It is a problem of rationalization of denominator of a irrational denominator. Which can be done by using the formula (a + b)(a - b) = (a^2 - b^2)nickhar130 wrote:If x > 0, then 1/[√(2x)+√x] =
A. 1/√(3x)
B. 1/[2√(2x)]
C. 1/(x√2)
D. (√2-1)/√x
E. (1+√2)/√x
Anurag@Gurome wrote:Anurag, I don't understand that final step... I got locked at [√(2x) - √x]/x. Where does that final division by √x come from?nickhar130 wrote: ... 1/[√(2x) + √x]
= [√(2x) - √x]/[(√(2x) + √x)*(√(2x) - √x)] ....... Multiply with [√(2x) - √x]
= [√(2x) - √x]/[2x - x]
= [√(2x) - √x]/x
= [√2 - 1]/√x ...................................................... Divide with √x
The correct answer is D.
Thanks!
aleph777 wrote:Anurag, I don't understand that final step... I got locked at [√(2x) - √x]/x. Where does that final division by √x come from?
Thanks![/quote
We can write √(2x) as √(2)*(√x), and x as (√x)*(√x)right?
Then [√(2x) - √x] = [√(2)*(√x) - (√x)] = (√x)*[√(2) - 1]
Now, [√(2x) - √x]/x = (√x)*[√(2) - 1]/[(√x)*(√x)] = [√(2) - 1]/(√x)
Hope it is clear now.
