Problem Solving

If x > 0 and y > 0, which of the following is equal to 1/(root x + root (x + y))?

(A) 1/y

(B) Root (2x + y)

(C) Root x/Root(x + y)

(D) (Root x - Root (x + y))/y

(E) (Root (x + y) - Root x)/y

OA: E

HG10 wrote:If x > 0 and y > 0, which of the following is equal to 1/(root x + root (x + y))?

(A) 1/y

(B) Root (2x + y)

(C) Root x/Root(x + y)

(D) (Root x - Root (x + y))/y

(E) (Root (x + y) - Root x)/y

OA: E

1/[root (x) + root (x + y)] = ?

Let's try to remove the roots from the denominator by multiplying by [root(x) - root(x+y)] / [root(x) - root(x+y)]

Formula becomes [root(x) - root(x+y)] / (x - x - y) = [root(x+y) - root(x)]/y

E

Thanks for the explanation. Please let me know why you multiplied [root x - root ( x + y)] / [root x - root (x + y)] to get rid of the denominator, instead of multiplying the equation by [root x + root ( x + y)] / [root x + root ( x + y)]? Shouldn't the denominator and numerator be multiplied by the expression in the denominator of the equation in question?

Thanks for your help.

HG10 wrote:Thanks for the explanation. Please let me know why you multiplied [root x - root ( x + y)] / [root x - root (x + y)] to get rid of the denominator, instead of multiplying the equation by [root x + root ( x + y)] / [root x + root ( x + y)]? Shouldn't the denominator and numerator be multiplied by the expression in the denominator of the equation in question?

Thanks for your help.

In short, because it was the only way to get rid of the roots.

Remember, (a+b)*(a-b) = a^2 - b^2
(a+b)(a+b) = a^2 + 2ab + b^2

The 2ab, in this case, will still have a root in it.