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## Algebra II (2018 edition)

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

Lesson 1: Solving square-root equations- Intro to square-root equations & extraneous solutions
- Intro to solving square-root equations
- Square-root equations intro
- Solving square-root equations
- Solving square-root equations: one solution
- Solving square-root equations: two solutions
- Solving square-root equations: no solution
- Square-root equations

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# Solving square-root equations

Practice some problems before going into the exercise.

## Introduction

In this article, we will solve more square-root equations. They're a little different than the equations you've solved before: they'll require more work for solving, and the problems will be more challenging problems with extraneous solutions.

### Practice question 1: Isolating the radical term

### Practice question 2: Two possible solutions

### Practice question 3

## Want to join the conversation?

- Why do we get extraneous solutions? At which point during the solution process causes this?(7 votes)
- Imagine this: There's an equation x = 2.

Right now, x is only equal to two.

Square both sides, and x^2 = 4.

For some reason, if you want to take the square root of both sides, and you get x= +/- 2, because -2 squared is still equal to four.

But, according to the original equation, x is only equal to 2.

Therefore -2 is an extraneous solution, and squaring both sides of the equation creates them.(35 votes)

- In the last problem how does it equal 30x and not 15x to begin with in step 2?(5 votes)
- Hey there friend!

So in Step 2 you get ((5+3x)^2 which is equivalent to (5+3x)*(5+3x).

After you distribute you get this string of numbers: 25+15x+15x+9x^2

From there you simplify to: 9x^2+30x+25.

I hope this helps! Have a good day! :D(6 votes)

- how do u work out the the square root of n equal square root of 2n plus 6(3 votes)
- sqrt(n) = sqrt(2n+6)

To undo a square root, you raise it to the 2nd power (square it). What you do on one side you must do on the other side.

So sqrt(n)^2 = n while sqrt(2n+6)^2 = 2n+6

This will leave you with n=2n+6 to solve for n.(5 votes)

- I don't know how to answer these questions.(4 votes)
- (2x + 5)(x + 5) = 0 How did you get this?(2 votes)
- Factor the quadratic using grouping. See the videos at this link: https://www.khanacademy.org/math/algebra/polynomial-factorization#factoring-quadratics-2(4 votes)

- Interesting, the videos didn't discuss (that I saw) there being NO solutions.(2 votes)
- That is covered in a later video in this lesson.(3 votes)

- *Is there a specific way to solve for square roots or is it always just going to be trial and error?*(2 votes)
- We can use a linear approximation to find a close estimate for the square root. √(x) ≈ (x + y) / (2 * √(y)) where y is a number that is "close to" x. Typically, you would choose y to be a perfect square to make the math easy.(3 votes)

- On the left sidebar this worksheet is after the exercise "Solve Square Root Equations (basic)", but the subtitle is "Practice Some Problems Before Going Into the Exercise." Is there a mix up here? I'm accessing the page via URL: https://www.khanacademy.org/math/algebra2/radical-equations-and-functions/solving-square-root-equations/a/solving-square-root-equations-advanced(3 votes)
- Hmm... I think it might be a typo.(1 vote)

- For practice question #1, why couldn't I square both sides- the squareroot of 2x-7 -3 and then the 4? Wouldn't that keep the equations equivalent?(2 votes)
- You could, but it doesn't eliminate the radical.

You may be thinking that if you square the left side that it becomes 2x-7+9. It doesn't. (a+b)^2 never equals a^2+b^2.

Here's what happens if you square both sides immediately

[sqrt(2x-7)-3]^2 = 4^2

The left side is a binomial squared, so use FOIL or extended distribution to square that side.

2x-7 -3sqrt(2x-7) - 3sqrt(2x-7) + 9 = 16

2x+2-6sqrt(2x-7) = 16

You have just made the equation more complicated, and it still has a radical.

This is why the correct 1st step is to isolate the radical**before**squaring both sides.

Hope this helps.(3 votes)

- For root(7x^2 + 15x) - 2x = 5 + x

Got solutions as x = -5 and x = -(5/2)

While checking for extraneous Solutions -

For x = -5 we get

root(100) + 10 = 0

The solution mentioned on the website considers only positive value of root(100) i.e. +10. Shouldn't it also consider -10 ?(1 vote)- There is a difference between taking the square root of a number which is always positive (√100=10) and solving x^2=100 which gives both a positive and negative answer. The first is finding a value on the square root function, the second is finding the x intercepts of an equation. Does this make sense?

I looked on DESMOS and they did not cross as a check. This means both answers are extraneous.(3 votes)