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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?
• 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.
• Why does sal not multiply the numbers(2,14,5) together?? to get square root of 140a^5... can someone please explain!
• He could do that and he kind of did when he put them all under one radical. But, he kept them in factored form because it is easier to see the perfect squares with factored form. If you muliply to 140a^5, you then need to split it up into perfect squares.
• Expand and simplify 5square root of 12 multiply square root of 6
• 5 sqrt(12) * sqrt(6) = 5 sqrt(72) = 30 sqrt(2).
• I don't get this!
• Is it all just a WHOLE LOT of simplifying, or am I missing something?
• what do we do with points?
• 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.