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?!
If x > 0, then 1/[v(2x)+vx] = ? ....Bizarre Question!
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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
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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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.
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.
Rich Zwelling
GMAT Instructor, Veritas Prep
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oh ok! Thanks!!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.
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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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)
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)
Rich Zwelling
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oh ok! makes sense but i would never guess that on my own...! thank you!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)