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M28-13

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Bunuel
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M28-13 [#permalink] New post   16 Sep 2014, 01:28
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35% (medium)

Question Stats:

76% (01:53) correct 24% (02:05) wrong based on 206 sessions
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If \(x^4 = 29x^2 - 100\), then which of the following is NOT a product of three possible values of \(x\)?

I. -50

II. 25

III. 50


A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(II\) only
E. \(I\) and \(III\) only
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B
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Bunuel
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Re M28-13 [#permalink] New post   16 Sep 2014, 01:28
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Official Solution:


If \(x^4 = 29x^2 - 100\), then which of the following is NOT a product of three possible values of \(x\)?

I. -50

II. 25

III. 50


A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(II\) only
E. \(I\) and \(III\) only


Rearrange and factor for \(x^2\): \((x^2-25)(x^2-4)=0\).

Thus, we find that \(x=5\), \(x=-5\), \(x=2\), or \(x=-2\).

For the given options:

I. \(-50 = 5*(-5)*2\);

III. \(50 = 5*(-5)*(-2)\);

However, 25 cannot be expressed as a product of three of these values. Thus, only 25 is NOT a product of three possible values of \(x\).


Answer: B
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Viktorijakm
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Re: M28-13 [#permalink] New post   17 Feb 2015, 13:25
I think this question is poor.
it is 29 in the question ,but factoring shows 25!!!
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Re: M28-13 [#permalink] New post   17 Feb 2015, 13:29
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Viktorijakm wrote:
I think this question is poor.
it is 29 in the question ,but factoring shows 25!!!


You are wrong.

Did you try to expand (x^2 - 25)(x^2 - 4) = 0? You'll get x^4 - 29x^2 + 100 = 0, which is the same as x^4 = 29x^2 - 100.
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culsivaji
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Re: M28-13 [#permalink] New post   22 Aug 2017, 12:25
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I think this is a high-quality question and I agree with explanation.
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Arro44
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Re: M28-13 [#permalink] New post   06 Jul 2018, 09:23
Viktorijakm wrote:
I think this question is poor.
it is 29 in the question ,but factoring shows 25!!!


The factoring is correct though, as you are looking for numbers which add to -29 and multiply to 100 respectively.

I think the question is of high quality and similar to stuff I have seen on the GMAC prep tests.
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ram1annepu
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Re: M28-13 [#permalink] New post   28 Aug 2018, 08:17
I think this is a high-quality question.
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RenanBragion
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Re: M28-13 [#permalink] New post   12 Jun 2020, 03:50
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Great question

An alternative approach:

\(x^4 - 29x^2 + 100 = 0\)

Create a variable \( y = x^2 \)
\(x^4 - 29x^2 + 100 \)
\(y^2-29y^2+100 = 0\)

y = 25 or y = 4

Substitute the values of y in the equation \( y = x^2 \)
\(x = y^2\)

Find X:
If y = 25 then x = 5 or x = -5
If y = 4 then x = 2 or x = -2
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Bunuel
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Re: M28-13 [#permalink] New post   14 Apr 2023, 08:38
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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: M28-13 [#permalink]
14 Apr 2023, 08:38
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