Question

(i) If the sum of the roots of the quadratic equation ax2 + bx+c = 0 is equal to the
sum of the squares of their reciprocals, then prove that 2a+c = c2b + b²a.​

Answer 1

Question

(i) If the sum of the roots of the quadratic equation ax2 + bx+c = 0 is equal to the

sum of the squares of their reciprocals, then prove that 2a+c = c2b + b²a.

Solution

Given equation ⁣           ax²+bx²+c=0

let p & q root of given equation then,

⁣⁣      A/Q

⁣     $\bold{p+q={\dfrac{1}{p²} + \dfrac{1}{q²}}}$ ⁣   ..1

now using property we know

[p+q=(-b/a)]⁣            & [pq=c/a]

from equation i we have

⁣            $\bold{p+q={\dfrac{p²+q²}{p²q²}}}$

⁣            $\bold{p+q={\dfrac{(p²+q²)-2pq}{p²-q²}}}$

on putting the value we get

-b/a= $\bold{\frac{(-b/a)²-2a/a)}{c²/a²}}$

=-bc²/a³= (ab²-2ac²)/a³

b²a+c²b=2a²c

⁣        proved