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New Algebra Set!!!

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Joined: 02 Sep 2009
Posts: 55272
Re: New Algebra Set!!!  [#permalink]

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New post 25 Nov 2018, 02:21
shiv17 wrote:
Bunuel wrote:
4. What is the maximum value of -3x^2 + 12x -2y^2 - 12y - 39 ?

A. -39
B. -9
C. 0
D. 9
E. 39

\(-3x^2 + 12x -2y^2 - 12y - 39=-3x^2 + 12x-12-2y^2 - 12y-18-9=-3(x^2-4x+4)-2(y^2+6y+9)-9=\)
\(=-3(x-2)^2-2(y+3)^2-9\).

So, we need to maximize the value of \(-3(x-2)^2-2(y+3)^2-9\).

Since, the maximum value of \(-3(x-2)^2\) and \(-2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+0-9=-9\).

Answer: B.



Can anybody help me with this solution. Why the maximum value of \(-3(x-2)^2\) and \(-2(y+3)^2\) is zero?


Please read the whole thread before posting: https://gmatclub.com/forum/new-algebra- ... l#p2020076
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Re: New Algebra Set!!!  [#permalink]

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New post 25 Nov 2018, 13:37
putting the value of x in eq -
3^2+3a+15=0
=> a= -8
Now putting a= -8 in x^2 + ax - b = 0
x^2−8x−b=0.

as roots are equal
=> b^2-4ac = 0
=> b = -16 (B)
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Re: New Algebra Set!!!  [#permalink]

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New post 25 Mar 2019, 07:36
Bunuel

I have a question to exercise 5: My solution was x < -5 and x > 3 and therefore my probability was 4/7. Even after reading this post https://gmatclub.com/forum/solving-quad ... 70528.html, I am not sure where I went wrong. Could you help me?
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Re: New Algebra Set!!!   [#permalink] 25 Mar 2019, 07:36

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