Question

Using Sridharacharyya's formula solve:
a(x^2+1) = (a^2+1)x, a is not equal to 0.

Answer 1

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\rm :\longmapsto\:a( {x}^{2} + 1) = ( {a}^{2} + 1)x$

$\rm :\longmapsto\:a{x}^{2} +a = x({a}^{2} + 1)$

$\rm :\longmapsto\:a{x}^{2} - x({a}^{2} + 1) + a = 0$

We know,

SriSridharacharyya's formula is

$\rm :\longmapsto\:x = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

Here,

$\rm :\longmapsto\:a = a$

$\rm :\longmapsto\:b = {a}^{2} + 1$

$\rm :\longmapsto\:c = a$

On substituting the values,

$\rm :\longmapsto\:x = \dfrac{ {a}^{2} + 1 \: \pm \: \sqrt{ { {( {a}^{2} + 1)} }^{2} - 4(a)(a)} }{2a}$

$\rm :\longmapsto\:x = \dfrac{ {a}^{2} + 1 \: \pm \: \sqrt{ { {( {a}^{2} + 1)} }^{2} - 4 {a}^{2} } }{2a}$

$\rm :\longmapsto\:x = \dfrac{ {a}^{2} + 1 \: \pm \: \sqrt{ { {( {a}^{2} - 1)} }^{2} } }{2a}$

$\rm :\longmapsto\:x = \dfrac{ {a}^{2} + 1 \: \pm \: { {( {a}^{2} - 1)} }}{2a}$

$\rm :\longmapsto\:x = \dfrac{ {a}^{2} + 1 + \: { { {a}^{2} - 1} }}{2a} \: \: or \: \: \dfrac{ {a}^{2} + 1 - { {{a}^{2} + 1} }}{2a}$

$\rm :\longmapsto\:x = \dfrac{2{a}^{2}}{2a} \: \: or \: \: \dfrac{ 2}{2a}$

$\rm :\longmapsto\:x = a \: \: \: or \: \: \: \dfrac{1}{a}$