SOLUTION: If the ratio of the roots of the equation x^2+bx+c=0 is as that of x^2+qx+r=0,then a)r^2b=qc^2 b)r^2c=qb^2 c)c^2r=q^2b d)b^2r=q^2c

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Question 973011: If the ratio of the roots of the equation x^2+bx+c=0 is as that of x^2+qx+r=0,then
a)r^2b=qc^2
b)r^2c=qb^2
c)c^2r=q^2b
d)b^2r=q^2c
Answer by anand429(138) About Me  (Show Source):
You can put this solution on YOUR website!
Let the roots of x^2+bx+c=0 be g and h, and
Let the roots of x^2+qx+r=0 be m and n
Let ratio of roots be given by g/h = R1 and m/n = R2
Now, g+h = -b ---(i)
and gh =c ---(ii)
Similarly,
m+n = -q ---(iii)
mn= r ----(iv)
Squaring eqn (i),
g%5E2%2Bh%5E2%2B2gh+=+b%5E2
=> g%5E2%2Bh%5E2+=+b%5E2-2c (using eqn. (ii)-----(v)
Divinding(v) by (ii)
%28g%5E2%2Bh%5E2%29%2Fgh+=+%28b%5E2-2c%29%2Fc
=> %28g%2Fh%29%2B%28h%2Fg%29+=+%28b%5E2-2c%29%2Fc
=> R1%2B1%2FR1+=%28b%5E2-2c%29%2Fc
Similarly,
R2%2B1%2FR2+=%28q%5E2-2r%29%2Fr
Since R1=R2 (as per ques.),
So, R1%2B1%2FR1+=+R2%2B1%2FR2
=> %28b%5E2-2c%29%2Fc=%28q%5E2-2r%29%2Fr
=> b%5E2r-2cr+=+q%5E2c-2cr
=> b%5E2r+=+q%5E2c
Hence option (iv) is correct.