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Find 'k' in the equation 2x^2 + kx + 32 = 0 if the

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Find 'k' in the equation 2x^2 + kx + 32 = 0 if the [#permalink] New post 30 Jan 2004, 10:41
Find 'k' in the equation 2x^2 + kx + 32 = 0 if the difference of the squares of the root is 60.

a. 10
b. 20
c. 16
d. 24
e. none of these
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Pls include reasoning along with all answer posts.
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 [#permalink] New post 30 Jan 2004, 11:34
k^2 = 496
I guess none of these.

I got
sqrt(k^2-2*32*4)/4 = 60
k^2 = 240+256 = 496
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Good question [#permalink] New post 30 Jan 2004, 11:57
Hi Sunni

I read in GMAC that they are looking for someone who can
come up with deadly questions. Your questions prove that you
are very well fit for the post :lol:

Back sloving helps to find the answer, but is there a way to
solve this!
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 [#permalink] New post 30 Jan 2004, 14:45
20 is the answer
I was stupid to solve the equation instead of substituting for K

if you make k=20 then
(-20+12)/4 is one root = -2
(-20-12)/4 is another root = -8
so 8^2-2^2 = 60
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 [#permalink] New post 30 Jan 2004, 15:05
Detailed solution

r1 = (-K + sqrt( k^2-256 ) )/4
r2 = (-K - sqrt( k^2-256 ) )/4

for 10 the roots are imaginary
for 16 the difference is 0
only 24 and 20 remain.
substitute 20 you will get the answer.
  [#permalink] 30 Jan 2004, 15:05
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