Solve the following equations by using formula:

a(x² + 1) = (a² + 1)x , a ≠ 0

Given,

a(x² + 1) = (a² + 1)x

First converting the equation in the form ax² + bx + c = 0.

$\Rightarrow ax^2 + a = a^2x + x \\[0.5em] \Rightarrow ax^2 - a^2x - x + a = 0 \\[0.5em] \Rightarrow ax^2 -(a^2 + 1)x + a = 0 \\[0.5em]$

Comparing it with ax² + bx + c = 0, we get
a = a , b = -(a² + 1) , c = a

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-(-(a^2 + 1)) ± \sqrt{(-(a^2 + 1))^2 - 4 \times a \times a}}{2 \times a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± \sqrt{(a^2 + 1)^2 - 4a^2}}{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± \sqrt{a^4 + 1 + 2a^2 - 4a^2}}{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± \sqrt{a^4 + 1 - 2a^2}}{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± \sqrt{(a^2 - 1)^2}}{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± (a^2 - 1) }{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 + a^2 - 1}{2a} \text{ or } \dfrac{a^2 + 1 - a^2 + 1}{2a} \\[1em] \Rightarrow \dfrac{2a^2}{2a} \text{ or } \dfrac{2}{2a} \\[1em] \Rightarrow a \text{ or } \dfrac{1}{a}$

Hence roots of the given equations are a , $\dfrac{1}{a}$.