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E
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Difficulty:
25%
(medium)
Question Stats:
79% (01:51) correct
21%
(02:18)
wrong
based on 396
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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
A. a/12
B. a/8
C. 6a
D. 8a
E. 12a
06 Mar 2013, 02:09
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alexpavlos
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.
General Discussion
05 Mar 2013, 12:57
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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
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
