Question

If $x+\frac{1}{x}=5$, find the value of $x^2+\frac{1}{x^2}$ and $x^4+\frac{1}{x^4}$

• A. $25,625$
• B. $23,527$
• C. $25,529$
• D. $23,531$

Answer 1

The correct option is B $23,527$

We are given that $x+\frac{1}{x}=5$, squaring both sides we get:

${(x+\frac{1}{x})}^2=5^2\Rightarrow x^2+\frac{1}{x^2}+(2\times x\times \frac{1}{x})=25(\because (a+b{)}^2=a^2+b^2+2ab)\Rightarrow x^2+\frac{1}{x^2}+2=25\Rightarrow x^2+\frac{1}{x^2}=25-2\Rightarrow x^2+\frac{1}{x^2}=23........\left(1\right)$

Again squaring both sides of equation 1, we get:

${(x^2+\frac{1}{x^2})}^2={23}^2\Rightarrow x^4+\frac{1}{x^4}+(2\times x^2\times \frac{1}{x^2})=529(\because (a+b{)}^2=a^2+b^2+2ab)\Rightarrow x^4+\frac{1}{x^4}+2=529\Rightarrow x^4+\frac{1}{x^4}=529-2\Rightarrow x^4+\frac{1}{x^4}=527$

Hence, $x^2+\frac{1}{x^2}=23,x^4+\frac{1}{x^4}=527$.