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### Course: Integrated math 1>Unit 13

Lesson 3: Simplifying square roots

# Simplifying square-root expressions

Worked examples of taking expressions with square roots and taking all of the perfect squares out of the square roots. For example, 2√(7x)⋅3√(14x²) can be written as 42x√(2x).

## Want to join the conversation?

• Solve this, please square root of (x+ 15) + square root of (x) = 15
• First, you are in the wrong section of lessons. You have a radical equation, not a radical expression.

For a problem more like yours, I would suggest you look at the 2nd problem at this link: http://www.purplemath.com/modules/solverad3.htm

I'll get you started on your equation: √(x+15) + √(x) = 15
1) I would move one radical to the other side. I think it is less confusing. The link above keeps them both on the same side.
Subtract √(x): √(x+15) = 15 - √(x)
2) Square both sides: [√(x+15)]^2 = [15 - √(x) ]^2
3) Simplify left side. FOIL or use extended distribution on the right side to eliminate the exponents
x + 15 = 225 - 30√(x) + x
4) Subtract x: 15 = 225 - 30√(x)
5) Subtract 225: -210 = - 30√(x)
6) Divide by -30: 7 = √(x)
7) Square both sides again: 7^2 = √(x)^2
8) Simplify: 49 = x
9) Check answer back in original equation to verify that it isn't an extraneous solution.

hope this helps.
• At , you state the answer: 6xz√2xz.
However, what happened to the rule that x or z can't be negative?
Did we assume all variables were greater than or equal to zero in the beginning?
I know we don't need absolute value of the variable if it is x^2, or x^4. But in this case, it was just 6xz√2xz. Both x and z were singular, and if one of them were negative(and the other positive), wouldn't this answer be incorrect without a restraint?
• Why has this question remained unanswered for five years? In the first example, why was the absolute value of X not necessary?
(1 vote)
• at , how do you solve the equation radical 75yz to the second power? i watched the video but i am still a little confused.
• What is to the second power, √(75yz^2)? I think this is what you mean, so 75 breaks down to 25z^2 (perfect square) and 3y (non-perfect square), √(25z^2)*√(3y) = 5z√3y.
The other choice is √(75yz)^2 in which case the square and square root cancel to give 75yz.
• Hello, I'm hoping that someone can show me the way!

I am working through the following square root simplification problem:

Square Root of 108a^6

108 can be prime factored into: 2*2*3*3*3

I then broke it down as:

Sq rt of 2^2
Sq rt of 3^3
Sq rt of a^6

I then came up with the following:

2*3*a^3
or further simplified: 6a^3

The Khan answer (which I of course presume is correct) came up with the following simplification:

6a^3 sq rt of 3

It looks like rather than combining like integers and adding exponents, Khan multiplied 2*2*3*3*3= 6^2*3

----
Is there a rule or a step that I'm missing here that brings me to an incorrect answer?

• Your approach is ok. But, the sqrt(3^3) will not equal 3. You need to look for perfect squares which would have an even exponent. Since this exponent is odd, regroup. Split the factors into sqrt(3^2) sqrt(3) = 3 sqrt(3)
2*3*a^3 sqrt(3)
Multiply factors outside to get:
6 a^3 sqrt(3)

Hope this helps.
• I don't get this!
• I understood everything, it was easy to understand
(1 vote)
• good for you bud
• i do not understand any of it, just since like you just make up numbers, how is any of is simplified?