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factorization

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factorization

by ddm » Tue Aug 19, 2008 9:27 pm
f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the number for which f(n) = 0, what is the value of n?


A)6
B)2
C)0
D)-2
E)-12
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Re: factorization

by sudhir3127 » Tue Aug 19, 2008 10:03 pm
ddm wrote:f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the number for which f(n) = 0, what is the value of n?


A)6
B)2
C)0
D)-2
E)-12
IMO B.2
Last edited by sudhir3127 on Wed Aug 20, 2008 5:12 am, edited 1 time in total.
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by parallel_chase » Wed Aug 20, 2008 5:05 am
I think its B.

f(x) = 12

f(-6) = 0

k = 0

f(2) = 0


Hence B.

Whats the OA?
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Re: factorization

by Stockmoose16 » Wed Aug 20, 2008 5:20 pm
ddm wrote:f(x) = x^2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the number for which f(n) = 0, what is the value of n?


A)6
B)2
C)0
D)-2
E)-12
Can someone rephrase this question in layman's terms? What's n got to do with anything? It's not even part of the original equation.
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by Stockmoose16 » Wed Aug 20, 2008 5:27 pm
parallel_chase wrote:I think its B.

f(x) = 12

f(-6) = 0

k = 0

f(2) = 0


Hence B.

Whats the OA?
How did you get F(2)=0, if you plug into the above equation, F(2)=12+k=12
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by parallel_chase » Wed Aug 20, 2008 5:43 pm
Here is the entire solution.

f(x)= x^2 + 4x + k =12

f(-6) = 36 -24 +k = 12

12 + k =12

k=0

f(n) = 0

f(2) = 4 + 8 + k = 12

12 + k = 12

f(2) = 12+0=12
f(2) =12-12=0
f(2) =0


Hence B is the answer. Let me know if you still have any doubts.
Last edited by parallel_chase on Wed Aug 20, 2008 5:51 pm, edited 1 time in total.
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by nervesofsteel » Wed Aug 20, 2008 5:44 pm
Can someone explain the solution in detail...

Thanks
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by nervesofsteel » Wed Aug 20, 2008 10:53 pm
sorry for last post .. didn't see the detailed solution
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