Solve the quadratic equation $a (x^{2} + 1) = x (a^{2} + 1)$ by using the quadratic formula.

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Solution

Step 1: Writing the given equation in the standard form of the quadratic equation
The standard form of the quadratic equation is $a x^{2} + b x + c = 0$ .
$a (x^{2} + 1) = x (a^{2} + 1) \Rightarrow a x^{2} + a = a^{2} x + x \Rightarrow a x^{2} - a^{2} x - x + a = 0 \Rightarrow a x^{2} - x (a^{2} + 1) + a = 0$
Step 2: Determining the value of $x$ by using the quadratic formula
The quadratic formula is $\begin{array}{rcl} x & = & \frac{- B \pm \sqrt{B^{2} - 4 A C}}{2 A} \end{array}$ (Considering $A, B, C$ for ease of calculation because the question has variable $a, b, c$ ).
For the given equation $a x^{2} - x (a^{2} + 1) + a = 0$ , the value of variables are $A = a, B = - (a^{2} + 1)$ and $C = a$ .
Substituting the values of $A, B$ and $C$ in quadratic formula,
$x = \frac{- B \pm \sqrt{B^{2} - 4 A C}}{2 A} \Rightarrow x = \frac{(a^{2} + 1) \pm \sqrt{{(a^{2} + 1)}^{2} - 4 a (a)}}{2 a} \Rightarrow x = \frac{(a^{2} + 1) \pm \sqrt{(a^{4} + 2 a^{2} + 1) - 4 a^{2}}}{2 a} \Rightarrow x = \frac{(a^{2} + 1) \pm \sqrt{a^{4} - 2 a^{2} + 1}}{2 a} \Rightarrow x = \frac{(a^{2} + 1) \pm \sqrt{{(a^{2} - 1)}^{2}}}{2 a} \Rightarrow x = \frac{(a^{2} + 1) \pm (a^{2} - 1)}{2 a} \Rightarrow x = \frac{(a^{2} + 1) - (a^{2} - 1)}{2 a}, \frac{(a^{2} + 1) + (a^{2} - 1)}{2 a} \Rightarrow x = \frac{a^{2} + 1 - a^{2} + 1}{2 a}, \frac{a^{2} + 1 + a^{2} - 1}{2 a} \Rightarrow x = \frac{2}{2 a}, \frac{2 a^{2}}{2 a} ∴ x = \frac{1}{a}, a$
Therefore, the value of $x$ is either $\frac{1}{a}$ or $a$ .

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