Question

If $x^2+\frac{1}{x^2}=18,$ find the values of $x+\frac{1}{x}\mathrm{and}x-\frac{1}{x}.$

Answer 1

Let us consider the following expression:

$x+\frac{1}{x}$
Squaring the above expression, we get:

$$\begin{gathered} {\left(x+\frac{1}{x}\right)}^2=x^2+2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=x^2-2+\frac{1}{x^2}\left[\right(a+b{)}^2=a^2+b^2+2ab] \\ \Rightarrow {\left(x+\frac{1}{x}\right)}^2=x^2+2+\frac{1}{x^2} \end{gathered}$$
$\Rightarrow {\left(x+\frac{1}{x}\right)}^2=20$ ($\because$ $x^2+\frac{1}{x^2}=18$)
$\Rightarrow x+\frac{1}{x}=\pm \sqrt{20}$ (Taking square root of both sides)

Now, let us consider the following expression:

$x-\frac{1}{x}$
Squaring the above expression, we get:

$$\begin{gathered} {\left(x-\frac{1}{x}\right)}^2=x^2-2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=x^2-2+\frac{1}{x^2}\left[\right(a-b{)}^2=a^2+b^2-2ab] \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^2=x^2-2+\frac{1}{x^2} \end{gathered}$$
$\Rightarrow {\left(x-\frac{1}{x}\right)}^2=16$ ($\because$ $x^2+\frac{1}{x^2}=18$)
$\Rightarrow x-\frac{1}{x}=\pm 4$ (Taking square root of both sides)