ORCCA Solving Radical Equations

Section 13.5 Solving Radical Equations

In this section, we will learn how to solve equations involving radicals.

Subsection 13.5.1 Solving Radical Equations

One common application of radicals is the Pythagorean Theorem. We already saw some examples in earlier sections. We will look at some other applications of radicals in this section.

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.

en.wikipedia.org/wiki/Christiaan_Huygens#Pendulums

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{align*} T\amp=2\pi\sqrt{\frac{L}{g}}\\ \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, the pendulum's length should be approximately \(24.82\) meters.

Remark 13.5.1.

The basic strategy to solve radical equations, where the radical is a square root, is to isolate the radical on one side of the equation and then square both sides to cancel the radical.

Remark 13.5.2.

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\text{false}\\ (1)^2\amp=(-1)^2\\ 1\amp=1\amp\text{true} \end{align*}

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

Example 13.5.3.

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. Substitute \(\substitute{10}\) into \(1+\sqrt{y-1}=4\text{,}\) and we have:

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

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

Example 13.5.4.

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. Substitute \(\substitute{4}\) into \(5+\sqrt{q}=3\text{,}\) and we have:

\begin{align*} 5+\sqrt{q}\amp=3\\ 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, \(\emptyset\text{.}\)

Remark 13.5.5.

In the previous example, it would be legitimate to observe earlier stages that there are no solutions. 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.

Example 13.5.6.

Solve for \(z\) in \(\sqrt{z}+2=z\text{.}\)

Explanation

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

\begin{align*} \sqrt{z}+2\amp=z\\ \sqrt{z}\amp=z-2\\ \left(\sqrt{z}\right)^{\highlight{2}}\amp=(z-2)^{\highlight{2}}\\ z\amp=z^2-4z+4\\ 0\amp=z^2-5z+4\\ 0\amp=(z-1)(z-4)\\ z-1=0\amp\text{ or }z-4=0\\ z=1\amp\text{ or }z=4 \end{align*}

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

Substitute \(\substitute{1}\) into \(\sqrt{z}+2=z\text{,}\) and we have:

\begin{align*} \sqrt{z}+2\amp=z\\ \sqrt{\substitute{1}}+2\amp\stackrel{?}{=}1\\ 1+2\amp\stackrel{?}{=}1\\ 3\amp\stackrel{\text{no}}{=}1 \end{align*}

It turned out \(1\) is an extraneous solution.

Next, we substitute \(\substitute{4}\) into \(\sqrt{z}+2=z\text{:}\)

\begin{align*} \sqrt{z}+2\amp=z\\ \sqrt{\substitute{4}}+2\amp\stackrel{?}{=}4\\ 2+2\amp\stackrel{?}{=}4\\ 4\amp\stackrel{\checkmark}{=}4 \end{align*}

So, \(4\) is the solution.

The equation has one solution: \(4\text{.}\) The solution set is \(\{4\}\text{.}\)

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

Example 13.5.7.

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}\\ \frac{-30}{-10}\amp=\frac{-10\sqrt{p}}{-10}\\ 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 by substituting \(\substitute{9}\) into \(\sqrt{p-5}=5-\sqrt{p}\text{,}\) and we have:

\begin{align*} \sqrt{p-5}\amp=5-\sqrt{p}\\ \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 \(\{9\}\text{.}\)

Example 13.5.8.

Solve the equation \(\sqrt{2n-6}=1+\sqrt{n-2}\) for \(n\text{.}\)

Explanation

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

\begin{align*} \sqrt{2n-6}\amp=1+\sqrt{n-2}\\ \left(\sqrt{2n-6}\right)^\highlight{2}\amp=\left(1+\sqrt{n-2}\right)^\highlight{2}\\ 2n-6\amp=\left(1+\sqrt{n-2}\right)\left(1+\sqrt{n-2}\right)\\ 2n-6\amp=1+\sqrt{n-2}+\sqrt{n-2}+n-2\\ 2n-6\amp=2\sqrt{n-2}+n-1 \text{ by combining like terms}\\ n-5\amp=2\sqrt{n-2}\\ \end{align*}

Note here that we can leave the coefficient \(2\) next to the radical in the squaring process. We will square the \(2\) also.

\begin{align*} (n-5)^\highlight{2}\amp=\left(2\sqrt{n-2}\right)^\highlight{2}\\ n^2-10n+25\amp=4(n-2)\\ n^2-10n+25\amp=4n-8\\ n^2-14n+33\amp=0\\ (n-11)(n-3)\amp=0\\ n-11=0\amp\text{ or }n-3=0\\ n=11\amp\text{ or } n=3 \end{align*}

So our two potential solutions are \(11\) and \(3\text{.}\) We should now verify that they truly are solutions.

