Question

Solve the given quadratic equation by using the formula method, $a(x^2+1)=x(a^2+1)$

Answer 1

The given equation $a(x^2+1)=x(a^2+1)$ can be simplified as shown below:

$a(x^2+1)=x(a^2+1)\Rightarrow a{x}^2+a=x(a^2+1)\Rightarrow a{x}^2-x(a^2+1)+a=0$

Thequadratic equation $a{x}^2-x(a^2+1)+a=0$ is in the form $a{x}^2+bx+c=0$ where $a=a,b=-(a^2+1)$ and $c=a$.

We know that the quadratic formula is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, therefore, substitute $a=a,b=-(a^2+1)$ and $c=a$ in $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ as follows:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-(a^2+1\left)\right)\pm \sqrt{(-(a^2+1){)}^2-(4\times a\times a)}}{2\times a}$

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

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

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

$x=\frac{(a^2+1)+(a^2-1)}{2a}=\frac{a^2+1+a^2-1}{2a}=\frac{2a^2}{2a}=a,x=\frac{(a^2+1)-(a^2-1)}{2a}=\frac{a^2+1-a^2+1}{2a}=\frac{2}{2a}=\frac{1}{a}$

Hence, $x=\frac{1}{a}$ or $x=a$.