Question:
If \[ x-\frac{1}{x}=\frac12 \] find:
\[ 4x^2+\frac{4}{x^2} \]
Solution:
Using identity:
\[ \left(x-\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}-2 \]
Substituting the given value:
\[ \left(\frac12\right)^2 = x^2+\frac{1}{x^2}-2 \]
\[ \frac14 = x^2+\frac{1}{x^2}-2 \]
\[ x^2+\frac{1}{x^2} = 2+\frac14 \]
\[ = \frac94 \]
Multiplying both sides by 4:
\[ 4\left(x^2+\frac{1}{x^2}\right) = 4\times\frac94 \]
\[ 4x^2+\frac{4}{x^2} = 9 \]