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In the equation ax^2 + bx + c = 0 a, b, and c are constants, [#permalink]
05 Mar 2013, 10:32

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Question Stats:

80% (02:05) correct
20% (01:10) wrong based on 80 sessions

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?

Re: In the equation ax^2 + bx + c = 0 [#permalink]
05 Mar 2013, 21:03

Expert's post

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.

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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This post received KUDOS

Expert's post

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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 \(x_1+x_2=-2+x_2=\frac{-b}{a}=\frac{-8a}{a}=-8\) --> \(x_2=-6\).

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