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M02-13

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Bunuel
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M02-13 [#permalink] New post   16 Sep 2014, 00:17
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If \(y = \sqrt{64}\) and \(x^2 - 10x = (-4y^3 + 64y)*\frac{1}{96}\), what is the minimum possible value of \(x\)?

A. 2
B. 4
C. 8
D. 12
E. 16
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A
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Bunuel
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Re: M02-13 [#permalink] New post   05 Apr 2015, 05:31
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healthjunkie wrote:
Couldn't Y also equal -8? I know that when you plug in y=-8 you end up getting x^2 - 10x -16 = 0 which you can't factor out, but I didnt realize that until after I plugged in -8 (I plugged in -8 since we were trying to find the smallest value of X, I figured a negative Y would give us smaller values for X)


No.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59


Hope it helps.
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Re M02-13 [#permalink] New post   16 Sep 2014, 00:17
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Official Solution:

If \(y = \sqrt{64}\) and \(x^2 - 10x = (-4y^3 + 64y)*\frac{1}{96}\), what is the minimum possible value of \(x\)?

A. 2
B. 4
C. 8
D. 12
E. 16


\(y = \sqrt{64}=8\) (recall that \(√\) symbol always means non-negative root). Plug this into the second equation:

\(x^2-10x = \frac{-4*8^3 + 64*8}{96} \)

\(x^2-10x = \frac{-4*8^3 + 8^3}{96}\)

\(x^2-10x = \frac{-3*8^3}{8*4*3}\)

\(x^2-10x = -16\)

\(x^2-10x +16 = 0\)

\((x-8)(x-2)=0\)

The possible values of \(x\) are 8 and 2. Therefore the minimum value of \(x\) is 2.


Answer: A
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Re: M02-13 [#permalink] New post   Updated on: 22 Apr 2015, 20:38
Couldn't Y also equal -8? I know that when you plug in y=-8 you end up getting x^2 - 10x -16 = 0 which you can't factor out, but I didnt realize that until after I plugged in -8 (I plugged in -8 since we were trying to find the smallest value of X, I figured a negative Y would give us smaller values for X)

Originally posted by healthjunkie on 04 Apr 2015, 11:52.
Last edited by healthjunkie on 22 Apr 2015, 20:38, edited 1 time in total.
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Re: M02-13 [#permalink] New post   09 Aug 2015, 07:08
Bunuel wrote:
healthjunkie wrote:
Couldn't Y also equal -8? I know that when you plug in y=-8 you end up getting x^2 - 10x -16 = 0 which you can't factor out, but I didnt realize that until after I plugged in -8 (I plugged in -8 since we were trying to find the smallest value of X, I figured a negative Y would give us smaller values for X)


No.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59


Hope it helps.


Hi Bunuel ,

Please correct me if i amwrong

Does this mean when gmat gives sqare root sign itself in question stem , then it means we should consider only positive root
and on the other hand gmat gave me a quadratic eqaution in question stem and when i solve it i should consider both +ve and - ve root ?

right ?
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Re: M02-13 [#permalink] New post   17 Aug 2015, 03:27
Expert Reply
adityadon wrote:
Bunuel wrote:
healthjunkie wrote:
Couldn't Y also equal -8? I know that when you plug in y=-8 you end up getting x^2 - 10x -16 = 0 which you can't factor out, but I didnt realize that until after I plugged in -8 (I plugged in -8 since we were trying to find the smallest value of X, I figured a negative Y would give us smaller values for X)


No.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59


Hope it helps.


Hi Bunuel ,

Please correct me if i amwrong

Does this mean when gmat gives sqare root sign itself in question stem , then it means we should consider only positive root
and on the other hand gmat gave me a quadratic eqaution in question stem and when i solve it i should consider both +ve and - ve root ?

right ?


\(\sqrt{}\) sign always means non-negative root.

In contrast x^2 = 4 gives two solutions: 2 and -2.
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Re: M02-13 [#permalink] New post   10 Sep 2016, 21:29
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I think this is a high-quality question and I agree with explanation.
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Re: M02-13 [#permalink] New post   23 Mar 2017, 17:06
Bunuel wrote:
Official Solution:

\(x\) and \(y\) are positive integers. If \(y = \sqrt{64}\) and \(x^2 - 10x = (-4y^3 + 64y)*\frac{1}{96}\), what is the minimum possible value of \(x\)?

A. 2
B. 4
C. 8
D. 12
E. 16


We know that \(y = 8\).

Plugging this into the second equation gives:
\(x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16\)
\(x^2-10x = -16\)
\(x^2-10x +16 = 0\)
\((x-8)(x-2)=0\)

The possible values of \(x\) are 8 and 2. Therefore the minimum value of \(x\) is 2.


Answer: A


How did you get The numerator to -3*8^3? ->>> \(x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16\)
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Re: M02-13 [#permalink] New post   24 Mar 2017, 04:55
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Prostar wrote:
Bunuel wrote:
Official Solution:

\(x\) and \(y\) are positive integers. If \(y = \sqrt{64}\) and \(x^2 - 10x = (-4y^3 + 64y)*\frac{1}{96}\), what is the minimum possible value of \(x\)?

A. 2
B. 4
C. 8
D. 12
E. 16


We know that \(y = 8\).

Plugging this into the second equation gives:
\(x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16\)
\(x^2-10x = -16\)
\(x^2-10x +16 = 0\)
\((x-8)(x-2)=0\)

The possible values of \(x\) are 8 and 2. Therefore the minimum value of \(x\) is 2.


Answer: A


How did you get The numerator to -3*8^3? ->>> \(x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16\)


It's basic manipulation: \(-4*8^3 + 8^3 = 8^3(-4+1) = 3*8^3\). The same way -4x + x = -3x.
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Re: M02-13 [#permalink] New post   05 Mar 2023, 13:59
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M02-13 [#permalink] New post   20 Apr 2024, 02:03
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Re: M02-13 [#permalink]
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