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For what value of 'k' will the sum of the equation 27x^2 +

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For what value of 'k' will the sum of the equation 27x^2 + [#permalink] New post 30 Jan 2004, 14:42
For what value of 'k' will the sum of the equation
27x^2 + (3k + 9)x - 43 + k = 0 be equal to the product of the roots?

a. 17/2
b. -9/2
c. -17/2
d. :dunnow
e. none of these
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 [#permalink] New post 30 Jan 2004, 14:46
sum of the roots = -(3k+9)/27
product of the roots = (k-43)/27

4k = 43-9 = 34
k = 17/2
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 [#permalink] New post 30 Jan 2004, 15:40
sunniboy...love your avatars...but not your problems...it makes me loose confidence :dunnow

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 [#permalink] New post 30 Jan 2004, 15:57
I guess sunniboy will get 100% of the votes polled. 8-) :-D
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 [#permalink] New post 30 Jan 2004, 16:19
Vivek,
DOn't lose confidence.
The solution is as follows:

Let x1 and x2 be the respective roots.
So, x1 + x2 = x1*x2....This is a given
When we have a ax^2 + bx + c = 0
=>Sum of the roots is equal to -b/a
=>The product of the roots is qual to c/a
so, in 27x^2 + (3k+9)x - 43 + k = 0
a = 27
b = 3k+9
c = -43 + k

(1)Then, -d/a must = -3k + 9/27
(2)and, c/a must = -43 + k/27

Set these two equations equal to each other and solve for 'k'
That will be your answer.
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 [#permalink] New post 30 Jan 2004, 16:27
If a and b are the roots of equation
then that euqation is (x-a)(x-b) = x^2+x(a+b)-ab = 0

Compare it to the given equation we have

27x^2 + (3k + 9)x - 43 + k = 0
or
x^2+x*(3k+9)/27-(43-k)/27 = 0
We have
sum of roots (a+b) = products of roots (ab)
(a+b) = (3k+9)/27 = ab = (43-k)/27
4k = 34 or k = 17/2
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 [#permalink] New post 30 Jan 2004, 17:14
Thanks buddy ....appreciate your help n encouragement.

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  [#permalink] 30 Jan 2004, 17:14
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