Question

For all x > 0 and y > 0, the radical expression:
√x divided by 3√x-√y is equivalent to.
3x + √xy divided by 9x-y.
How did they get this answer?

Answer 1

they just multiplied and divided by 3√x+√y

Answer 2

When you have the sum or differences of two square roots in your denominator, generally you "multiply by the conjugate" to eliminate them.

so if you have say, 1 / ( sqrt(a) + sqrt(b) ) you would multiply THAT by the fraction

sqrt(a) - sqrt(b) over sqrt(a) - sqrt(b)

(note the minuses!)

The reverse is true if you have say, 1 / ( sqrt(a) - sqrt(b) ) you would multiply THAT by the fraction

sqrt(a) + sqrt(b) over sqrt(a) + sqrt(b)

Answer 3

yea conjugate :D

Answer 4

I'm sorry but I still don't understand.

Answer 5

Where did the rest of the numbers come from?

Answer 6

do u know, (a-b)(a+b) = a^2 - b^2

Answer 7

No

Answer 8

Well, it is. They hv used it.

√x / 3√x-√y = (√x )* (3√x+√y) / (3√x-√y )* (3√x+√y)
= (3x + √xy) / (9x-y)

Answer 9

Oooh ok thank you very much....i understand now!

Answer 10

You're welcome! :)