Question

Evaluate $9x^2+\frac{1}{9x^2}$ if $(3x+\frac{1}{3x})=\frac{8}{3}$

Answer 1

$(3x+\frac{1}{3x})=\frac{8}{3}$

Squaring both sides

$(3x+\frac{1}{3x}{)}^2=(\frac{8}{3}{)}^2$

$(3x{)}^2+({\frac{1}{3x}}^2)+2(3x\left)\right(\frac{1}{3x})=\frac{64}{9}$

$9x^2+\frac{1}{9x^2}+2=\frac{64}{9}$

$9x^2+\frac{1}{9x^2}=\frac{64}{9}-2$

$9x^2+\frac{1}{9x^2}=\frac{64-18}{9}$

$9x^2+\frac{1}{9x^2}=\frac{46}{9}$