# 11.3: Radical Equations

**At Grade**Created by: CK-12

## Learning Objectives

- Solve a radical equation.
- Solve radical equations with radicals on both sides.
- Identify extraneous solutions.
- Solve real-world problems using square root functions.

## Introduction

When the variable in an equation appears inside a radical sign, the equation is called a **radical equation.** To solve a radical equation, we need to eliminate the radical and change the equation into a polynomial equation.

A common method for solving radical equations is to isolate the most complicated radical on one side of the equation and raise both sides of the equation to the power that will eliminate the radical sign. If there are any radicals left in the equation after simplifying, we can repeat this procedure until all radical signs are gone. Once the equation is changed into a polynomial equation, we can solve it with the methods we already know.

We must be careful when we use this method, because whenever we raise an equation to a power, we could introduce false solutions that are not in fact solutions to the original problem. These are called **extraneous solutions.** In order to make sure we get the correct solutions, we must always check all solutions in the original radical equation.

## Solve a Radical Equation

Let’s consider a few simple examples of radical equations where only one radical appears in the equation.

**Example 1**

*Find the real solutions of the equation* \begin{align*}\sqrt{2x-1}=5\end{align*}

**Solution**

Since the radical expression is already isolated, we can just square both sides of the equation in order to eliminate the radical sign:

\begin{align*}\left(\sqrt{2x-1}\right)^2=5^2\end{align*}

\begin{align*}\text{Remember that} \ \sqrt{a^2}=a \ \text{so the equation simplifies to:} && 2x-1& =25\\ \text{Add one to both sides:} && 2x& =26\\ \text{Divide both sides by 2:} &&& \underline{\underline{x=13}}\end{align*}

Finally we need to plug the solution in the original equation to see if it is a valid solution.

\begin{align*}\sqrt{2x-1}=\sqrt{2(13)-1}=\sqrt{26-1}=\sqrt{25}=5\end{align*}**The solution checks out.**

**Example 2**

*Find the real solutions of* \begin{align*}\sqrt[3]{3-7x}-3=0\end{align*}

**Solution**

\begin{align*}\text{We isolate the radical on one side of the equation:} && \sqrt[3]{3-7x}& =3\\ \text{Raise each side of the equation to the third power:} && \left(\sqrt[3]{3-7x}\right)^3& =3^3\\ \text{Simplify:} && 3-7x& =27\\ \text{Subtract 3 from each side:} && -7x& =24\\ \text{Divide both sides by -7:} &&& \underline{\underline{x=-\frac{24}{7}}}\end{align*}

**Check:** \begin{align*}\sqrt[3]{3-7x}-3=\sqrt[3]{3-7 \left(-\frac{24}{7}\right)}-3=\sqrt[3]{3+24}-3=\sqrt[3]{27}-3=3-3=0\end{align*}

**Example 3**

*Find the real solutions of* \begin{align*}\sqrt{10-x^2}-x=2\end{align*}

**Solution**

\begin{align*}\text{We isolate the radical on one side of the equation:} && \sqrt{10-x^2}& =2+x\\ \text{Square each side of the equation:} && \left(\sqrt{10-x^2}\right)^2& =(2+x)^2\\ \text{Simplify:} && 10-x^2& =4+4x+x^2\\ \text{Move all terms to one side of the equation:} && 0& =2x^2+4x-6\\ \text{Solve using the quadratic formula:} && x& =\frac{-4 \pm \sqrt{4^2-4(2)(-6)}}{4}\\ \text{Simplify:} && x& =\frac{-4 \pm \sqrt{64}}{4}\\ \text{Re-write} \ \sqrt{24} \ \text{in simplest form:} && x& =\frac{-4 \pm 8}{4}\\ \text{Reduce all terms by a factor of 2:} && x& =1 \ \text{or} \ x=-3\end{align*}

**Check:** \begin{align*}\sqrt{10-1^2}-1=\sqrt{9}-1=3-1=2\end{align*}

\begin{align*}\sqrt{10-(-3)^2}-(-3)=\sqrt{1}+3=1+3=4\end{align*}

The equation has only one solution, \begin{align*}\underline{\underline{x=1}}\end{align*}

## Solve Radical Equations with Radicals on Both Sides

Often equations have more than one radical expression. The strategy in this case is to start by isolating the most complicated radical expression and raise the equation to the appropriate power. We then repeat the process until all radical signs are eliminated.

