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If r = (3p + q)/2 and s = p q, for which of the following

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If r = (3p + q)/2 and s = p q, for which of the following [#permalink] New post 26 Oct 2007, 13:00
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If r = (3p + q)/2 and s = p – q, for which of the following values of p would r^2 = s^2 ?

a) q/5
b) 10 - (3/2q)
c) q - 1
d) 3q
e) (9/2q) - 9
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 [#permalink] New post 26 Oct 2007, 13:19
(A) for me :)

r^2 = s^2
<=> ((3p + q)/2)^2 = (p – q)^2
<=> 9*p^2 + 6*p*q + q^2 = 4*p^2 - 8*p*q + 4*q^2
<=> 5*p^2 + 14*p*q - 3*q^2 = 0

From here, I set p = k*q where k is a real number to obtain an expression such that (something) * q^2 = 0.

5*k^2*q^2 + 14*k*q^2 - 3*q^2 = 0
<=> (5*k^2 + 14*k - 3)*q^2 = 0

=> 5*k^2 + 14*k - 3 = 0
-3 is a root...

So we have,
5*(k+3)*(k-1/5) = 0

Finally,
p could be -3*q or q/5
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 [#permalink] New post 26 Oct 2007, 13:26
the answer is (A)

((3p + q)/2)^2 = (p - q)(p - q)

(3p + q)(3p + q) = 4(p - q)(p - q)

-----------------------------------------

p = q/5

(3*q/5 + q) = 8q/5

(q/5 - q) = 4q/5

-----------------------------------------

(8q/5)*(8q/5) = 4*(4q/5*4q/5)

64q/5 = 64q/5

:)
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 [#permalink] New post 26 Oct 2007, 13:33
Other way...

r^2 = s^2
<=> ((3p + q)/2)^2 = (p – q)^2
<=> ((3p + q)/2)^2 - (p – q)^2 = 0
<=> ((3p+q)/2 - (p-q)) * ((3p+q)/2 + (p-q)) = 0
<=> (p/2 + 3*q/2) * (5*p/2 - q/2) = 0
<=> (p + 3*q)*(5*p - q) = 0
<=> p = -3q or p = q/5

:)
  [#permalink] 26 Oct 2007, 13:33
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If r = (3p + q)/2 and s = p q, for which of the following

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