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# Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c =

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Math Expert
Joined: 02 Sep 2009
Posts: 43831
Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c = [#permalink]

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06 Nov 2014, 08:23
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Difficulty:

25% (medium)

Question Stats:

76% (01:24) correct 24% (01:36) wrong based on 188 sessions

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Tough and Tricky questions: Algebra.

Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c =

A. 18
B. 15
C. -15
D. -21
E. -24

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c = [#permalink]

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06 Nov 2014, 08:38
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f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c =

f(6) = 0= 36+6b+c--- Taking 36 to the other side

-> 6b+c= -36

f(-3) = 0= 9-3b+c --- taking -3b+c to the other side

-> 3b-c=9

When we add these 2 equations, we get 9b= -27---> b= -3

And while substituting b= -3 we get c= -18.

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Re: Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c = [#permalink]

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10 Nov 2014, 00:37
1
KUDOS
$$f(x) = x^2 + bx + c$$

f(6) = 36 + 6b + c = 0

f(-3) = 9 - 3b + c = 0

c = -18; b = -3

b+c = -18-3 = -21

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Director
Joined: 23 Jan 2013
Posts: 601
Schools: Cambridge'16
Re: Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c = [#permalink]

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17 Aug 2015, 05:45
6 and -3 are roots of quadratic equation x^2+bx+c=0

c=6*(-3)= -18
b=6+(-3)=3*-1=-3
b+c=-18+(-3)=-21

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Re: Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c = [#permalink]

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02 Dec 2017, 23:01
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c =   [#permalink] 02 Dec 2017, 23:01
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