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

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Senior Manager
Joined: 30 Aug 2003
Posts: 330
Location: BACARDIVILLE
Followers: 1

Kudos [?]: 1 [0], given: 0

Find 'k' in the equation 2x^2 + kx + 32 = 0 if the [#permalink]  30 Jan 2004, 09:41
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
_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

SVP
Joined: 30 Oct 2003
Posts: 1794
Location: NewJersey USA
Followers: 5

Kudos [?]: 43 [0], given: 0

k^2 = 496
I guess none of these.

I got
sqrt(k^2-2*32*4)/4 = 60
k^2 = 240+256 = 496
Director
Joined: 03 Jul 2003
Posts: 654
Followers: 2

Kudos [?]: 32 [0], given: 0

Good question [#permalink]  30 Jan 2004, 10:57
Hi Sunni

I read in GMAC that they are looking for someone who can
are very well fit for the post

Back sloving helps to find the answer, but is there a way to
solve this!
SVP
Joined: 30 Oct 2003
Posts: 1794
Location: NewJersey USA
Followers: 5

Kudos [?]: 43 [0], given: 0

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
SVP
Joined: 30 Oct 2003
Posts: 1794
Location: NewJersey USA
Followers: 5

Kudos [?]: 43 [0], given: 0

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.
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