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tricky

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tricky

by PAB2706 » Sat May 16, 2009 6:23 pm
If root ( 3y+ 4 ) = y then the product of all possible solution(s) for y is
-4
-2
0
4
6
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by myohmy » Sat May 16, 2009 7:18 pm
I am not a quant jock but here's my best shot.

√(3y + 4) = y
(√(3y + 4))^2=y^2
3y + 4 = y^2
y^2 - 3y - 4 = 0
(y+1) (y-4) = 0
Y=-1,4

-1 * 4 = -4
So I would go with choice A.
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by fleshins » Sat May 16, 2009 7:23 pm
y = -1 isn't a solution to the equation since a square root can't be negative. The answer should be 4 (D).
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by myohmy » Sat May 16, 2009 7:27 pm
Whoops, good point. D it is.
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Re: tricky

by Vemuri » Sat May 16, 2009 7:41 pm
PAB2706 wrote:If root ( 3y+ 4 ) = y then the product of all possible solution(s) for y is
-4
-2
0
4
6
Good question !! IMO 4
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by fleshins » Sat May 16, 2009 10:22 pm
What's the OA?
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