqrt[3]{2-q}\right)^{\highlight{3}}\amp=3^{\highlight{3}}\\ 2-q\amp=27\\ -q\amp=25\\ q\amp=-25 \end{align*}

Unlike squaring both sides of an equation, raising both sides of an equation to the 3rd power will not create extraneous solutions. It's still good practice to check solution, though. This part is left as exercise.

permalinkFor summary reference, here is the general procedure for solving a radical equation.

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Process 6.4.10. Solving Radical Equations.

A basic strategy to solve radical equations is to take the following steps:

Isolate a radical on one side of the equation.
Raise both sides of the equation to a power to cancel the radical.
If there is still a radical in the equation, repeat the isolation and raising to a power.
Once the remaining equation has no radicals, solve it.
Check any and all solutions. Be aware that there may be “extraneous solutions”.

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Subsection 6.4.2 Solving a Radical Equation with More Than One Variable

permalinkWe also need to be able to solve radical equations with other variables, like in the next example. The strategy is the same: isolate the radical, and then raise both sides to a certain power to cancel the radical.

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Example 6.4.11.

Solve for $L$ in the formula $T=2\pi\sqrt{\frac{L}{g}}\text{.}$ (This is the formula for a the period $T$ of a swinging pendulum whose length is $L\text{,}$ on earth where the acceleration from earth's gravity is $g\text{.}$ )

Explanation

\begin{align*} T\amp=2\pi\sqrt{\frac{L}{g}}\\ \multiplyleft{\frac{1}{2\pi}}T\amp=\multiplyleft{\frac{1}{2\pi}}2\pi\sqrt{\frac{L}{g}}\\ \frac{T}{2\pi}\amp=\sqrt{\frac{L}{g}}\\ \left(\frac{T}{2\pi}\right)^2\amp=\left(\sqrt{\frac{L}{g}}\right)^2\\ \frac{T^2}{4\pi^2}\amp=\frac{L}{g}\\ \multiplyleft{g}\frac{T^2}{4\pi^2}\amp=\multiplyleft{g}\frac{L}{g}\\ \frac{T^2g}{4\pi^2}\amp=L \end{align*}

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Example 6.4.12.

The study of black holes has resulted in some interesting mathematics. One fundamental concept about black holes is that there is a distance close enough to the black hole that not even light can escape, called the Schwarzschild radius ² or the event horizon radius. To find the Schwarzschild radius, $R_s\text{,}$ we set the formula for the escape velocity equal to the speed of light, $c\text{,}$ and we get $c=\sqrt{\frac{2GM}{R_s}}$ which we need to solve for $R_s\text{.}$ Note that $G$ is a constant, and $M$ is the mass of the black hole.

Explanation

We will start by taking the equation $c=\sqrt{\frac{2GM}{R_s}}$ and applying our standard radical-equation-solving techniques. Isolate the radical and square both sides:

\begin{align*} c\amp=\sqrt{\frac{2GM}{R_s}}\\ c^{\highlight{2}}\amp=\left(\sqrt{\frac{2GM}{R_s}}\right)^{\highlight{2}}\\ c^2\amp=\frac{2GM}{R_s}\\ \multiplyleft{R_s}c^2\amp=\multiplyleft{R_s}\frac{2GM}{R_s}\\ R_s c^2\amp=2GM\\ \divideunder{R_s c^2}{c^2}\amp=\divideunder{2GM}{c^2}\\ R_s\amp=\frac{2GM}{c^2} \end{align*}

So, the Schwarzschild radius can be found using the formula $R_s=\frac{2GM}{c^2}\text{.}$

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Reading Questions 6.4.3 Reading Questions

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1.

What is the basic approach to solving a radical equation?

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2.

What is it called when doing algebra leads you to find a number that could be a solution to an equation, but is not actually a solution?

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Exercises 6.4.4 Exercises

Review and Warmup

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1.

Solve the equation.

$\displaystyle{ {-9n+8} = {-n-8} }$

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2.

Solve the equation.

$\displaystyle{ {-8p+4} = {-p-24} }$

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3.

Solve the equation.

$\displaystyle{ {18}={-3\!\left(8-2x\right)} }$

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4.

Solve the equation.

$\displaystyle{ {66}={-2\!\left(2-5y\right)} }$

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5.

Solve the equation.

$\displaystyle{ {15}={8-7\!\left(t-8\right)} }$

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6.

Solve the equation.

$\displaystyle{ {144}={4-10\!\left(a-8\right)} }$

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7.

Solve the equation.

$\left(x - 1\right)^2 = 4$

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8.

Solve the equation.

$\left(x+2\right)^2 = 81$

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9.

Solve the equation.

${x^{2}+x-20}= 0$

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10.

Solve the equation.

${x^{2}+19x+84}= 0$

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11.

Solve the equation.

${x^{2}+13x+12}= -18$

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12.

Solve the equation.

${x^{2}-17x+59}= -1$

Solving Radical Equations

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13.

Solve the equation.

$\displaystyle{ {\sqrt{x}} = {12} }$

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14.

Solve the equation.

$\displaystyle{ {\sqrt{x}} = {8} }$

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15.

Solve the equation.

$\displaystyle{ {\sqrt{2y}} = {8} }$

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16.

Solve the equation.

$\displaystyle{ {\sqrt{5y}} = {10} }$

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17.

Solve the equation.

$\displaystyle{ {4\sqrt{r}} = {16} }$

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18.

Solve the equation.

$\displaystyle{ {2\sqrt{r}} = {10} }$

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19.

Solve the equation.

$\displaystyle{ {-5\sqrt{t}} = {15} }$

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20.

Solve the equation.

$\displaystyle{ {-4\sqrt{t}} = {20} }$

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21.

