Question

If \[x - \frac{1}{x} = 3,\]  find the values of \[x^2 + \frac{1}{x^2}\] and \[x^4 + \frac{1}{x^4} .\]

Answer 1

Let us consider the following equation: \[x - \frac{1}{x} = 3\]

Squaring both sides, we get:

\[\left( x - \frac{1}{x} \right)^2 = \left( 3 \right)^2 = 9\]

\[ \Rightarrow \left( x - \frac{1}{x} \right)^2 = 9\]

\[ \Rightarrow x^2 - 2 \times x \times \frac{1}{x} + \left( \frac{1}{x} \right)^2 = 9\]

\[ \Rightarrow x^2 - 2 + \frac{1}{x^2} = 9\]

\[\Rightarrow x^2 + \frac{1}{x^2} = 11\]            (Adding 2 to both sides)

Squaring both sides again, we get:

\[\left( x^2 + \frac{1}{x^2} \right)^2 = \left( 11 \right)^2 = 121\]

\[ \Rightarrow \left( x^2 + \frac{1}{x^2} \right)^2 = 121\]

\[ \Rightarrow \left( x^2 \right)^2 + 2\left( x^2 \right)\left( \frac{1}{x^2} \right) + \left( \frac{1}{x^2} \right)^2 = 121\]

\[ \Rightarrow x^4 + 2 + \frac{1}{x^4} = 121\]

\[\Rightarrow x^4 + \frac{1}{x^4} = 119\]