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 xx.
x=x+2+7x=\sqrt{x+2}+7
His first few steps are given below.
x7=x+2(x7)2=(x+2)2x214x+49=x+2\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?
Choose 1 answer:
Choose 1 answer:
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 reversible operation. For example, if a=ba=b, then a2=b2a^2=b^2, but if a2=b2a^2=b^2, it is not necessary that a=ba=b.
For instance, consider the equation x=1\sqrt{x}=-1. If we square both sides of the equation, we get x=1x=1, but 11\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 xx.
3x53+2=7\sqrt[\Large3]{3x-5}+2=7
Her first few steps are given below.
3x53=5(3x53)3=(5)33x5=125\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?
Choose 1 answer:
Choose 1 answer:
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 reversible operation. For example, if a=ba=b, then a3=b3a^3=b^3, and if a3=b3a^3=b^3, then a=ba=b.
For instance, consider the equation x3=1\sqrt[\Large3]{x}=-1. If we cube both sides of the equation, we get x=1x=-1, and 13=1\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.
2x1=8x2x-1=\sqrt{8-x}
What extraneous solution does Addison obtain?
x=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

2x1=8x(2x1)2=(8x)24x24x+1=8x4x23x7=0(4x7)(x+1)=0\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=1x=-1 and x=74x=\dfrac{7}{4}.

Step 2: Check for extraneous solutions

Now let's check if any of them is extraneous.
The original equation is 2x1=8x2x-1=\sqrt{8-x}.
When x=1\maroonD{x}=\maroonD{-1}, we get:
2x1=8x2(1)1=?8(1)3=?933\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=1x=-1 is not a solution. It is an extraneous solution.
When x=74\maroonD{x}=\maroonD{\dfrac{7}{4}}, we get:
2x1=8x2(74)1=?8(74)52=?25452=52\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=74x=\dfrac{7}{4} is a solution.

Conclusion

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

Practice question 4

Which value for the constant dd makes x=1x=-1 an extraneous solution in the following equation?
8x=2x+d\sqrt{8-x}=2x+d
d=d=
In order to solve the original equation, we would have to square both sides of the equation:
8x=2x+d(8x)2=(2x+d)28x=(2x+d)2\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 if and only if operation. For example, if a=ba=b, then a2=b2a^2=b^2, but if a2=b2a^2=b^2, it is not necessary that a=ba=b.
When we have an equation of the form a2=b2a^2=b^2, we can only conclude that a=b a = b or a=ba=-b.
Let's plug x=1\blueD x=\blueD{-1} into the last equation we obtained:
8x=(2x+d)28(1)=(2(1)+d)29=(2+d)2\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=3-2+d=3 and when 2+d=3-2+d=-3.
However, the original equation is not correct for 2+d=3-2+d=-3, since this way we obtain 9=3\sqrt{9}=-3.
Therefore, an extraneous solution is obtained for the dd-value that makes 2+d-2+d equal 3-3, which is d=1d=-1.
Substituting this back into the original equation gives 8x=2x1\sqrt{8-x}=2x-1. You can now solve this for xx and see for yourselves that x=1x=-1 is indeed extraneous.
In conclusion, the solution is d=1d=-1.