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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Roots problem 2
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[email protected]
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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
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
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!
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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killer1387
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i proceeded as following (took me around 15 sec):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
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.
