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

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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

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[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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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.

b+c= -21--- Answer D
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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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New post 10 Nov 2014, 00:37
\(f(x) = x^2 + bx + c\)

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

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

c = -18; b = -3

b+c = -18-3 = -21

Answer = D
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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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New post 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

D

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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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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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