Question
Solve the following equation by using formula :
$a (x^2 + 1) = (a^2+ 1) x , a \neq 0$
$a (x^2 + 1) = (a^2+ 1) x , a \neq 0$
✓
Answer
$
\begin{aligned}
& a\left(x^2+1\right)=\left(a^2+1\right) x \\
& a x^2-\left(a^2+1\right) x+a=0 \\
& \text { Here } a=a, b=-\left(a^2+1\right), c=a \\
& D=b^2-4 a c \\
& =\left[-\left(a^2+1\right)\right]^2-4 \times a x a \\
& =a^4+2 a^2+1-4 a^2 \\
& =a^4-2 a+1 \\
& =\left(a^2-1\right)^2 \\
& \because x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{\left(a^2+1\right) \pm \sqrt{\left(a^2-1\right)^2}}{2 a} \\
& =\frac{\left(a^2+1\right)+\left(a^2-1\right)}{2 a} \\
& \therefore x_1=\frac{a^2+1+a^2-1}{2 a} \\
& =\frac{2 a^2}{2 a} \\
& =a \\
& x_2=\frac{a^2+1-a^2+1}{2 a} \\
& =\frac{2}{2 a} \\
& =\frac{1}{a}
\end{aligned}
$
Hence $x =a, \frac{1}{a}$.
\begin{aligned}
& a\left(x^2+1\right)=\left(a^2+1\right) x \\
& a x^2-\left(a^2+1\right) x+a=0 \\
& \text { Here } a=a, b=-\left(a^2+1\right), c=a \\
& D=b^2-4 a c \\
& =\left[-\left(a^2+1\right)\right]^2-4 \times a x a \\
& =a^4+2 a^2+1-4 a^2 \\
& =a^4-2 a+1 \\
& =\left(a^2-1\right)^2 \\
& \because x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{\left(a^2+1\right) \pm \sqrt{\left(a^2-1\right)^2}}{2 a} \\
& =\frac{\left(a^2+1\right)+\left(a^2-1\right)}{2 a} \\
& \therefore x_1=\frac{a^2+1+a^2-1}{2 a} \\
& =\frac{2 a^2}{2 a} \\
& =a \\
& x_2=\frac{a^2+1-a^2+1}{2 a} \\
& =\frac{2}{2 a} \\
& =\frac{1}{a}
\end{aligned}
$
Hence $x =a, \frac{1}{a}$.
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