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In the equation ax^2 + bx + c = 0 a, b, and c are constants,
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Updated on: 06 Mar 2013, 00:44
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79% (01:39) correct 21% (02:09) wrong based on 145 sessions
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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?
Re: In the equation ax^2 + bx + c = 0
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05 Mar 2013, 21:03
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,
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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 \(x_1+x_2=-2+x_2=\frac{-b}{a}=\frac{-8a}{a}=-8\) --> \(x_2=-6\).
Re: In the equation ax^2 + bx + c = 0 a, b, and c are constants,
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14 Jul 2018, 18:40
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alex1233 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
We can substitute -2 for x and 8a for b to solve for c:
a(-2)^2 + (8a)(-2) + c = 0
4a - 16a + c = 0
-12a + c = 0
c = 12a
Answer: E
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79% (01:39) correct 
