Question

Solve the following quadratic equation using quadratic formula: $a(x^2+1)=x(a^2+1)$

• A. $\{\frac{1}{a},a\}$
• B. $\{\frac{1}{a},1\}$
• C. $\{\frac{1}{a},a^2\}$
• D. None of these

Answer 1

The correct option is A $\{\frac{1}{a},a\}$

Given equation is $a(x^2+1)=x(a^2+1)$

$\Rightarrow$ $a{x}^2+a=a^{2}x+x$

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

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

$\Rightarrow$ $a=a,b=-(a^2+1),c=a$

$\Rightarrow$ $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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

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

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

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

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

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

$\Rightarrow$ $x=\frac{2a^2}{2a}$ and $x=\frac{2}{2a}$

$\therefore$ $x=a$ and $x=\frac{1}{a}$