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If x > 0, then 1/[v(2x)+vx] = ? ....Bizarre Question!

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If x > 0, then 1/[v(2x)+vx] = ? ....Bizarre Question!

by mitzwillrockgmat » Sat Jun 12, 2010 1:46 pm
Hi can someone solve this for me! In my opinion, the answer should be A but D is given. But I don't see how!

If x > 0, then 1/[v(2x)+vx] =

A. 1/v(3x)
B. 1/[2v(2x)]
C. 1/(xv2)
D. (v2-1)/vx
E. (1+v2)/vx

This is what I did:

1/[v(2x)+vx] => 1/[vx(2+1)] = 1/vx3 & this is A!

HOW can get reach D?! Is this a typo or does the "v" stand for some function?!
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by clock60 » Sat Jun 12, 2010 2:18 pm
my try
i hope that problem looks like
1/((2x)^1/2+(x)^1/2)

multiply both nominator and denomimator with (2x)^1/2-(x)^1/2
let us attack denominator at first

{(2x)^1/2+(x)^1/2}*{(2x)^1/2-(x)^1/2}=2x-x=x

and nominator=(2x)^1/2-x^1/2=x^1/2*(2^(1/2)-1)

and divide fraction (x^1/2)/x=x^(-1/2)-put it into denominator (as a^-1=1/a)

so final result

(2^1/2-1)/(x^1/2)

so D
Last edited by clock60 on Sat Jun 12, 2010 2:20 pm, edited 1 time in total.
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by Rich@VeritasPrep » Sat Jun 12, 2010 2:20 pm
Hi Mitz,

Yes, the 'v' in this problem is actually a square root. It should read:

If x > 0, then 1 / [sqrt(2x) + sqrt(x)] =

Break up the denominator:

= 1 / [sqrt(2)*sqrt(x) + sqrt(x)] =

= 1 / [(sqrt(2)+1)*sqrt(x)]

Solve using conjugates:

= 1 / [(sqrt(2)+1)*sqrt(x)] * ( [sqrt(2) - 1] / [sqrt(2) - 1] ) <---essentially multiply everything by 1

= [sqrt(2) - 1] / [ (2-1)*sqrt(x)]

= [sqrt(2) - 1] / sqrt(x)

which is D.
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by mitzwillrockgmat » Sat Jun 12, 2010 2:41 pm
raz1024 wrote:Hi Mitz,

Yes, the 'v' in this problem is actually a square root. It should read:

If x > 0, then 1 / [sqrt(2x) + sqrt(x)] =

Break up the denominator:

= 1 / [sqrt(2)*sqrt(x) + sqrt(x)] =

= 1 / [(sqrt(2)+1)*sqrt(x)]

Solve using conjugates:

= 1 / [(sqrt(2)+1)*sqrt(x)] * ( [sqrt(2) - 1] / [sqrt(2) - 1] ) <---essentially multiply everything by 1

= [sqrt(2) - 1] / [ (2-1)*sqrt(x)]

= [sqrt(2) - 1] / sqrt(x)

which is D.
oh ok! Thanks!! :)

one more question though how come when you conjugated

you did the above & not, 1 / [(sqrt(2)+1)*sqrt(x)] * [(sqrt(2)-1)*sqrt(x)] / [(sqrt(2)-1)*sqrt(x)] ?

i mean how come you didn't include the sqrt(x) in the num & den when multiplying?

want to know just in case there's a shortcut because conjugation takes time! so if there's a way to cut down time, why not?
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by Rich@VeritasPrep » Sat Jun 12, 2010 5:05 pm
You could include sqrt(x) in the multiplication, but that will add an extra step at the end. You'd end up with sqrt(x) / x as part of the final term, and that would become the 1 / sqrt(x) that you see as part of answer choice D.

Just to clarify why these are equal:

sqrt(x) / x

= x^(1/2) / x

= x^( 1/2 - 1)

= x^(-1/2)

= 1 / x^(1/2)

= 1 / sqrt(x)
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by mitzwillrockgmat » Sun Jun 13, 2010 4:58 am
raz1024 wrote:You could include sqrt(x) in the multiplication, but that will add an extra step at the end. You'd end up with sqrt(x) / x as part of the final term, and that would become the 1 / sqrt(x) that you see as part of answer choice D.

Just to clarify why these are equal:

sqrt(x) / x

= x^(1/2) / x

= x^( 1/2 - 1)

= x^(-1/2)

= 1 / x^(1/2)

= 1 / sqrt(x)
oh ok! makes sense but i would never guess that on my own...! thank you!
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