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