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Quad - Find the relationship between roots

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by gmatrant » Wed Jun 23, 2010 9:38 am
If the ratio of roots of the equation x^2 +px+ q =0 be equal to the
ratio of roots of equation x^2 +bx+c then
1)p^2 q =b^2 c
2)p^2 c =b^2 q
3)p^3 c =b^2 q
4)p^2 c =b^3 q
5)p^3 c =b^3 q

Can you let me know how to solve this without back substitution of values.

OA is B
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by outreach » Wed Jun 23, 2010 10:09 am
lengthy solution
Let the roots of x^2+px+q=0 be A and B

Let the roots of x^2+bx+c=0 be C and D

B satisfies x^2+px+q=0. Therefore B^2+pB+q=0 or B^2=-(pB+q)

D satisfies x^2+bx+c=0. Therefore D^2+pD+q=0 or D^2=-(bD+c)

A+B=-p and AB=q

C+D=-b and CD=c

A/B=C/D(given)

(A/B)+1=(C/D)+1

(A+ B)/B=(C+ D)/D

=> -p/B=-b/D

=> B/p= D/b

Squaring both sides,
B^2/p^2= D^2/b^2

-(pB+q)/p^2= -(bD+c)/b^2

-(B/p)-(q/p^2)= -(D/b)-(c/b^2)

(q/p^2)=(c/b^2)

p^2c= b^2q
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by odod » Thu Jun 24, 2010 4:43 pm
insane...does anyone have a way that won't take me 2 hours?
ODOD
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