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product of all possible solutions [#permalink]
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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============================================== Do not answer without sharing the reasoning behind ur choice ----------------------------------------------------------- Working on my weakness : GMAT Verbal ------------------------------------------------------------ Ask: Why, What, How, When, Where, Who ==============================================

Re: product of all possible solutions [#permalink]
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

_________________

============================================== Do not answer without sharing the reasoning behind ur choice ----------------------------------------------------------- Working on my weakness : GMAT Verbal ------------------------------------------------------------ Ask: Why, What, How, When, Where, Who ==============================================

Re: product of all possible solutions [#permalink]
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. Why dont u re-try this question and find out yourself In case you want the solution let me know.

BTW, I have already posted the OA with the qestion.

_________________

============================================== Do not answer without sharing the reasoning behind ur choice ----------------------------------------------------------- Working on my weakness : GMAT Verbal ------------------------------------------------------------ Ask: Why, What, How, When, Where, Who ==============================================

Re: product of all possible solutions [#permalink]
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