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A,B are roots of equation (x)(x)-px+q=0. A,C are roots of

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A,B are roots of equation (x)(x)-px+q=0. A,C are roots of [#permalink] New post 30 Oct 2004, 12:11
A,B are roots of equation (x)(x)-px+q=0. A,C are roots of equation (x)(x)-qx+p=0. Find the equation whose roots are B and C.
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 [#permalink] New post 30 Oct 2004, 20:33
This may not be a GMAT type question. However, here's how to solve.
Note: quadratic equation can be represented as a1x+b1x+c1 = 0

The common root between the two quadratic is given as A.

Common root (here A) is given by (c1a2-c2a1)/(a1b2-a2b1) = (q-p)/(-q+p) = -1.

From equations we have
AB = q and AC = p (Note: sum of roots = -b/a. Product of roots = c/a)
=>B=-q (since A=-1) and C = -p.
We have the roots B and C.

Also, in any quadratic the euation is essentially

x^2 -(sum of roots)x + product of roots = 0

sum of roots = -b/a. Product of roots = c/a

So, the new quadratic equation is
x^2 -(p+q)x +pq = 0.
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 [#permalink] New post 31 Oct 2004, 02:38
answer should be:
x^2 + (p+q)x + pq

I did this one by solving two equations on A.
A = -1 is correct
so
B = -q and C = -p => B + C = -p-q = -(p+q) & BC = pq

So equations with B anc C as a roots is:

x^2 - (B+C)x + BC = x^2 + (p+q)x + pq

cheers,
Dharmin
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 [#permalink] New post 31 Oct 2004, 03:41
Dharmin wrote:
answer should be: x^2 + (p+q)x + pq

Agree. Silly mistake from my end.
  [#permalink] 31 Oct 2004, 03:41
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