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## Algebra (all content)

### Course: Algebra (all content) > Unit 12

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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# Intro to solving square-root equations

CCSS.Math:

Practice some problems before going into the exercise.

## Introduction

In this article, we will practice the ways we can solve equations with square-roots in them.

### Practice question 1

### Practice question 2

### Practice question 3

## Want to join the conversation?

- wow! i still dont get it(71 votes)
- When I solve a square-root equation, what is it exactly what I'm finding? In what situations would I use the number that I found? (excuse my broken english)(24 votes)
- You are finding the possible solutions of x. The use of the number you found will make the equation on both sides equal if you plugged in x on both sides.(8 votes)

- So, do all quadratic equations have extraneous solutions? Help pleaseeee. Does it also mean that we should always check the solutions whenever we solve for quadratics?(7 votes)
- Extraneous solutions are often introduced whenever you square both sides of an equation. After all, you're turning something that was NOT a quadratic into a quadratic. If you start out with a quadratic, you will not get extraneous solutions.

Another situation where extraneous solutions are possible is when you're dealing with rational expressions. After cancelling denominators and simplifying down to a quadratic, you'll sometimes introduce solutions which result in a 0 in a denominator when they're plugged back in. All you have to check for in this situation is division by 0.(13 votes)

- My question is "How can you do the problem without making one mistake in this long process of answer-finding?"(5 votes)
- Why does squaring work on both sides of an equation? You're multiplying each side by a different amount!

Please clear this up for me.(3 votes)- No, you aren't multiplying a different amount.

Remember, the equations says that the both sides are equal to start with.

If you square both sides, the sides are still equal.

For example, start with: sqrt(9) = 3

These sides are currently equal, just in a different form. Now, square both sides.

[sqrt(9)]^2= 3^2

You will get 9 = 9. They are still equal.

Hope this helps.(6 votes)

- Why does

sqrt(x-2)^2=sqrt(1) equal x-2 = +/-1 and not +/-(x-2) = +/-1 ?(2 votes)- Because some of the options create the same solution
`+(x-2) = +1`

is identical to`-(x-2) = -1`

`-(x-2) = +1`

is identical to`+(x-2) = -1`

For the equation versions that match, multiply one by -1 and you get the other.

Thus, there is no need to do 4 different sign combinations.

Hope this makes sense.(3 votes)

- what if you get a desimal do you do the abc formula(1 vote)
- Yes a decimal is just a number and can be treated a such though it may be harder(Hey if u could upvote this so i get a badge that would be great thx)(5 votes)

- I did not get anything! who knows where I can find radical equations for beginners(3 votes)
- This is the intro video. Try this site as an alternative and see if it helps: http://www.purplemath.com/modules/solverad.htm(0 votes)

- Just trying to solve the final practice question, and a bit stuck on something:

How in the solution did the equation come to (x)squared−4x+4? I don't understand how it came to that?(3 votes)- (x-2)^2=(x-2)(x-2)

=x(x-2)-2(x-2) distribute the terms

=x^2-(2.2.x)+2^2

=x^2- 4x+4(1 vote)

- How would you solve square root of (x-5) + square root of (x) = 5?(1 vote)
- So we begin with
`sqrt(x-5)+sqrt(x)=5`

. We can subtract`sqrt(x)`

from both sides to get`sqrt(x-5)=5-sqrt(x)`

. Then, squaring both sides, we have`x-5=25-10sqrt(x)+x`

. We can subtract`x`

to get`-5=25-10sqrt(x)`

. Subtracting`25`

and dividing by`-10`

gives`3=sqrt(x)`

. Squaring both sides again gives`x=9`

. Of course, you should also check to see if`x=9`

is an extraneous solution by plugging it into the original equation and seeing if it works, which it does.

Most square root equations can be solved in a way similar to this, by trying to get a least one square root term along on a side, then squaring it. You sometimes have to end up being lucking with the squaring the binomial if you have a term like`5-sqrt(x)`

. Hope this helps!(3 votes)