## Extraneous solutions of radical equations

# Extraneous solutions of radical equations

CCSS Math: HSA.REI.A.2

Practice some problems before going into the exercise.

## Introduction

#### In this article we will practice some problems that involve thinking about the conditions for obtaining extraneous solutions while solving radical equations.

### Practice question 1

Caleb is solving the following equation for $x$.

$x=\sqrt{x+2}+7$

His first few steps are given below.

$\begin{aligned}x-7&=\sqrt{x+2}\\
\\
(x-7)^2&=(\sqrt{x+2})^2\\
\\
x^2-14x+49&=x+2
\end{aligned}$

**Is it necessary for Caleb to check his answers for extraneous solutions?**

An

**extraneous solution**is a solution that arises from the solving process that is not really a solution at all!Specifically, whenever you raise both sides of an equation to an

**even power**, you must check for extraneous solutions. This is because raising both sides of an equation to an**even power**is not a*operation. For example, if $a=b$, then $a^2=b^2$, but if $a^2=b^2$, it is not necessary that $a=b$.***reversible**For instance, consider the equation $\sqrt{x}=-1$. If we square both sides of the equation, we get $x=1$, but $\sqrt{1}\neq -1$.

Since Caleb squared both sides of the equation in his second step, it is necessary for him to check for extraneous solutions.

**In conclusion, the answer is Yes.**

### Practice question 2

Elena is solving the following equation for $x$.

$\sqrt[\Large3]{3x-5}+2=7$

Her first few steps are given below.

$\begin{aligned}\sqrt[\Large3]{3x-5}&=5\\
\\
\left(\sqrt[\Large3]{3x-5}\right)^3&=(5)^3\\
\\
3x-5&=125
\end{aligned}$

**Is it necessary for Elena to check her answers for extraneous solutions?**

An

**extraneous solution**is a solution that arises from the solving process that is not really a solution at all!Whenever you raise both sides of an equation to an

**even power**, you must check for extraneous solutions.However, raising both sides of an equation to an

**odd power**does not introduce the possibility of extraneous solutions. This is because raising both sides of an equation to an**odd power**is a*operation. For example, if $a=b$, then $a^3=b^3$, and if $a^3=b^3$, then $a=b$.***reversible**For instance, consider the equation $\sqrt[\Large3]{x}=-1$. If we cube both sides of the equation, we get $x=-1$, and $\sqrt[\Large3]{-1}= -1$.

Additionally, addition, subtraction, multiplication, and division with non-zero real numbers are all reversible operations that do not produce extraneous solutions.

Since Elena used all reversible operations in her solution process, she did not introduce the possibility of extraneous solutions.

**In conclusion, the answer is No.**

### Practice question 3

Addison solves the equation below by first squaring both sides of the equation.

$2x-1=\sqrt{8-x}$

**What extraneous solution does Addison obtain?**

$x=$

### The strategy

To find the extraneous solution, let's start by solving the equation. The solution that arises from the solving process that does not satisfy the original equation is the extraneous solution.

### Step 1: Solve the equation

$\begin{aligned} 2x-1&=\sqrt{8-x}\\\\
\left(2x-1\right)^\blueD2&=\left(\sqrt{8-x}\right)^{\blueD{2}}\\\\
4x^2-4x+1&=8-x\\\\
4x^2-3x-7&=0\\\\
(4x-7)(x+1)&=0
\end{aligned}$

We obtained two possible solutions. $x=-1$ and $x=\dfrac{7}{4}$.

### Step 2: Check for extraneous solutions

Now let's check if any of them is extraneous.

The original equation is $2x-1=\sqrt{8-x}$.

When $\maroonD{x}=\maroonD{-1}$, we get:

$\begin{aligned}2\maroonD{x}-1&=\sqrt{8-\maroonD{x}}\\\\
2(\maroonD{-1})-1&\stackrel{?}{=} \sqrt{8-(\maroonD{-1})}\\\\
-3&\stackrel{?}=\sqrt{9}\\\\
-3&\neq3
\end{aligned}$

So $x=-1$ is not a solution. It is an extraneous solution.

When $\maroonD{x}=\maroonD{\dfrac{7}{4}}$, we get:

$\begin{aligned}2\maroonD{x}-1&=\sqrt{8-\maroonD{x}}\\\\
2\left(\maroonD{\dfrac{7}{4}}\right)-1&\stackrel{?}{=} \sqrt{8-\left(\maroonD{\dfrac{7}{4}}\right)}\\\\
\dfrac{5}{2}&\stackrel{?}=\sqrt{\dfrac{25}{4}}\\\\
\dfrac{5}{2}&=\dfrac{5}{2}
\end{aligned}$

So $x=\dfrac{7}{4}$ is a solution.

### Conclusion

**The extraneous solution that arises from solving the equation is $x=-1$.**

### Practice question 4

**Which value for the constant $d$ makes $x=-1$ an extraneous solution in the following equation?**

$\sqrt{8-x}=2x+d$

$d=$

In order to solve the original equation, we would have to square both sides of the equation:

$\begin{aligned}\sqrt{8-x}&=2x+d\\\\
(\sqrt{8-x})^2&=(2x+d)^2\\\\
8-x&=(2x+d)^2\end{aligned}$

However, squaring both sides of an equation can create extraneous solutions!

When solving radical equations, extraneous solutions may occur because squaring both sides of an equation is not an

*operation. For example, if $a=b$, then $a^2=b^2$, but if $a^2=b^2$, it is not necessary that $a=b$.***if and only if**When we have an equation of the form $a^2=b^2$, we can only conclude that $a = b$ or $a=-b$.

Let's plug $\blueD x=\blueD{-1}$ into the last equation we obtained:

$\begin{aligned}8-\blueD x&=(2\blueD x+d)^2\\\\
8-(\blueD{-1})&=(2(\blueD{-1})+d)^2\\\\
9&=(-2+d)^2\end{aligned}$

*This equation*is correct, both when $-2+d=3$

*when $-2+d=-3$.*

**and**However,

*the original equation*is*correct for $-2+d=-3$, since this way we obtain $\sqrt{9}=-3$.***not**Therefore, an extraneous solution is obtained for the $d$-value that makes $-2+d$ equal $-3$, which is $d=-1$.

Substituting this back into the original equation gives $\sqrt{8-x}=2x-1$. You can now solve this for $x$ and see for yourselves that $x=-1$ is indeed extraneous.

**In conclusion, the solution is $d=-1$.**