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## Algebra II (2018 edition)

### Course: Algebra II (2018 edition)>Unit 5

Lesson 2: Extraneous solutions of radical equations

# Extraneous solutions of radical equations

Practice some problems before going into the exercise.

## Introduction

### Practice question 1

Caleb is solving the following equation for x.
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?

### Practice question 2

Elena is solving the following equation for x.
cube root of, 3, x, minus, 5, end cube root, plus, 2, equals, 7
Her first few steps are given below.
\begin{aligned}\sqrt[3]{3x-5}&=5\\ \\ \left(\sqrt[3]{3x-5}\right)^3&=(5)^3\\ \\ 3x-5&=125 \end{aligned}
Is it necessary for Elena to check her answers for extraneous solutions?

### Practice question 3

Addison solves the equation below by first squaring both sides of the equation.
2, x, minus, 1, equals, square root of, 8, minus, x, end square root
What extraneous solution does Addison obtain?
x, equals

### Practice question 4

Which value for the constant d makes x, equals, minus, 1 an extraneous solution in the following equation?
square root of, 8, minus, x, end square root, equals, 2, x, plus, d
d, equals

## Want to join the conversation?

• is 0 an extraneous solution
• It doesn't have to be. For example, x^2+4x=0 has solutions 0 and -4 (try them)
• In the last situation, why did they plug -1 in the equation? was it a random number?
• This got me at first too. In the original question, it specifically asks you to use x=-1
• So take the following as an example.

sqrt(4x + 41) = x+5

Originally, x must be greater than -41/4 to be valid. When we solve this equation algebraically, we get the following two solutions: X = -8 and X = 2.

When you plug X = -8 back into the equation, you end up with sqrt(9) = -3.
This is extraneous because we are supposed to use the principle root.

I have three questions.
#1: Why do we have to use the principle root? It seems logical that the sqrt(9) should be equal to both +3 and -3.
#2: Why is this extraneous solution still within the original domain of the function, as it is > -41/4.
#3: What happened mathematically to produce this extraneous solution. If you graph the two equations separately, they only intersect at x = 2, therefore x = 8 is not a solution. But why is this? What happened to produce this extra answer?
• 1. In your example, we use the principal square root because the original problem statement says so. It says √(4x + 41), and not -√(4x + 41). By definition, the radical notation without a minus or ± sign in front of it means “the principal root”, which is always positive. The solutions to Power Equations and the solutions to Radical Equations are different things. For example, x^2 = 4 has two solutions: x = ±2, while x = √4 has only one solution: x = 2, and x = -√4 also has one solution: x = -2. We define the square root as a function, so it must have only one output for each input. If we define the function y = √x as having two solutions, then it is no longer a function.

2. The solution x = -8 is extraneous to the original equation √(4x + 41) = x + 5. However, it is the solution to the equation -√(4x + 41) = x + 5. The expression under the radical is same in both equations, so in terms of keeping the radicand non-negative, the value -8 is OK. If we take the function y = √(4x + 41), then -8 would be a valid input value for x. However, for the equation √(4x + 41) = x + 5, the value x = -8 is not a solution, because it leads to an invalid statement 3 = -3, which is not true.

3. As mentioned above, x = -8 is the solution to the equation -√(4x + 41) = x + 5. So, if we graph -√(4x + 41) instead of √(4x + 41), it will intersect with x + 5 at x = -8.
• I still don't understand why I should care about extraneous solution. It's outside the domain, not a solution, a wrong answer. Or is there any use of finding it?
• Extraneous solutions are not necessarily outside the domain. But they can appear as extra solutions when we square both sides of an equation, because when we square an equation, we would get the same result whether the original equation was positive or negative. So one solution corresponds to the positive equation, and the other to the negative equation. But both of them will fall out of the algebra. We need to be able to tell which solution is extraneous and which works.
• I am having a really hard time with this unit. I have watched all the videos several times and I am still very confused. I would really appreciate some help.
• The key idea for solving square root equations is to isolate a radical on one side and square both sides. If there's still a radical in the equation, then this process would need to be performed a second time. (By the way, don't forget to include the middle term when squaring a binomial. Many students forget to do this.)

After you have solved the resulting linear or quadratic equation for x, remember that you're not finished yet! Because every positive number has a positive and a negative square root, but the radical symbol denotes only the positive (principal) square root, the act of squaring both sides can create invalid (extraneous) solutions! So plug in your solutions to the original equation to determine which solutions work and which solutions must be discarded.

Have a blessed, wonderful day!
• square root 9 should be either 3 or -3.......why should -3 be extraneous then?
• Hi I don't understand why √9 would not equal -3 when it equals 3. In practice question 3, why wouldn't x=-1 be a correct solution??