Question

If $3x-\frac{1}{3x}=5$, then find $9x^2+\frac{1}{9x^2}$.

• A. 25
• B. 26
• C. 27
• D. 28

Answer 1

The correct option is C 27

Applying the formula, ${(a-b)}^2=a^2+b^2-2ab$
${(3x-\frac{1}{3x})}^2={3x}^2+{\left(\frac{1}{3x}\right)}^2-2\times 3x\times \frac{1}{3x}=9x^2+\frac{1}{9x^2}-2$
$=>9x^2+\frac{1}{9x^2}={(3x-\frac{1}{3x})}^2+2$
Substituting $3x-\frac{1}{3x}=5$,
$=>9x^2+\frac{1}{9x^2}=5^2+2=25+2=27$