GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Director
Joined: 16 May 2014
Posts: 592
Given Kudos: 0
GRE 1: Q165 V161
GRE Math Challenge #60- If f(x) = 2x - 1 and g(x) = x^2
[#permalink]
28 Feb 2015, 02:40
28 Feb 2015, 02:40
Expert Reply
5
Bookmarks
00:00
Question Stats:
58% (02:12) correct
41% (01:46) wrong based on 84 sessions
Hide Show timer Statistics
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
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
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
