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In the equation ax^2 + bx + c = 0 a, b, and c are constants,

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In the equation ax^2 + bx + c = 0 a, b, and c are constants, [#permalink] New post 05 Mar 2013, 10:32
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In the equation ax^2 + bx + c = 0 a, b, and c are constants, and abc # 0. If one root of the equation is -2, and b = 8a then which of the following is c equal to?

A. a/12
B. a/8
C. 6a
D. 8a
E. 12a
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Mar 2013, 00:44, edited 1 time in total.
Renamed the topic and edited the question.
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Re: In the equation ax^2 + bx + c = 0 [#permalink] New post 05 Mar 2013, 11:57
8a=b a=1 to make things simple
x^2+8x+c=0
given one of the roots are -2 we can factor and find the other root (x+2)(x+6)
c=12
c/a=12/1
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Re: In the equation ax^2 + bx + c = 0 [#permalink] New post 05 Mar 2013, 21:03
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alexpavlos wrote:
In the equation \(ax^2 + bx + c = 0\) a, b, and c are constants, and abc # 0. If one root of the equation is -2, and b = 8a then which of the following is c equal to?

a) a/12
b) a/8
c) 6a
d) 8a
e) 12a


The sum of the roots is = -b/a = -8a/a = -8. Let the other root be x. Thus, x-2 = -8

x = -6. Again, the product of the roots is -2*-6 = 12. Thus, c/a = 12. c = 12a.

E.
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Re: In the equation ax^2 + bx + c = 0 a, b, and c are constants, [#permalink] New post 06 Mar 2013, 01:09
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alexpavlos wrote:
In the equation ax^2 + bx + c = 0 a, b, and c are constants, and abc # 0. If one root of the equation is -2, and b = 8a then which of the following is c equal to?

A. a/12
B. a/8
C. 6a
D. 8a
E. 12a


Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).


Thus according to the above \(x_1+x_2=-2+x_2=\frac{-b}{a}=\frac{-8a}{a}=-8\) --> \(x_2=-6\).

Also, \(x_1*x_2=-2*(-6)=\frac{c}{a}\) --> \(c=12a\).

Answer: E.

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Hope it helps.
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Re: In the equation ax^2 + bx + c = 0 a, b, and c are constants,   [#permalink] 06 Mar 2013, 01:09
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In the equation ax^2 + bx + c = 0 a, b, and c are constants,

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