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Roots problem 2

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Master | Next Rank: 500 Posts
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by Troika » Wed Mar 07, 2012 6:32 pm
If root (3 - x) = 1 - root (2x), then 9x^2 =

(A) 5

(B) 8

(C) 2 - 3x

(D) 12 - 4x

(E) 20x - 4

OA: E
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by [email protected] » Wed Mar 07, 2012 7:14 pm
Root(3-x)=1-root(2x)
3-x=(1-root(2x))^2
3-x=2x+1-2*root(2x)
3x=2*root(2x)+2 ---Eq 1

So 9x^2=(2*root(2x)+2)^2
=(4+8*root(2x)+8x)
=4(1+2x+2*root(2x))
=4(1+2x+3x-2) --substituting from Eq 1
=4(5x-1)
=20x-4

OA is E
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by kjallow » Thu Mar 08, 2012 4:22 pm
someone please tell me this is a 700-800 level question! how is this to be done in less than 2 mins??

Good explanation though..i was able to work it to this point: 4(1+2x+2*root(2x)) but didn't think to re-substitute from eqn 1. Thanks a bunch!
[email protected] wrote:Root(3-x)=1-root(2x)
3-x=(1-root(2x))^2
3-x=2x+1-2*root(2x)
3x=2*root(2x)+2 ---Eq 1

So 9x^2=(2*root(2x)+2)^2
=(4+8*root(2x)+8x)
=4(1+2x+2*root(2x))
=4(1+2x+3x-2) --substituting from Eq 1
=4(5x-1)
=20x-4

OA is E
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by Troika » Thu Mar 08, 2012 5:14 pm
Thank you for the very helpful solution!
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by killer1387 » Thu Mar 08, 2012 6:43 pm
HG10 wrote:If root (3 - x) = 1 - root (2x), then 9x^2 =

(A) 5

(B) 8

(C) 2 - 3x

(D) 12 - 4x

(E) 20x - 4

OA: E
i proceeded as following (took me around 15 sec):
root (3 - x) = 1 - root (2x),(square both sides)
=>3-x = 1+ 2x- 2 root(2x),
=> 2 root(2x)= 3x-2, (square both sides)
=> 8x = 9x^2+ 4- 12x,
=> 9x^2=20x-4

hence E.

HTH.
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