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Question

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


Solution

$$9\left(x^2+\dfrac{1}{x^2}\right)-9\left(x+\dfrac{1}{x}\right)=52$$
Let $$\left(x+\dfrac{1}{x}\right)=t$$
$$\left(x+\dfrac{1}{x}\right)^2=t^2$$
$$\Rightarrow x^2+\dfrac{1}{x^2}+2=t^2$$
$$\Rightarrow x^2+\dfrac{1}{x^2}=t^2-2$$
$$9(t^2-2)-9(t)=52$$
$$9t^2-9t-70=0$$ (factorising)
$$\Rightarrow t=\dfrac{10}{3}$$ or $$t=\dfrac{-7}{3}$$
$$t=\dfrac{10}{3}$$                                                              $$t=\dfrac{-7}{3}$$
$$\Rightarrow x+\dfrac{1}{x}=\dfrac{10}{3}$$                                             $$x+\dfrac{1}{x}=\dfrac{-7}{3}$$
$$3(x^2+1)=10x$$                                            $$3(x^2+1)=-7x$$
$$3x^2+3=10x$$                                               $$3x^2+7x+3=0$$
$$3x^2-10x+3=0$$                                       $$b^2-4ac=49-4\times 3\times 3$$
$$3x^2-9x-x+3=0$$                                  $$=13$$
$$3x(x-3)+(x-3)=0$$                              $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$(3x-1)=0$$ or $$x-3=0$$                             $$=\dfrac{-7\pm\sqrt{13}}{6}$$
$$x=\dfrac{1}{3}; x=3$$.         

Mathematics

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