Question

Solve the following equation for $x$,
$9(x^2+\frac{1}{x^2})-9(x+\frac{1}{x})-52=0$.

Answer 1

$9(x^2+\frac{1}{x^2})-9(x+\frac{1}{x})=52$

Let $(x+\frac{1}{x})=t$

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

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

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

$9(t^2-2)-9\left(t\right)=52$

$9t^2-9t-70=0$ (factorising)

$\Rightarrow t=\frac{10}{3}$ or $t=\frac{-7}{3}$

$t=\frac{10}{3}$ $t=\frac{-7}{3}$

$\Rightarrow x+\frac{1}{x}=\frac{10}{3}$ $x+\frac{1}{x}=\frac{-7}{3}$

$3(x^2+1)=10x$ $3(x^2+1)=-7x$

$3x^2+3=10x$ $3x^2+7x+3=0$

$3x^2-10x+3=0$ $b^2-4ac=49-4\times 3\times 3$

$3x^2-9x-x+3=0$ $=13$

$3x(x-3)+(x-3)=0$ $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$(3x-1)=0$ or $x-3=0$ $=\frac{-7\pm \sqrt{13}}{6}$

$x=\frac{1}{3};x=3$.