10.6: Radical Equations

Here we look at equations with radicals. As you might expect, to clear a radical we can raise both sides to an exponent. Recall, the roots of radicals can be thought of reversing an exponent. Hence, to reverse a radical, we will use exponents.

Solving Radical Equations

If \(x\geq 0\) and \(a\geq 0\), then \[\sqrt{x}=a\text{ if and only if }x=a^2\nonumber\]

In general, if \(x\geq 0\) and \(a\) is a real number, then \[\sqrt[n]{x}=a\text{ if and only if }x=a^n\nonumber\]

We assume in this chapter that all variables are greater than or equal to zero.

Recall, if \(n\) is even, then \(x\) must be greater than or equal to zero. If \(n\) is odd, then \(x\) can be any real number.

We can apply the following method to solve equations with radicals.

Steps for solving radical equations

Step 1. Isolate the radical.

Step 2. Raise both sides of the equation to the power of the root (index).

Step 3. Solve the equation as usual.

Step 4. Verify the solution(s). (Recall, we will omit any extraneous solutions.)

Radical Equations with Square Roots

Example 10.6.1

Solve for \(x\): \(\sqrt{7x+2}=4\)

Solution

Step 1. Isolate the radical. Notice the radical is already isolated for us on the left, with no coefficients: \[\sqrt{7x+2}=4\nonumber\]

Step 2. Raise both sides of the equation to the power of the root (index). \[\begin{array}{rl}\sqrt{7x+2}=4&\text{Raise each side to the power of }2 \\ (\sqrt{7x+2})^2=4^2 &\text{Evaluate} \\ 7x+2=16\end{array}\nonumber\]

Step 3. Solve the equation as usual. \[\begin{array}{rl}7x+2=16&\text{Isolate the variable term} \\ 7x=14&\text{Solve for }x \\ x=2&\text{Solution}\end{array}\nonumber\]

Step 4. Verify the solution(s). (Recall, we will omit any extraneous solutions.) \[\begin{aligned}\sqrt{7x+2}&\stackrel{?}{=}4 \\ \sqrt{7\color{blue}{(2)}\color{black}{+2}}&\stackrel{?}{=}4 \\ \sqrt{16}&\stackrel{?}{=}4 \\ 4&=4\end{aligned}\]

Thus, \(x=2\) is, in fact, a solution.

Example 10.6.2

Solve for \(x\): \(\sqrt{x+5}=-1\)

Solution

Step 1. Isolate the radical. Notice the radical is already isolated for us on the left, with no coefficients: \[\sqrt{x+5}=-1\nonumber\]

Step 2. Raise both sides of the equation to the power of the root (index). \[\begin{array}{rl}\sqrt{x+5}=-1 &\text{Raise each side to the power of }2 \\ (\sqrt{x+5})^2=(-1)^2&\text{Evaluate} \\ x+5=1\end{array}\nonumber\]

Step 3. Solve the equation as usual. \[\begin{array}{rl}x+5=1&\text{Solve for }x \\ x=-4&\text{Solution}\end{array}\nonumber\]

Step 4. Verify the solution(s). (Recall, we will omit any extraneous solutions.) \[\begin{aligned}\sqrt{x+5}&\stackrel{?}{=}-1 \\ \sqrt{\color{blue}{(-4)}\color{black}{+5}}&\stackrel{?}{=}-1 \\ \sqrt{1}&\stackrel{?}{=}-1 \\ 1&\neq -1\end{aligned}\]

Oh no! When verifying the solution, we obtained a false statement. Thus, this equation has no solution and \(x = −4\) is an extraneous solution

Isolate the Radical

Example 10.6.3

Solve for \(x\): \(x+\sqrt{4x+1}=5\)

Solution

Step 1. Isolate the radical. Let’s isolate the radical on the left by moving the \(x\) to the right side. \[\begin{aligned}x+\sqrt{4x+1}&=5 \\ \sqrt{4x+1}&=5-x\end{aligned}\]

Step 2. Raise both sides of the equation to the power of the root (index). \[\begin{array}{rl}\sqrt{4x+1}=5-x &\text{Raise each side to the power of }2 \\ (\sqrt{4x+1})^2=(5-x)^2&\text{Evaluate} \\ 4x+1=25-10x+x^2\end{array}\nonumber\]

