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Director
Joined: 16 May 2014
Posts: 590
Given Kudos: 0
GRE 1: Q165 V161
GRE Math Challenge #60- If f(x) = 2x - 1 and g(x) = x^2
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28 Feb 2015, 02:40
28 Feb 2015, 02:40
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Question Stats:
58% (02:15) correct
41% (01:45) wrong based on 92 sessions
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If \(f(x) = 2x - 1\) and \(g(x) = x^2\), then what is the product of all values of n for which \(f(n^2)=g(n+12)\) ?
A. -145
B. -24
C. 24
D. 145
E. None of the above
A. -145
B. -24
C. 24
D. 145
E. None of the above
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Posts: 6218
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Re: GRE Math Challenge #60
[#permalink]
16 May 2015, 15:27
16 May 2015, 15:27
1
soumya1989 wrote:
If \(f(x) = 2x - 1\) and \(g(x) = x^2\), then what is the product of all values of n for which \(f(n^2)=g(n+12)\) ?
A. -145
B. -24
C. 24
D. 145
E. None of the above
A. -145
B. -24
C. 24
D. 145
E. None of the above
Let's plug n² and n+12 into their respective functions and see what we get.
f(n²) = 2(n²) - 1 = 2n² - 1
g(n+12) = (n+12)² = n² + 24n +144
We want f(n²) = g(n+12)
So, we have 2n² - 1 = n² + 24n +144
Subtract n² from both sides to get: n² - 1 = 24n +144
Subtract 24n from both sides to get: n² - 24n - 1 = 144
Subtract 144 from both sides to get: n² - 24n - 145 = 0
Factor: (n + 5)(n - 29) = 0
So, we have two possible solutions: n = -5 and n = 29
We want the PRODUCT of all possible solutions.
So, (-5)(29) = -145
Answer:
Show: ::
A
Cheers,
Brent
