substitution

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by albatross86 » Tue Jun 29, 2010 4:44 am
r^2 = s^2

=> (3p + q)^2 / 4 = (p-q)^2

=> 9p^2 + 6pq +q^2 = 4p^2 - 8pq + 4q^2

=> 5p^2 +14pq - 3q^2 = 0

=> 5p^2 + 15pq - pq - 3q^2 = 0

=> 5p(p+3q) - q(p-3q) = 0

=> (5p - q)*(p-3q) = 0

=> p = q/5 OR p = 3q

Pick D

I'm not sure what A is, but I think you may have mistyped it?

EDIT: The equation should be (5p - q)*(p + 3q) MY BAD! Thanks selango!
Last edited by albatross86 on Tue Jun 29, 2010 4:51 am, edited 1 time in total.
~Abhay

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by selango » Tue Jun 29, 2010 4:49 am
albatross86,

the equation is (5p-q)* (p+3q)=0

p=q/5 and p=-3q

Hence A

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by albatross86 » Tue Jun 29, 2010 4:51 am
selango wrote:albatross86,

the equation is (5p-q)* (p+3q)=0

p=q/5 and p=-3q

Hence A
You're right! Thanks!
~Abhay

Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide