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## Algebra (all content)

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

Lesson 1: Solving square-root equations

# Intro to solving square-root equations

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

Solve the following equation for x.
square root of, 2, x, minus, 5, end square root, equals, 7
x, equals

### Practice question 2

Solve the following equation for x.
square root of, 10, minus, 3, x, end square root, equals, square root of, 2, x, plus, 5, end square root
x, equals

### Practice question 3

Solve the following equation for x.
square root of, 5, x, minus, 4, end square root, equals, x, minus, 2
x, equals

## Want to join the conversation?

• wow! i still dont get it
• 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)
• 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.
• 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?
• 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.
• My question is "How can you do the problem without making one mistake in this long process of answer-finding?"
• 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.
• 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.
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.
• Why does
sqrt(x-2)^2=sqrt(1) equal x-2 = +/-1 and not +/-(x-2) = +/-1 ?
• 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.
• 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)
• I did not get anything! who knows where I can find radical equations for beginners
• 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?
• 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!