Question

If $x+\frac{1}{x}=9$, then find the value of $x-\frac{1}{x}$.

Answer 1

Consider $x+\frac{1}{x}=9$, squaring both sides we get,

${(x+\frac{1}{x})}^2=9^2\Rightarrow x^2+{\left(\frac{1}{x}\right)}^2+(2\times x\times \frac{1}{x})=81(\because (a+b{)}^2=a^2+b^2+2ab)\Rightarrow x^2+\frac{1}{x^2}+2=81\Rightarrow x^2+\frac{1}{x^2}=81-2\Rightarrow x^2+\frac{1}{x^2}=79$

We know that $(a-b{)}^2=a^2+b^2-2ab$, substitute $a=x$ and $b=\frac{1}{x}$ as shown below:

${(x-\frac{1}{x})}^2=x^2+{\left(\frac{1}{x}\right)}^2-(2\times x\times \frac{1}{x})=79-2(\because x^2+\frac{1}{x^2}=79)=77$

Therefore,

${(x-\frac{1}{x})}^2=77\Rightarrow x-\frac{1}{x}=\pm \sqrt{77}$

Hence, $x-\frac{1}{x}=\pm \sqrt{77}$.