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product of all possible solutions

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Manager
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product of all possible solutions [#permalink] New post 02 Jun 2009, 03:59
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If y = (3y+4)^(1/2) then the product of all possible solution(s) for y is
-4
-2
0
4
6

D
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Re: product of all possible solutions [#permalink] New post 02 Jun 2009, 16:05
If y = (3y+4)^(1/2) then the product of all possible solution(s) for y is

y^2=3y+4
y^2-3y-4
(y-4)(y+1)

And it looks like i get the wrong answer.....

I would have thought it was -4
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Re: product of all possible solutions [#permalink] New post 02 Jun 2009, 20:32
mkrump wrote:
If y = (3y+4)^(1/2) then the product of all possible solution(s) for y is

y^2=3y+4
y^2-3y-4
(y-4)(y+1)

And it looks like i get the wrong answer.....

I would have thought it was -4



This is not a straight algebra question
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Re: product of all possible solutions [#permalink] New post 03 Jun 2009, 01:40
-4 is the answer..

the equation has 2 possible solutions.. 4 and -1..
and the product is -4...

Also this an straight Algebra question.. i donot why 'mbamission' said it is not straight algebra ..

Whats the OA?
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Re: product of all possible solutions [#permalink] New post 03 Jun 2009, 08:27
Neochronic wrote:
-4 is the answer..

the equation has 2 possible solutions.. 4 and -1..
and the product is -4...

Also this an straight Algebra question.. i donot why 'mbamission' said it is not straight algebra ..

Whats the OA?


Hey dude, trust me when i say this question is not a straight algebra. :-D
Why dont u re-try this question and find out yourself :wink:
In case you want the solution let me know.

BTW, I have already posted the OA with the qestion.
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Re: product of all possible solutions [#permalink] New post 03 Jun 2009, 20:48
mbaMission wrote:
This is not a straight algebra question


Initially I also thought the answer is -4 but on second look I think you are right.

Because plugging in solution -1 in equation will give

-1 = [3*(-1) + 4]^1/2
-1 = 1^1/2 and
-1 = 1 which is not correct

But still not sure how to get the correct answer. Please post the solution.
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Re: product of all possible solutions [#permalink] New post 04 Jun 2009, 00:17
Sqroot of 1 can be -1 and 1..

still feel that the answer is -4..

any other explanations..
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Re: product of all possible solutions [#permalink] New post 29 Jun 2009, 12:20
I also get A. mbaMission could you please post your solution so we can learn from this question?

thank you.
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Re: product of all possible solutions [#permalink] New post 29 Jun 2009, 12:44
We have
y = (3y + 4)^{\frac{1}{2}} = \sqrt{3y + 4}

From this, we can immediately conclude two things:

* 3y + 4 > 0, since it's under a square root. Thus, y > -4/3;
* y > 0, since y = \sqrt{\text{something}} (the square root function can't produce a negative result)

So when we solve the equation by squaring both sides (to find y = 4 or -1), we need to discard the negative solution. The only legitimate solution is y = 4.
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Re: product of all possible solutions   [#permalink] 29 Jun 2009, 12:44
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