Question

If $x+\frac{1}{x}=6$ find $x-\frac{1}{x}$.

Answer 1

Compute the required value.

It is given that, $x+\frac{1}{x}=6$.

On squaring both sides, we get;

$$\begin{gathered} {\left(x+\frac{1}{x}\right)}^2=36 \\ \Rightarrow x^2+\frac{1}{x^2}+2=36 \\ \Rightarrow x^2+\frac{1}{x^2}=34 \end{gathered}$$ {Since, ${(a+b)}^2=a^2+b^2+2ab$}

Subtract $2\left(x\times \frac{1}{x}\right)$ from both sides.

$$\begin{gathered} \Rightarrow x^2+\frac{1}{x^2}-2\left(x\times \frac{1}{x}\right)=34-2\left(x\times \frac{1}{x}\right) \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^2=34-2[\mathrm{Since},{(\mathrm{a}-\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}] \\ \mathrm{Taking}\mathrm{square}\mathrm{root}\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}; \\ \Rightarrow x-\frac{1}{x}=\sqrt{32} \\ \Rightarrow x-\frac{1}{x}=\sqrt{4^2\times 2} \end{gathered}$$

Therefore, the value of $x-\frac{1}{x}$ is $4\sqrt{2}$.