Urgent! Save Temulin's Math Grade!

[USSRGirl]

Okay, doin' an online math program and if I don't answer this question in about fifteen min it will time me out and give me a new one. Soooo, could some kind saintly soul show me how to do this radical-expression-torture-device-thing? Please? I'll give you a small third world country to rule over in return.

Simplify the following expression:

[Doubleshadow]

It's probably way too late, but here it is:

When the number under the radical doesn't have an easy square root, break it down into factors that do. So,

SQR(24) = SQR(4*6)

because we know the square root of 4 is 2, we can move it to the outside and get

2SQR(6)

The same is true for variables with exponents.

SQR(x^5) = SQR(x^4 *x)

Since the square root of x^4 is x^2, it can be moved out side the radical and we get

(x^2)SQR(x)

In the case of y^2, its just the square root of y^2 which is y. Remember, the square root of anything squared is itself.

After that, group common factors so that

(x^2)*y*SQR(6x)*(2-3)

Since 2-3=-1, the answer is

-(x^2)*y*SQR(6x)

[Mononoke]

I read the first line of the 2nd response and grade 12 principles math flooded back to me in a wave of horror.