Roots of Quadratic

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dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

by dtweah » Sun May 31, 2009 2:56 pm

What is the sum of the square of the roots of the quadratic equation X^2 -x +1/1004 =0

A. 405/502

B. 501/502

C. 400/502

D. 3/4

E. 1/2

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Source: Beat The GMAT — Problem Solving | Sun May 31, 2009 2:56 pm

ontherocks27: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Mon Sep 01, 2008 8:13 pm; Thanked: 3 times; GMAT Score:700

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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drian: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Sat May 30, 2009 8:33 pm

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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ontherocks27: 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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