Substitute \(\substitute{11}\) into \(\sqrt{2n-6}=1+\sqrt{n-2}\text{,}\) and we have:

\begin{align*} \sqrt{2n-6}\amp=1+\sqrt{n-2}\\ \sqrt{2(\substitute{11})-6}\amp\stackrel{?}{=}1+\sqrt{\substitute{11}-2}\\ \sqrt{16}\amp\stackrel{?}{=}1+\sqrt{9}\\ 4\amp\stackrel{?}{=}1+3\\ 4\amp\stackrel{\checkmark}{=}4 \end{align*}

So, \(11\) is the solution.

Next, we substitute \(\substitute{3}\) into \(\sqrt{2n-6}=1+\sqrt{n-2} \text{,}\) and we have:

\begin{align*} \sqrt{2n-6}\amp=1+\sqrt{n-2}\\ \sqrt{2(\substitute{3})-6}\amp\stackrel{?}{=}1+\sqrt{\substitute{3}-2}\\ \sqrt{0}\amp\stackrel{?}{=}1+\sqrt{1}\\ 0\amp\stackrel{\text{no}}{=}2 \end{align*}

So, \(3\) is not a solution.

So \(11\) is the solution. The solution set is \(\{11\}\text{.}\)

We also need the ability to solve radical equations with multiple variables, like in the next example that comes from physics. The strategy is the same: isolating the radical, and then raise both sides to a certain power to cancel the radical.

Example 13.5.9.

The study of black holes has resulted in some interesting science. 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 \(c=\sqrt{\frac{2GM}{R_s}}\text{,}\) which we need to solve for \(R_s\text{.}\) Note that \(G\) is a constant, and \(M\) is the mass of the black hole.

en.wikipedia.org/wiki/Schwarzschild_radius

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=\frac{2GM}{R_s}\multiplyright{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 \(R_s=\frac{2GM}{c^2}\text{.}\)

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

Example 13.5.10.

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.

Exercises 13.5.2 Exercises

Review and Warmup

1.

Solve the equation.

\(\displaystyle{ {-7n+7} = {-n-5} }\)

2.

Solve the equation.

\(\displaystyle{ {-4p+2} = {-p-25} }\)

3.

Solve the equation.

\(\displaystyle{ {-225}={-5\!\left(5-4x\right)} }\)

4.

Solve the equation.

\(\displaystyle{ {-156}={-4\!\left(9-3y\right)} }\)

5.

Solve the equation.

\(\displaystyle{ {82}={2-4\!\left(t-10\right)} }\)

6.

Solve the equation.

\(\displaystyle{ {49}={9-8\!\left(a-10\right)} }\)

7.

Solve the equation.

\(\left(x - 1\right)^2 = 49\)

8.

Solve the equation.

\(\left(x+1\right)^2 = 9\)

9.

Solve the equation.

\({x^{2}+11x+28}= 0\)

10.

Solve the equation.

\({x^{2}+6x-7}= 0\)

11.

Solve the equation.

\({x^{2}+17x+65}= -5\)

12.

Solve the equation.

\({x^{2}-6x-85}= -13\)

Solving Radical Equations

13.

Solve the equation.

\(\displaystyle{ {\sqrt{x}} = {6} }\)

14.

Solve the equation.

\(\displaystyle{ {\sqrt{x}} = {2} }\)

15.

Solve the equation.

\(\displaystyle{ {\sqrt{4y}} = {12} }\)

16.

Solve the equation.

\(\displaystyle{ {\sqrt{3y}} = {15} }\)

17.

Solve the equation.

\(\displaystyle{ {2\sqrt{r}} = {6} }\)

18.

Solve the equation.

\(\displaystyle{ {4\sqrt{r}} = {16} }\)

19.

Solve the equation.

\(\displaystyle{ {-3\sqrt{t}} = {6} }\)

20.

Solve the equation.

\(\displaystyle{ {-5\sqrt{t}} = {20} }\)

21.

Solve the equation.

\(\displaystyle{ {\sqrt{9-x}+6} = {13} }\)

22.

Solve the equation.

\(\displaystyle{ {-2\sqrt{-3-x}+6} = {0} }\)

23.

Solve the equation.

\(\displaystyle{ {\sqrt{7x+30}} = {x} }\)

24.

Solve the equation.

