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:

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

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

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

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

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

Now

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

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

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

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

\[ \Rightarrow x - \frac{1}{x} = \pm \sqrt{140} \left( \text { Taking square root } \right)\]