Question

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

Answer 1

It is given that $3x+\frac{1}{3x}=3$, by squaring both sides we get:

${(3x+\frac{1}{3x})}^2=3^2$

$\Rightarrow {\left(3x\right)}^2+{\left(\frac{1}{3x}\right)}^2+(2\times 3x\times \frac{1}{3x})=9(\because (a+b{)}^2=a^2+b^2+2ab)$

$\Rightarrow 9x^2+\frac{1}{9x^2}+2=9$

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

$\Rightarrow 9x^2+\frac{1}{9x^2}=7$

Hence, $9x^2+\frac{1}{9x^2}=7$.