Step 3. Solve the equation as usual. \[\begin{array}{rl} 4x+1=25-10x+x^2&\text{Notice the }x^2\text{ term; solve by factoring} \\ x^2-14x+24=0&\text{Factor} \\ (x-12)(x-2)=0&\text{Apply zero product rule} \\ x-12=0\quad\text{or}\quad x-2=0&\text{Solve} \\ x=12\quad\text{or}\quad x=2&\text{Solutions}\end{array}\nonumber\]

Step 4. Verify the solution(s). (Recall, we will omit any extraneous solutions.)

\[\begin{array}{rl}\color{blue}{12}\color{black}{+}\sqrt{4\color{blue}{(12)}\color{black}{+1}}\stackrel{?}{=}5 &\color{blue}{5}\color{black}{+}\sqrt{4\color{blue}{(2)}\color{black}{+1}}\stackrel{?}{=} 5 \\ 12+\sqrt{49}\stackrel{?}{=}5 &2+\sqrt{9}\stackrel{?}{=}5 \\ 12+7\stackrel{?}{=}5& 2+3\stackrel{?}{=}5 \\ 19\neq 5 &5=5\end{array}\nonumber\]

Since \(x = 12\) gives a false statement, then \(x = 12\) is an extraneous solution. Thus, \(x = 2\) is, in fact, the solution.

Note

The Babylonians were the first known culture to solve quadratics in radicals as early as 2000 BC.

Radical Equations with Two Square Roots

Example 10.6.4

Solve for \(x\): \(\sqrt{2x+1}-\sqrt{x}=1\)

Solution

Step 1. Isolate the radical. Since there are two radicals in the equation, we will isolate only one of them. \[\begin{aligned}\sqrt{2x+1}-\sqrt{x}&=1 \\ \sqrt{2x+1}&=1+\sqrt{x}\end{aligned}\]

Step 2. Raise both sides of the equation to the power of the root (index). \[\begin{array}{rl}\sqrt{2x+1}=1+\sqrt{x}&\text{Raise each side to the power of }2 \\ (\sqrt{2x+1})^2=(1+\sqrt{x})^2&\text{Evaluate} \\ 2x+1=1+2\sqrt{x}+x\end{array}\nonumber\] Notice there is \(\sqrt{x}\) that still remains in the equation even after squaring each side. Hence, we should repeat steps 1 and 2 again to obtain an equation without radicals. \[\begin{array}{rl}2x+1=1+2\sqrt{x}+x&\text{Isolate the radical} \\ x=2\sqrt{x} \\ \dfrac{x}{2}=\sqrt{x}&\text{Raise each side to the power of }2 \\ \left(\dfrac{x}{2}\right)^2=(\sqrt{x})^2&\text{Evaluate} \\ \dfrac{x^2}{4}=x\end{array}\nonumber\]

Step 3. Solve the equation as usual. \[\begin{array}{rl}\dfrac{x^2}{4}=x&\text{Notice the }x^2\text{ term; solve by factoring} \\ \dfrac{x^2}{4}-x=0&\text{Multiply each term by LCD }4 \\ \color{blue}{4}\color{black}{\cdot}\dfrac{x^2}{4}-\color{blue}{4}\color{black}{\cdot}x=\color{blue}{4}\color{black}{\cdot }0&\text{Simplify}\end{array}\nonumber\]

\[\begin{array}{rl}x^2-4x=0&\text{Factor} \\ x(x-4)=0&\text{Apply zero product rule} \\ x=0\quad\text{or}\quad x-4=0&\text{Solve} \\ x=0\quad\text{or}\quad x=4&\text{Solutions}\end{array}\nonumber\]

Step 4. Verify the solution(s). (Recall, we will omit any extraneous solutions.) \[\begin{array}{rl}\sqrt{2\color{blue}{(0)}\color{black}{+1}}-\sqrt{\color{blue}{0}}\color{black}{}\stackrel{?}{=}1&\sqrt{2\color{blue}{(4)}\color{black}{+1}}-\sqrt{4}\stackrel{?}{=}1 \\ \sqrt{1}\stackrel{?}{=}1&\sqrt{9}-2\stackrel{?}{=}1 \\ 1=1&3-2\stackrel{?}{=}1 \\ 1=1&1=1\end{array}\nonumber\]

Since \(x = 0\) and \(x = 4\) both give true statements, then \(x = 0\) and \(x = 4\) are, in fact, the solutions.

