# Extraneous solutions of radical equations

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.

$x=\sqrt{x+2}+7$x, equals, square root of, x, plus, 2, end square root, plus, 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 $a=b$a, equals, b, then $a^2=b^2$a, start superscript, 2, end superscript, equals, b, start superscript, 2, end superscript, but if $a^2=b^2$a, start superscript, 2, end superscript, equals, b, start superscript, 2, end superscript, it is not necessary that $a=b$a, equals, b.***reversible*For instance, consider the equation $\sqrt{x}=-1$square root of, x, end square root, equals, minus, 1. If we square both sides of the equation, we get $x=1$x, equals, 1, but $\sqrt{1}\neq -1$square root of, 1, end square root, does not equal, minus, 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 $x$x.

Her first few steps are given below.

**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 $a=b$a, equals, b, then $a^3=b^3$a, start superscript, 3, end superscript, equals, b, start superscript, 3, end superscript, and if $a^3=b^3$a, start superscript, 3, end superscript, equals, b, start superscript, 3, end superscript, then $a=b$a, equals, b.***reversible*For instance, consider the equation . If we cube both sides of the equation, we get $x=-1$x, equals, minus, 1, and .

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}$2, x, minus, 1, equals, square root of, 8, minus, x, end square root

**What extraneous solution does Addison obtain?**

$x=$x, equals

**Your answer should be**- an integer, like $6$6
- a
*simplified proper*fraction, like $3/5$3, slash, 5 - a
*simplified improper*fraction, like $7/4$7, slash, 4 - a mixed number, like $1\ 3/4$1, space, 3, slash, 4
- an
*exact*decimal, like $0.75$0, point, 75 - a multiple of pi, like $12\ \text{pi}$12, space, p, i or $2/3\ \text{pi}$2, slash, 3, space, p, i

### 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. $x=-1$x, equals, minus, 1 and $x=\dfrac{7}{4}$x, equals, start fraction, 7, divided by, 4, end fraction.

### 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}$2, x, minus, 1, equals, square root of, 8, minus, x, end square root.

When $\maroonD{x}=\maroonD{-1}$start color maroonD, x, end color maroonD, equals, start color maroonD, minus, 1, end color maroonD, 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 $x=-1$x, equals, minus, 1 is not a solution. It is an extraneous solution.

When $\maroonD{x}=\maroonD{\dfrac{7}{4}}$start color maroonD, x, end color maroonD, equals, start color maroonD, start fraction, 7, divided by, 4, end fraction, end color maroonD, 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 $x=\dfrac{7}{4}$x, equals, start fraction, 7, divided by, 4, end fraction is a solution.

### Conclusion

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

### Practice question 4

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

$\sqrt{8-x}=2x+d$square root of, 8, minus, x, end square root, equals, 2, x, plus, d

$d=$d, equals

**Your answer should be**- an integer, like $6$6
- a
*simplified proper*fraction, like $3/5$3, slash, 5 - a
*simplified improper*fraction, like $7/4$7, slash, 4 - a mixed number, like $1\ 3/4$1, space, 3, slash, 4
- an
*exact*decimal, like $0.75$0, point, 75 - a multiple of pi, like $12\ \text{pi}$12, space, p, i or $2/3\ \text{pi}$2, slash, 3, space, p, i

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 $a=b$a, equals, b, then $a^2=b^2$a, start superscript, 2, end superscript, equals, b, start superscript, 2, end superscript, but if $a^2=b^2$a, start superscript, 2, end superscript, equals, b, start superscript, 2, end superscript, it is not necessary that $a=b$a, equals, b.***if and only if*When we have an equation of the form $a^2=b^2$a, start superscript, 2, end superscript, equals, b, start superscript, 2, end superscript, we can only conclude that $a = b$a, equals, b or $a=-b$a, equals, minus, b.

Let's plug $\blueD x=\blueD{-1}$start color blueD, x, end color blueD, equals, start color blueD, minus, 1, end color blueD 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 $-2+d=3$minus, 2, plus, d, equals, 3

**when $-2+d=-3$minus, 2, plus, d, equals, minus, 3.**

*and*However,

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

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

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