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help me solve this

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help me solve this

by nickhar130 » Mon Jan 10, 2011 11:43 am
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
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by Anurag@Gurome » Mon Jan 10, 2011 12:04 pm
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
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)

... 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.
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by aleph777 » Mon Jan 10, 2011 1:14 pm
Anurag@Gurome wrote:
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.
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!
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by Anurag@Gurome » Mon Jan 10, 2011 1:33 pm
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.
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by aleph777 » Mon Jan 10, 2011 1:51 pm
Of course! Those inverted sq rts alway get me! I didn't think to extract (sqrt of x * sqrt of x) in the denominator!

Thanks!
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