Question

If $x+\frac{1}{x}=12,$ find the value of $x-\frac{1}{x}.$

Answer 1

Let us consider the following equation:

$x+\frac{1}{x}=12$

Squaring both sides, we get:

$$\begin{gathered} {\left(x+\frac{1}{x}\right)}^2={\left(12\right)}^2=144 \\ \Rightarrow {\left(x+\frac{1}{x}\right)}^2=144 \\ \Rightarrow x^2+2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=144[(a+b{)}^2=a^2+b^2+2ab] \\ \Rightarrow x^2+2+\frac{1}{x^2}=144 \end{gathered}$$
$\Rightarrow x^2+\frac{1}{x^2}=142$ (Subtracting 2 from both sides)

Now

$$\begin{gathered} {\left(x-\frac{1}{x}\right)}^2=x^2-2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=x^2-2+\frac{1}{x^2}\left[\right(a-b{)}^2=a^2+b^2-2ab] \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^2=x^2-2+\frac{1}{x^2} \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^2=142-2(\because x^2+\frac{1}{x^2}=142) \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^2=140 \\ \Rightarrow x-\frac{1}{x}=\pm \sqrt{140}\left(\text{Taking square root}\right) \end{gathered}$$