Question

The roots of the equation x^3+px^2+qx+r=0 are kb, b, b/k. where all alphabet, except x, are non-zero real constants. Prove that b=-q/p, hence deduce that rp^3=q^3

Answer 1

some one help me please!

Answer 2

\[\huge x^3+Px^2 +Qx +R=0~and~kb,~b,~\frac{b}{k}~are~roots\\therefore\]
\[\huge kb+b+\frac{b}{k}=P\\\huge kb*b+kb*\frac{b}{k}+b*\frac{b}{k}=Q\\\huge kb*b*\frac{b}{k}=R\]

Answer 3

from the first two, you solve for b, and then combine with the last one to get the answer for the last part.
Hope this help

Answer 4

thanks!!!!! you saved my life!

Answer 5

yw