Radical Equations with Higher Roots

Example 10.6.5

Solve for \(n\): \(\sqrt[3]{n-1}=-4\)

Solution

Step 1. Isolate the radical. Notice the radical is already isolated for us on the left, with no coefficients: \[\sqrt[3]{n-1}=-4\nonumber\]

Step 2. Raise both sides of the equation to the power of the root (index). Notice the root here is 3; hence, we will raise each side to the third power. \[\begin{array}{rl}\sqrt[3]{n-1}=-4&\text{Raise each side to the power of }3 \\ (\sqrt[3]{n-1})^3=(-4)^3&\text{Evaluate} \\ n-1=-64 \end{array}\nonumber\]

Step 3. Solve the equation as usual. \[\begin{array}{rl}n-1=-64&\text{Isolate the variable term} \\ n=-63&\text{Solution}\end{array}\nonumber\]

Step 4. Verify the solution(s). (Recall, we will omit any extraneous solutions.) \[\begin{aligned} \sqrt[3]{n-1}&\stackrel{?}{=}-4 \\ \sqrt[3]{\color{blue}{-63}\color{black}{-1}}&\stackrel{?}{=}-4 \\ \sqrt[3]{-64}&\stackrel{?}{=}-4 \\ -4&=-4\end{aligned}\]

Thus, \(n=-63\) is, in fact, a solution.

Applications with Radical Functions

Example 10.6.6

A person’s Body Mass Index (BMI) is a measure of body fat based on height and weight. If a person’s BMI is above \(25\) and below \(30\), he/she is classified as overweight. A person’s height in terms of his/her weight in pounds, \(w\), and body mass index (BMI), \(b\), is given by \[H(w)=\sqrt{\dfrac{703w}{b}}\nonumber\]

1. How tall is a person weighing \(250\) pounds and has a BMI of \(25\)? Round your answer to one decimal place.
2. If a person is \(71\) inches tall and has a BMI of \(25\), what is the person’s weight? Round your answer to one decimal place.

Solution

We apply the formula to answer both parts. Since this is a function with a square root, we use the techniques from above to solve.

1. Since the person weighs \(250\) pounds and has BMI \(25\), then \(w=250\) and \(b=25\). Let's plug-n-chug these in to find the person's height in inches. \[\begin{aligned}H(w)&=\sqrt{\dfrac{703w}{b}} \\ H(\color{blue}{250}\color{black}{)}&=\sqrt{\dfrac{703(\color{blue}{250}\color{black}{)}}{\color{blue}{25}}} \\ H(250)&=\sqrt{7030} \\ H(250)&\approx 83.5\end{aligned}\] Thus, a person whose weight is \(250\) pounds with a BMI of \(25\) is about \(83.5\) inches tall, which is nearly \(7\) feet tall!
2. Since the person is \(71\) inches tall and has BMI \(25\), then \(H = 71\) and \(b = 25\). Let’s plug-n-chug this into the function and solve for \(w\), the person’s weight in pounds. \[\begin{aligned}H(w)&=\sqrt{\dfrac{703w}{b}} \\ \color{blue}{71}\color{black}{}&=\sqrt{\dfrac{703w}{\color{blue}{25}}} \\ 71^2&=\dfrac{703w}{25} \\ 71^2\cdot 25&=703w \\ \dfrac{71^2\cdot 25}{703}&=w \\ w&\approx 179.3\end{aligned}\] Thus, a person whose height is \(71\) inches with a BMI of \(25\), weights about \(179.3\) pounds.

Example 10.6.7

The time it takes for a pendulum to swing back and forth one time can be represented by the function \[S(x)=2\pi\sqrt{\dfrac{x}{32}}\nonumber\] where \(S(x)\) is the time in seconds and \(x\) is the length of the pendulum in feet.

