Question

If sum of roots of the quadratic equation ax2 + bx + c = 0 is equal to the sum of the cubes of their reciprocals, then prove that ab2 = 3a2c + c3

Answer 1

let the roots be α and β

Answer 2

$\boxed{\underline{\textsf{Solution :}}}$

$\textsf{The given quadratic equation is}$

$\bold{ax^{2}+bx+c=0}\:\textsf{---(i)}$

$\textsf{If p and q be the roots of (i)}$

$\bold{p+q=-\frac{b}{a}}$

$\&\:\bold{pq=\frac{c}{a}}$

$\textsf{Now, given that}$

$\bold{p+q=\frac{1}{p^{3}}+\frac{1}{q^{3}}}$

$\to \bold{p+q=\frac{p^{3}+q^{3}}{p^{3}q^{3}}}$

$\to \bold{p+q=\frac{(p+q)^{3}-3pq(p+q)}{(pq)^{3}}}$

$\to \bold{1 = \frac{(p+q)^{2}-3pq}{(pq)^{3}}}$

$\textsf{since,}\:\bold{p+q \neq 0}$

$\to \bold{(pq)^{3}=(p+q)^{2}-3pq}$

$\to \bold{(\frac{c}{a})^{3}=(-\frac{b}{a})^{2}-3\frac{c}{a}}$

$\to \bold{\frac{c^{3}}{a^{3}} = \frac{b^{2}}{a^{2}}-\frac{3c}{a}}$

$\to \bold{c^{3} = ab^{2}-3a^{2}c}$

$\implies \boxed{\bold{ab^{2}=3a^{2}c+c^{3}}}$

$\textsf{which is the required relation.}$