\(\displaystyle{ {\sqrt{-3y+54}} = {y} }\)

25.

Solve the equation.

\(\displaystyle{ {\sqrt{y}+6} = {y} }\)

26.

Solve the equation.

\(\displaystyle{ {\sqrt{r}+56} = {r} }\)

27.

Solve the equation.

\(\displaystyle{ {r} = {\sqrt{r-4}+6} }\)

28.

Solve the equation.

\(\displaystyle{ {t} = {\sqrt{t+3}-1} }\)

29.

Solve the equation.

\(\displaystyle{ {\sqrt{t+8}} = {\sqrt{t}+2} }\)

30.

Solve the equation.

\(\displaystyle{ {\sqrt{x-15}} = {\sqrt{x}-3} }\)

31.

Solve the equation.

\(\displaystyle{ {\sqrt{x+8}} = {4-\sqrt{x}} }\)

32.

Solve the equation.

\(\displaystyle{ {\sqrt{x+8}} = {-2-\sqrt{x}} }\)

33.

Solve the equation.

\(\displaystyle{ {\sqrt{6y}} = {10} }\)

34.

Solve the equation.

\(\displaystyle{ {\sqrt{2y}} = {5} }\)

35.

Solve the equation.

\(\displaystyle{ \sqrt[3]{r-8} = {9} }\)

36.

Solve the equation.

\(\displaystyle{ \sqrt[3]{r-5} = {3} }\)

37.

Solve the equation.

\(\displaystyle{ {\sqrt{2t+7}+3} = {10} }\)

38.

Solve the equation.

\(\displaystyle{ {\sqrt{8t+2}+10} = {17} }\)

39.

Solve the equation.

\(\displaystyle{ {\sqrt{t}+42} = {t} }\)

40.

Solve the equation.

\(\displaystyle{ {\sqrt{x}+12} = {x} }\)

41.

Solve the equation.

\(\displaystyle{ \sqrt[3]{x-9} = {-5} }\)

42.

Solve the equation.

\(\displaystyle{ \sqrt[3]{y-4} = {4} }\)

43.

Solve the equation.

\(\displaystyle{ {y} = {\sqrt{y+1}+131} }\)

44.

Solve the equation.

\(\displaystyle{ {r} = {\sqrt{r+3}+69} }\)

45.

Solve the equation.

\(\displaystyle{ {\sqrt{40-r}} = {r+2} }\)

46.

Solve the equation.

\(\displaystyle{ {\sqrt{146-t}} = {t+10} }\)

Solving Radical Equations with Variables

47.

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

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

\(R =\) .

48.

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.

\(a =\) .

49.

In a coordinate system, the distance \(r\) from a point \((x,y)\) to the origin \((0,0)\) is given by the following equation:

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

Solve the equation for the coordinate \(y\text{.}\) Assume that \(y\) is positive.

\(y =\) .

50.

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

\(\displaystyle{{f} = {\frac{1}{2\pi \sqrt{LC}}}}\)

Solve the equation for the capacitance \(C\text{.}\)

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

If you need to use a square root, type sqrt(23) for \(\sqrt{23}\) (for example). To enter \(\pi\text{,}\) type pi.

\(C =\) .

51.

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.

\(L =\) .

52.

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

53.

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 \({17\ {\rm ft}}\) and one leg is \({15\ {\rm ft}}\) long, how long is the third side of the triangle?

The third side of the triangle is long.

54.

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 \({25\ {\rm ft}}\) and one leg is \({24\ {\rm ft}}\) long, how long is the third side of the triangle?

The third side of the triangle is long.

55.

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 \({25\ {\rm cm}}\) away from the origin and its x coordinate is \({24\ {\rm cm}}\text{,}\) what is its \(y\) coordinate? Assume that \(y\) is positive.

\(y =\) .

56.

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 \({41\ {\rm cm}}\) away from the origin and its x coordinate is \({40\ {\rm cm}}\text{,}\) what is its \(y\) coordinate? Assume that \(y\) is positive.

\(y =\) .

57.

A pendulum has length \(L\text{,}\) measured in feet. The time period \(T\) that it takes to swing back and forth one time is \({10\ {\rm s}}\text{.}\) The following formula from physics relates \(T\) to \(L\text{.}\)

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

Use this formula to find the length of the pendulum.

58.

A pendulum has length \(L\text{,}\) measured in feet. The time period \(T\) that it takes to swing back and forth one time is \({2\ {\rm s}}\text{.}\) The following formula from physics relates \(T\) to \(L\text{.}\)

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

Use this formula to find the length of the pendulum.