Question

For what values of $a$ $ϵR$ does the equation $x^2+1=\frac{x}{a}$ possess two distinct real roots $|x_1^2-x_2^2|>\frac{1}{a}$.?

Answer 1

$x^2+1=\frac{x}{a}$

$a{x}^2-x+a=0$

Roots are distinct and real

$\therefore 1-4a^2\ge 0$

$4a^2-1\le 0$

$(2a-1)(2a+1)\le 0$

$a\in (-\frac{1}{2},\frac{1}{2})$

$x_1+x_2=\frac{-1}{a}$

$x_1{x}_2=1$

$|x_1^2-x_2^2|>\frac{1}{a}$

Squaring both sides

$(x_1+x_2{)}^2(x_1-x_2{)}^2>\frac{1}{a^2}$

$\left(\frac{-1}{a}{)}^2\right[(x_1+x_2{)}^2-4x_1{x}_2{]}^2>\frac{1}{a^2}$

$\frac{1}{a^2}[\frac{1}{a^2}-4{]}^2>\frac{1}{a^2}$

As $a\in R$

$(\frac{1}{a}-2{)}^2(\frac{1}{a}+2{)}^2>0$

$(2a-1{)}^2(2a+1{)}^2>0$

$a\in R-\{\frac{1}{2},\frac{-1}{2}\}$

$a\in (\frac{-1}{2},\frac{1}{2})$