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Roots of Quadratic

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Junior | Next Rank: 30 Posts
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by ontherocks27 » Sun May 31, 2009 7:07 pm
For a quadratic equation ax^2+bx+c = 0

Sum of roots
r1 + r2 = (-b)/a

Product of roots
r1r2 = c/a

For the given equation

r1+r2 = (- (-1))/1 = 1
r1r2 = 1/1004

Now, r1^2 +r2^2 = (r1+r2)^2 - 2r1r2 = 1 - 2(1/1004) = 1-(1/502) = 501/502

Hence B
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sum of roots

by drian » Sun May 31, 2009 8:24 pm
Please help, unsure as to - 2r1r2 part of equation,

Thanks

[quote="ontherocks27"]For a quadratic equation ax^2+bx+c = 0

Sum of roots
r1 + r2 = (-b)/a

Product of roots
r1r2 = c/a

For the given equation

r1+r2 = (- (-1))/1 = 1
r1r2 = 1/1004

Now, r1^2 +r2^2 = (r1+r2)^2 - 2r1r2 = 1 - 2(1/1004) = 1-(1/502) = 501/502

Hence B[/quote]
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Junior | Next Rank: 30 Posts
Posts: 16
Joined: Mon Sep 01, 2008 8:13 pm
Thanked: 3 times
GMAT Score:700

by ontherocks27 » Tue Jun 02, 2009 8:07 am
We know the identity relation:

(r1+r2)^2 = r1^2 + r2^2 +2r1r2

Manipulating the above relation we get:

=> r1^2 + r2^2 = (r1+r2)^2 - 2r1r2

Hope that helps.
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