**Example 4**

*Find the real roots of the equation* \begin{align*}\sqrt{2x+1}-\sqrt{x-3}=2\end{align*}

**Solution**

\begin{align*}\text{Isolate one of the radical expressions:} && \sqrt{2x+1}& =2+\sqrt{x-3}\\ \text{Square both sides:} && \left(\sqrt{2x+1}\right)^2& =\left(2+\sqrt{x-3}\right)^2\\ \text{Eliminate parentheses:} && 2x+1& =4+4\sqrt{x-3}+x-3\\ \text{Simplify:} && x& =4 \sqrt{x-3}\\ \text{Square both sides of the equation:} && x^2& =\left(4 \sqrt{x-3} \right)^2\\ \text{Eliminate parentheses:} && x^2& =16(x-3)\\ \text{Simplify:} && x^2& =16x-48\\ \text{Move all terms to one side of the equation:} && x^2-16x+48& =0\\ \text{Factor:} && (x-12)(x-4)& =0\\ \text{Solve:} && x& =12 \ \text{or} \ x=4\end{align*}

**Check:** \begin{align*}\sqrt{2(12)+1}-\sqrt{12-3}=\sqrt{25}-\sqrt{9}=5-3=2\end{align*}. The solution checks out.

\begin{align*}\sqrt{2(4)+1}-\sqrt{4-3}=\sqrt{9}-\sqrt{1}=3-1=2\end{align*} The solution checks out.

The equation has two solutions: \begin{align*}x=12\end{align*} and \begin{align*}x=4\end{align*}.

## Identify Extraneous Solutions to Radical Equations

We saw in Example 3 that some of the solutions that we find by solving radical equations do not check out when we substitute (or “plug in”) those solutions back into the original radical equation. These are called **extraneous solutions.** It is very important to check the answers we obtain by plugging them back into the original equation, so we can tell which of them are real solutions.

**Example 5**

*Find the real solutions of the equation* \begin{align*}\sqrt{x-3}-\sqrt{x}=1\end{align*}.

**Solution**

\begin{align*}\text{Isolate one of the radical expressions:} && \sqrt{x-3}&=\sqrt{x}+1\\ \text{Square both sides:} && \left(\sqrt{x-3}\right)^2& =\left(\sqrt{x}+1\right)^2\\ \text{Remove parenthesis:} && x-3& =\left(\sqrt{x}\right)^2+2\sqrt{x}+1\\ \text{Simplify:} && x-3& =x+2\sqrt{x}+1\\ \text{Now isolate the remaining radical:} && -4& =2\sqrt{x}\\ \text{Divide all terms by 2:} && -2& =\sqrt{x}\\ \text{Square both sides:} && x& =4\end{align*}

**Check:** \begin{align*}\sqrt{4-3}-\sqrt{4}=\sqrt{1}-2=1-2=-1\end{align*} The solution does not check out.

** The equation has no real solutions.** \begin{align*}x=4\end{align*} is an extraneous solution.

## Solve Real-World Problems using Radical Equations

Radical equations often appear in problems involving areas and volumes of objects.

**Example 6**

*Anita’s square vegetable garden is 21 square feet larger than Fred’s square vegetable garden. Anita and Fred decide to pool their money together and buy the same kind of fencing for their gardens. If they need 84 feet of fencing, what is the size of each garden?*

**Solution**

**Make a sketch:**

**Define variables:** Let Fred’s area be \begin{align*}x\end{align*}; then Anita’s area is \begin{align*}x+21\end{align*}.

**Find an equation:**

Side length of Fred’s garden is \begin{align*}\sqrt{x}\end{align*}

Side length of Anita’s garden is \begin{align*}\sqrt{x+21}\end{align*}

The amount of fencing is equal to the combined perimeters of the two squares:

\begin{align*}4\sqrt{x}+4\sqrt{x+21}=84\end{align*}

**Solve the equation:**

\begin{align*}\text{Divide all terms by 4:} && \sqrt{x}+\sqrt{x+21}& =21\\ \text{Isolate one of the radical expressions:} && \sqrt{x+21}& =21-\sqrt{x}\\ \text{Square both sides:} && \left(\sqrt{x+21}\right)^2& =\left(21-\sqrt{x}\right)^2\\ \text{Eliminate parentheses:} && x+21& =441-42\sqrt{x}+x\\ \text{Isolate the radical expression:} && 42\sqrt{x}& =420\\ \text{Divide both sides by 42:} && \sqrt{x}& =10\\ \text{Square both sides:} && x& =100 \ ft^2\end{align*}

**Check:** \begin{align*}4\sqrt{100}+4\sqrt{100+21}=40+44=84\end{align*}. **The solution checks out.**

Fred’s garden is \begin{align*}10 \ ft \times 10 \ ft = 100 \ ft^2\end{align*} and Anita’s garden is \begin{align*}11 \ ft \times 11 \ ft = 121 \ ft^2\end{align*}.

**Example 7**

*A sphere has a volume of \begin{align*}456 \ cm^3\end{align*}. If the radius of the sphere is increased by 2 cm, what is the new volume of the sphere?*

**Solution**

**Make a sketch:**

**Define variables:** Let \begin{align*}R =\end{align*} the radius of the sphere.

**Find an equation:** The volume of a sphere is given by the formula \begin{align*}V=\frac{4}{3}\pi R^3\end{align*}.

**Solve the equation:**

\begin{align*}\text{Plug in the value of the volume:} && 456& =\frac{4}{3} \pi R^3\\ \text{Multiply by 3:} && 1368& =4 \pi R^3\\ \text{Divide by} \ 4 \pi: && 108.92& =R^3\\ \text{Take the cube root of each side:} && R& =\sqrt[3]{108.92} \Rightarrow R=4.776 \ cm\\ \text{The new radius is 2 centimeters more:} && R& =6.776 \ cm\\ \text{The new volume is:} && V & =\frac{4}{3} \pi (6.776)^3=\underline{\underline{1302.5}} \ cm^3\end{align*}

**Check:** Let’s plug in the values of the radius into the volume formula:

\begin{align*}V=\frac{4}{3} \pi R^3=\frac{4}{3} \pi (4.776)^3=456 \ cm^3\end{align*}. **The solution checks out.**

**Example 8**

*The kinetic energy of an object of mass \begin{align*}m\end{align*} and velocity \begin{align*}v\end{align*} is given by the formula: \begin{align*}KE=\frac{1}{2} mv^2\end{align*}. A baseball has a mass of 145 kg and its kinetic energy is measured to be 654 Joules \begin{align*}(kg \cdot m^2/s^2)\end{align*} when it hits the catcher’s glove. What is the velocity of the ball when it hits the catcher’s glove?*

**Solution**

\begin{align*}\text{Start with the formula:} && KE& =\frac{1}{2} mv^2\\ \text{Plug in the values for the mass and the kinetic energy:} && 654 \frac{kg \cdot m^2}{s^2}& =\frac{1}{2}(145\ kg)v^2\\ \text{Multiply both sides by 2:} && 1308 \frac{kg \cdot m^2}{s^2}& =145 \ kg \cdot v^2\\ \text{Divide both sides by 145} \ kg: && 9.02 \frac{m^2}{s^2}& =v^2\\ \text{Take the square root of both sides:} && v& =\sqrt{9.02} \sqrt{\frac{m^2}{s^2}}=3.003 \ m/s\end{align*}

Check: Plug the values for the mass and the velocity into the energy formula:

\begin{align*}KE=\frac{1}{2}mv^2=\frac{1}{2}(145 \ kg)(3.003 \ m/s)^2=654 \ kg \cdot m^2/s^2\end{align*}

## Review Questions

Find the solution to each of the following radical equations. Identify extraneous solutions.

- \begin{align*}\sqrt{x+2}-2=0\end{align*}
- \begin{align*}\sqrt{3x-1}=5\end{align*}
- \begin{align*}2 \sqrt{4-3x}+3=0\end{align*}
- \begin{align*}\sqrt[3]{x-3}=1\end{align*}
- \begin{align*}\sqrt[4]{x^2-9}=2\end{align*}
- \begin{align*}\sqrt[3]{-2-5x}+3=0\end{align*}
- \begin{align*}\sqrt{x^2-3}=x-1\end{align*}
- \begin{align*}\sqrt{x}=x-6\end{align*}
- \begin{align*}\sqrt{x^2-5x}-6=0\end{align*}
- \begin{align*}\sqrt{(x+1)(x-3)}=x\end{align*}
- \begin{align*}\sqrt{x+6}=x+4\end{align*}
- \begin{align*}\sqrt{x}=\sqrt{x-9}+1\end{align*}
- \begin{align*}\sqrt{x}+2=\sqrt{3x-2}\end{align*}
- \begin{align*}\sqrt{3x+4}=-6\end{align*}
- \begin{align*}5 \sqrt{x}=\sqrt{x+12}+6\end{align*}
- \begin{align*}\sqrt{10-5x}+\sqrt{1-x}=7\end{align*}
- \begin{align*}\sqrt{2x-2}-2\sqrt{x}+2=0\end{align*}
- \begin{align*}\sqrt{2x+5}-3\sqrt{2x-3}=\sqrt{2-x}\end{align*}
- \begin{align*}3\sqrt{x}-9=\sqrt{2x-14}\end{align*}
- \begin{align*}\sqrt{x+7}=\sqrt{x+4}+1\end{align*}
- The area of a triangle is \begin{align*}24 \ in^2\end{align*} and the height of the triangle is twice as long as the base. What are the base and the height of the triangle?
- The length of a rectangle is 7 meters less than twice its width, and its area is \begin{align*}660 \ m^2\end{align*}. What are the length and width of the rectangle?
- The area of a circular disk is \begin{align*}124 \ in^2\end{align*}. What is the circumference of the disk? \begin{align*}(\text{Area} = \pi R^2, \text{Circumference} =2 \pi R)\end{align*}.
- The volume of a cylinder is \begin{align*}245 \ cm^3\end{align*} and the height of the cylinder is one third of the diameter of the base of the cylinder. The diameter of the cylinder is kept the same but the height of the cylinder is increased by 2 centimeters. What is the volume of the new cylinder? \begin{align*}(\text{Volume} =\pi R^2 \cdot h)\end{align*}
- The height of a golf ball as it travels through the air is given by the equation \begin{align*}h=-16t^2+256\end{align*}. Find the time when the ball is at a height of 120 feet.

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Feb 23, 2012## Last Modified:

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