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## Algebra 1

### Course: Algebra 1>Unit 11

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? • at , how do you solve the equation radical 75yz to the second power? i watched the video but i am still a little confused. • 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? • Why does sal not multiply the numbers(2,14,5) together?? to get square root of 140a^5... can someone please explain! • Expand and simplify 5square root of 12 multiply square root of 6 • Is it all just a WHOLE LOT of simplifying, or am I missing something? • So none of the videos here show why anything beyond the 5th power for a variable has a perfect square that equals x to the power of 4 squared. So one of the questions in the foundations part lists y to the 9th. And the hint (I do not see this explained anywhere in either exponents or in simplifying square roots variables at all) shows that written out as y to the 4th squared with a remaining y but gives no explanation for WHY y to the 4th is a perfect square. • First, consider what happens when you square variables...
y*y = y^2
y^2 * y^2 = y^4
y^3 * y^3 = y^6
y^4 * y^4 = y^8
Notice, all the resulting exponents are even numbers.

Now, let's look at sqrt(y^9). The exponent is odd. So, y^9 is not a perfect square. However, just like with numbers, we can split the factors apart to find perfect squares. There are more than one way to do this.
Sal split y^9 into y^8 * y. y^8 has an even exponent. So it is a perfect square. It can split it into 2 factor groups: y^4*y^4 * y = (y^4)^2 * y. When you do sqrt(y^8), you get the y^4.

An alternative method is to split y^9 int pairs of y's. Each pair is a perfect square:
y^9 = y^2 * y^2 * y^2 * y^2 * y
You can then do the square root of each pair, which will give you y*y*y*y sqrt(y) = y^4 sqrt(y).

Hope this helps.  