Question

Find the value of $a$ for which both the roots of the equation $(1-a^2)x^2+2ax-1=0$ lie between $0$ and $1$.

Answer 1

Given $(1-a^2)x^2+2ax-1=0$

Usind sri dharacharya formula

$x=\frac{-2a\pm \sqrt{4a^2+4(1-a^2)}}{2(1-a^2)}$

$x=\frac{-a\pm 1}{(1+a)(1-a)}$

$x=\frac{1}{a-1}$ or $x=\frac{1}{a+1}$

If $a>1$ the roots will both be positive, but we require $a>2$ for both roots to be less than 1.

Therefore $a>2$