Question

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

Answer 1

Let us consider the following expression: \[x + \frac{1}{x}\]

Squaring the above expression, we get:

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

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

\[\Rightarrow \left( x + \frac{1}{x} \right)^2 = 20\] (\[\because\] \[x^2 + \frac{1}{x^2} = 18\])

\[\Rightarrow x + \frac{1}{x} = \pm \sqrt{20}\]               (Taking square root of both sides)

Now, let us consider the following expression: 

\[x - \frac{1}{x}\] 

Squaring the above expression, we get: 

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

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

\[\Rightarrow \left( x - \frac{1}{x} \right)^2 = 16\] (\[\because\] \[x^2 + \frac{1}{x^2} = 18\])

\[\Rightarrow x - \frac{1}{x} = \pm 4\]                  (Taking square root of both sides)