Solve the equation.

$\displaystyle{ {-5\sqrt{-5-x}+2} = {-8} }$

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22.

Solve the equation.

$\displaystyle{ {3\sqrt{3-x}+2} = {29} }$

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23.

Solve the equation.

$\displaystyle{ {\sqrt{x-12}} = {\sqrt{x}-2} }$

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24.

Solve the equation.

$\displaystyle{ {\sqrt{y+3}} = {\sqrt{y}+1} }$

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25.

Solve the equation.

$\displaystyle{ {\sqrt{y+9}} = {-1-\sqrt{y}} }$

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26.

Solve the equation.

$\displaystyle{ {\sqrt{r+9}} = {-1-\sqrt{r}} }$

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27.

Solve the equation.

$\displaystyle{ {\sqrt{2r}} = {8} }$

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28.

Solve the equation.

$\displaystyle{ {\sqrt{8t}} = {3} }$

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29.

Solve the equation.

$\displaystyle{ \sqrt[3]{t-5} = {7} }$

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30.

Solve the equation.

$\displaystyle{ \sqrt[3]{x-2} = {10} }$

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31.

Solve the equation.

$\displaystyle{ {\sqrt{8x+5}+4} = {10} }$

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32.

Solve the equation.

$\displaystyle{ {\sqrt{4x+9}+2} = {8} }$

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33.

Solve the equation.

$\displaystyle{ \sqrt[3]{y-12} = {-5} }$

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34.

Solve the equation.

$\displaystyle{ \sqrt[3]{y-8} = {3} }$

Solving Radical Equations with Variables

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35.

Solve the equation for $R\text{.}$ Assume that $R$ is positive.

$\begin{equation*} {Z} = {\sqrt{L^{2}+R^{2}}} \end{equation*}$

$R =$ .

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36.

According to the Pythagorean Theorem, the length $c$ of the hypothenuse of a rectangular triangle can be found through the following equation:

$\begin{equation*} {c} = {\sqrt{a^{2}+b^{2}}} \end{equation*}$

Solve the equation for the length $a$ of one of the triangle’s legs.

$a =$ .

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37.

In an electric circuit, resonance occurs when the frequency $f\text{,}$ inductance $L\text{,}$ and capacitance $C$ fulfill the following equation:

$\begin{equation*} {f} = {\frac{1}{2\pi \sqrt{LC}}} \end{equation*}$

Solve the equation for the inductance $L\text{.}$

The frequency is measured in Hertz, the inductance in Henry, and the capacitance in Farad.

$L =$ .

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38.

A pendulum has the length $L\text{.}$ The time period $T$ that it takes to once swing back and forth can be found with the following formula:

$\begin{equation*} {T} = {2\pi \sqrt{\frac{L}{32}}} \end{equation*}$

Solve the equation for the length $L\text{.}$

The length is measured in feet and the time period in seconds.

$L =$ .

Radical Equation Applications

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39.

According to the Pythagorean Theorem, the length $c$ of the hypothenuse of a rectangular triangle can be found through the following equation.

$\begin{equation*} {c} = {\sqrt{a^{2}+b^{2}}} \end{equation*}$

If a rectangular triangle has a hypothenuse of ${5\ {\rm ft}}$ and one leg is ${4\ {\rm ft}}$ long, how long is the third side of the triangle?

The third side of the triangle is long.

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40.

According to the Pythagorean Theorem, the length $c$ of the hypothenuse of a rectangular triangle can be found through the following equation.

$\begin{equation*} {c} = {\sqrt{a^{2}+b^{2}}} \end{equation*}$

If a rectangular triangle has a hypothenuse of ${5\ {\rm ft}}$ and one leg is ${4\ {\rm ft}}$ long, how long is the third side of the triangle?

The third side of the triangle is long.

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41.

In a coordinate system, the distance $r$ of a point $(x,y)$ from the origin $(0,0)$ is given by the following equation.

$\begin{equation*} {r} = {\sqrt{x^{2}+y^{2}}} \end{equation*}$

If a point in a coordinate system is ${13\ {\rm cm}}$ away from the origin and its x coordinate is ${12\ {\rm cm}}\text{,}$ what is its $y$ coordinate? Assume that $y$ is positive.

$y =$ .

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42.

In a coordinate system, the distance $r$ of a point $(x,y)$ from the origin $(0,0)$ is given by the following equation.

$\begin{equation*} {r} = {\sqrt{x^{2}+y^{2}}} \end{equation*}$

If a point in a coordinate system is ${13\ {\rm cm}}$ away from the origin and its x coordinate is ${12\ {\rm cm}}\text{,}$ what is its $y$ coordinate? Assume that $y$ is positive.

$y =$ .

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43.

A pendulum has the length $L$ ft. The time period $T$ that it takes to once swing back and forth is $6$ s. Use the following formula to find its length.

$\begin{equation*} {T} = {2\pi \sqrt{\frac{L}{32}}} \end{equation*}$

The pendulum is long.

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44.

A pendulum has the length $L$ ft. The time period $T$ that it takes to once swing back and forth is $6$ s. Use the following formula to find its length.

$\begin{equation*} {T} = {2\pi \sqrt{\frac{L}{32}}} \end{equation*}$

The pendulum is long.

Challenge

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45.

Solve for $x\text{.}$

$\begin{equation*} \sqrt{1+\sqrt{7}}=\sqrt{2+\sqrt{\frac{1}{\sqrt{x}}}-1} \end{equation*}$

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46.

Solve for $x\text{.}$

$\begin{equation*} \sqrt{1+\sqrt{8}}=\sqrt{2+\sqrt{\frac{1}{\sqrt{x}}}-1} \end{equation*}$