Question

If $\left(x-\frac{1}{x}\right)=8$, find the value of
(i)$\left(x+\frac{1}{x}\right)$
(ii)$\left(x^2-\frac{1}{x^2}\right)$

Answer 1

It is given that $x-\frac{1}{x}=8$, by squaring both sides we get:

${(x-\frac{1}{x})}^2=8^2\Rightarrow {\left(x\right)}^2+{\left(\frac{1}{x}\right)}^2-(2\times x\times \frac{1}{x})=64(\because (a-b{)}^2=a^2+b^2-2ab)\Rightarrow x^2+\frac{1}{x^2}-2=64\Rightarrow x^2+\frac{1}{x^2}=64+2\Rightarrow x^2+\frac{1}{x^2}=66....(1)$

Now, consider ${(x+\frac{1}{x})}^2$ then we have:

${(x+\frac{1}{x})}^2={\left(x\right)}^2+{\left(\frac{1}{x}\right)}^2+(2\times x\times \frac{1}{x})(\because (a+b{)}^2=a^2+b^2+2ab)=x^2+\frac{1}{x^2}+2=66+2(Fromeqn1)=68$

$\Rightarrow {(x+\frac{1}{x})}^2=68\Rightarrow x+\frac{1}{x}=\sqrt{68}\Rightarrow x+\frac{1}{x}=2\sqrt{17}$

Now, consider the product of $x-\frac{1}{x}$ and $x+\frac{1}{x}$ as shown below:

$(x-\frac{1}{x})(x+\frac{1}{x})=8\times 2\sqrt{17}=16\sqrt{17}\Rightarrow (x^2-\frac{1}{x^2})=16\sqrt{17}(\because a^2-b^2=(a-b\left)\right(a+b\left)\right)$

Hence, $(x+\frac{1}{x})=2\sqrt{17}$ and $(x^2-\frac{1}{x^2})=16\sqrt{17}$.