1. How many seconds will take for a \(7\)-foot pendulum to swing back and forth one time? Round your answer to one decimal place.
2. If it takes \(4\) seconds for a pendulum to swing back and forth one time, what is the length of the pendulum? Round your answer to one decimal place.

Solution

We apply the formula to answer both parts. Since this is a function with a square root, we use the techniques from above to solve.

1. Since it is given that the pendulum is \(7\) feet, then this implies \(x = 7\). Let’s plug-n-chug \(x = 7\) into \(S\) to obtain the time it takes for the pendulum to swing back and forth one time. \[\begin{aligned} S(x)&=2\pi\sqrt{\dfrac{x}{32}} \\ S(\color{blue}{7}\color{black}{)}&=2\pi\sqrt{\dfrac{\color{blue}{7}}{\color{black}{32}}} \\ S(7)&\approx 2.9\end{aligned}\] Thus, it will take about \(2.9\) seconds for a \(7\)-foot pendulum to swing back and forth one time.
2. If we are given that it takes \(4\) seconds for a pendulum to swing back and forth one time, then this means \(S = 4\). Let’s plug-n-chug this into the function to find \(x\), the length of the pendulum in feet. \[\begin{aligned}S(x)&=2\pi\sqrt{\dfrac{x}{32}} \\ \color{blue}{4}\color{black}{}&=2\pi\sqrt{\dfrac{x}{32}} \\ \dfrac{4}{2\pi}&=\sqrt{\dfrac{x}{32}} \\ \left(\dfrac{4}{2\pi}\right)^2&=\dfrac{x}{32} \\ 32\cdot\dfrac{16}{4\pi ^2}&=x \\ x&\approx 13.0\end{aligned}\] Thus, a \(13\)-foot pendulum will take \(4\) seconds to swing back and forth one time.

Radical Equations Homework

Solve. Be sure to verify all solutions.

Exercise 10.6.1

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

Exercise 10.6.2

\(\sqrt{6x-5}-x=0\)

Exercise 10.6.3

\(3+x=\sqrt{6x+13}\)

Exercise 10.6.4

\(\sqrt{3-3x}-1=2x\)

Exercise 10.6.5

\(\sqrt{4x+5}-\sqrt{x+4}=2\)

Exercise 10.6.6

\(\sqrt{2x+4}-\sqrt{x+3}=1\)

Exercise 10.6.7

\(\sqrt{2x+6}-\sqrt{x+4}=1\)

Exercise 10.6.8

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

Exercise 10.6.9

\(\sqrt{5x+1}-4=0\)

Exercise 10.6.10

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

Exercise 10.6.11

\(x-1=\sqrt{7-x}\)

Exercise 10.6.12

\(\sqrt{2x+2}=3+\sqrt{2x-1}\)

Exercise 10.6.13

\(\sqrt{3x+4}-\sqrt{x+2}=2\)

Exercise 10.6.14

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

Exercise 10.6.15

\(\sqrt{4x-3}-\sqrt{3x+1}=1\)

Exercise 10.6.16

\(\sqrt{2-3x}-\sqrt{3x+7}=3\)

Exercise 10.6.17

A person’s Body Mass Index (BMI) is a measure of body fat based on height and weight. If a person’s BMI is above \(25\) and below \(30\), he/she is classified as overweight. A person’s height in terms of his/her weight in pounds, \(w\), and body mass index (BMI), \(b\), is given by \[H(w) =\sqrt{\dfrac{703w}{b}}\nonumber\]

1. How tall is a person if he/she weighs \(225\) pounds and has a BMI of \(25\)? Round your answer to one decimal place.
2. If a person is \(64\) inches tall and has a BMI of \(25\), what is his/her weight? Round your answer to one decimal place.

Exercise 10.6.18

The time it takes for a pendulum to swing back and forth one time can be represented by the function \[S(x) = 2\pi\sqrt{\dfrac{x}{32}}\nonumber\] where \(S(x)\) is the time in seconds and \(x\) is the length of the pendulum in feet.

1. How many seconds will take for a \(5\)-foot pendulum to swing back and forth one time? Round your answer to one decimal place.
2. If it takes \(2\) seconds for a pendulum to swing back and forth one time, what is the length of the pendulum? Round your answer to one decimal place.