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Re: GRE Math Challenge #57-If x^4 = 29x^2 - 100 [#permalink]
1
soumya1989 wrote:
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 & II only
E. I & III only



Given: x⁴ = 29x² - 100
Rewrite as: x⁴ - 29x² + 100 = 0
Factor to get: (x² - 25)(x² - 4) = 0
Factor more to get: (x + 5)(x - 5)(x + 2)(x - 2) = 0
So, x = -5, 5, -2 or 2

(-5)(5)(2) = -50, so statement I is possible - ELIMINATE A, D, E
(-5)(5)(-2) = 50, so statement III is possible - ELIMINATE C

Answer: B

In other words, it is NOT possible to get a product of 25 using 3 of the possible x-values, the correct answer is B

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GRE Math Challenge #57-If x^4 = 29x^2 - 100 [#permalink]
1
Solution:

Substituting\( x^2\) as y
\(y^2\)=29y-100
i.e. \(y^2\)-29y+100=0

Factorizing above gives us;
(y-4)(y-25)=0
y=4 or y=25

Resubstituting y as \(x^2\)
\(x^2\)= 4 & 25
x = -2, +2, -5 & +5
Only B is not a possible combination. As we don not have two positive 5's.

IMO B

Hope this helps!
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GRE Math Challenge #57-If x^4 = 29x^2 - 100 [#permalink]
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