# Solve the following

Question:

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

Solution:

Let us consider the following equation:

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

Squaring both sides, we get:

$\left(x+\frac{1}{x}\right)^{2}=(12)^{2}=144$

$\Rightarrow\left(x+\frac{1}{x}\right)^{2}=144$

$\Rightarrow x^{2}+2 \times x \times \frac{1}{x}+\left(\frac{1}{x}\right)^{2}=144 \quad\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$\Rightarrow x^{2}+2+\frac{1}{x^{2}}=144$

$\Rightarrow x^{2}+\frac{1}{x^{2}}=142$             (Subtracting 2 from both sides)

Now

$\left(x-\frac{1}{x}\right)^{2}=x^{2}-2 \times x \times \frac{1}{x}+\left(\frac{1}{x}\right)^{2}=x^{2}-2+\frac{1}{x^{2}} \quad\left[(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

$\Rightarrow\left(x-\frac{1}{x}\right)^{2}=x^{2}-2+\frac{1}{x^{2}}$

$\Rightarrow\left(x-\frac{1}{x}\right)^{2}=142-2 \quad\left(\because x^{2}+\frac{1}{x^{2}}=142\right)$

$\Rightarrow\left(x-\frac{1}{x}\right)^{2}=140$

$\Rightarrow x-\frac{1}{x}=\pm \sqrt{140}$     (Taking square root)