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## Algebra II (2018 edition)

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

Lesson 1: Solving square-root equations

# 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

Solve the following equation for $x$.
$\sqrt{2x-7}-3=4$
$x=$

### Practice question 2: Two possible solutions

What are the solutions to the following equation?
$6+3x=\sqrt{2x+12}+2x$

### Practice question 3

What are the solutions to the following equation?
$\sqrt{7{x}^{2}+15x}-2x=5+x$

## Want to join the conversation?

• Why do we get extraneous solutions? At which point during the solution process causes this?
• 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.
• In the last problem how does it equal 30x and not 15x to begin with in step 2?
• 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
• how do u work out the the square root of n equal square root of 2n plus 6
• 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.
• I don't know how to answer these questions.
• (2x + 5)(x + 5) = 0 How did you get this?
• Interesting, the videos didn't discuss (that I saw) there being NO solutions.
• That is covered in a later video in this lesson.
• *Is there a specific way to solve for square roots or is it always just going to be trial and error?*
• 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.
• 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
• 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?
• 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.