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New Algebra Set!!! [#permalink]
18 Mar 2013, 06:56

30

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54

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The next set of medium/hard PS algebra questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.

1. If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:

Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 08:45

4) -3x^2 + 12x -2y^2 - 12y - 39 MAX -3x^2 + 12x has max in (2,12) -2y^2 - 12y - 39 has max in (-6,-3) -3x^2 + 12x -2y^2 - 12y - 39 MAX = 12 - 3 =9 D _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 10:25

1

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5. If x^2 + 2x -15 = -m, where m is an integer from -10 and 10, inclusive, what is the probability that m is greater than zero?

A. 2/7 B. 1/3 C. 7/20 D. 2/5 E. 3/7

\(x^2 + 2x -15 = -m\) \(=> x^2 + 2x + 1 -16 = -m\) \(=> (x+1)^2 = 16 -m\) \(=> 16 -m >=0\) and a square number

If m varies between [-10,10]

\(=> 16 - 10 <= 16 - m <= 16 - (-10)\) \(=> 6 <= 16 - m <= 26\)

Between 6 and 16, there are 3 square numbers i.e. 9, 16, 25

For \(16 - m = 9 => m = 7\) For\(16 - m = 16 => m = 0\) For\(16 - m = 25 => m = -9\)

So number of possible values of m = 3 i.e. [7, 0, -9] And for m > 0, number of possible values = 1 i.e only for m=7

So probability = 1/3 Ans = B _________________

"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well." ― Voltaire Press Kudos, if I have helped. Thanks! shit-happens-my-journey-to-172475.html#p1372807

Last edited by ConnectTheDots on 20 Mar 2013, 07:40, edited 1 time in total.

Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 10:55

2

This post received KUDOS

@Bunuel, for Q5, shouldn't the original question also say that we are only looking for solutions where x is an integer?

For this to be satisfied, in x^2 + 2x -15 = -m, 4 - 4(m-15)>0 and 4-4(m-15) must be a perfect square => 4 (16-m) must be a perfect square and >0 => 16-m must be a perfect square and >0 The only values that satisfy this for -10<=m<=10 are m=-9,0,7 of which only 7 is positive => probability = 1/3 _________________

Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 12:05

1

This post received KUDOS

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2. The equation x^2 + ax - b = 0 has equal roots, and one of the roots of the equation x^2 + ax + 15 = 0 is 3. What is the value of b?

A. -64 B. -16 C. -15 D. -1/16 E. -1/64

Let the roots of \(x^2 + ax +b = 0\) are h and h

So sum of the roots i.e \(2h = -a\) and product of the roots i.e \(h^2 = -b\) Now one root of \(x^2 + ax + 15=0\) is 5 So the other root must be 5 as the product is 15. So \(a = - (5+3) = -8\)

So, \(8 = 2h\) or, \(h=4\) So \(b = -4^2 = -16\) B it is _________________

Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 12:33

2

This post received KUDOS

Expert's post

5. If x^2 + 2x -15 = -m, where m is an integer from -10 and 10, inclusive, what is the probability that m is greater than zero?

A. 2/7 B. 1/3 C. 7/20 D. 2/5 E. 3/7

x^2 + 2x - 15 =-m or x^2 + 2x + (m-15) = 0 or (x+5)(x-3) = -m Now since m is an integer so x must also be an integer. x=1 m= 12 NO x=2 m= 7 YES x=3 m= 0 YES x=4 m= -9 YES x=5 m= -20 NO No further values will satisfy! So m has 3 values so far! x= 0 m= 15 NO x= -1 m= 16 NO x= -2 m= 15 NO x= -3 m= 12 NO x= -4 m= 7 YES x= -5 m= 0 YES x= -6 m= -9 YES x= -7 m= -20 NO

SO m can take 3 values 7, 0 and -9 and only ONE of them is greater than 0 SO the probability is 1/3 B _________________

Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 12:54

Expert's post

Hey Bunuel! I touched quant after 3 months and I am really out of shape and its 2:30 in the morning! So If i answer incorrectly its partly my brain's fault! Oh well, when it isn't! _________________

Re: New Algebra Set!!! [#permalink]
19 Mar 2013, 00:15

souvik101990 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

differentiating with respect to x and equating with 0

\(-6x + 12 = 0, OR x=2\)

differentiating with respect to y and equating with 0

\(-4y - 12 =0, OR y=-3\)

So max value of the expression \(-3x^2 + 12x -2y^2 - 12y - 39\) is at (2, -3) i.e -12+24-18+36-39 = -9 B

Hi Souvik I liked ur approach cn u pls expand on it...i mean is there any hard and fast rule that u r following.....completely missed this question....

Re: New Algebra Set!!! [#permalink]
19 Mar 2013, 00:48

1

This post received KUDOS

Expert's post

4. -3x^2 + 12x -2y^2 - 12y - 39 = -3x^2+12x-12-2y^2-12y-18-39+18+12 = -3(x-2)^2-2(y+3)^2 -9. Now a negative sign is attached to both the squares. Thus the maximum value is at x=2 and y=-3 and equals -9.

Re: New Algebra Set!!! [#permalink]
19 Mar 2013, 00:57

Expert's post

Archit143 wrote:

souvik101990 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

differentiating with respect to x and equating with 0

\(-6x + 12 = 0, OR x=2\)

differentiating with respect to y and equating with 0

\(-4y - 12 =0, OR y=-3\)

So max value of the expression \(-3x^2 + 12x -2y^2 - 12y - 39\) is at (2, -3) i.e -12+24-18+36-39 = -9 B

Hi Souvik I liked ur approach cn u pls expand on it...i mean is there any hard and fast rule that u r following.....completely missed this question....

Archit

I used calculus maxima minima. However you are free to use the "completing the square" method where you are left with 2 perfect squares and a -9 as the guy above me did. _________________

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