Question

If $x^2+\frac{1}{{}^{x^2}}=18$, then the value of $(x+\frac{1}{x})$ is ?

• A. $1$
• B. $3$
• C. $\sqrt{20}$
• D. $6$

Answer 1

The correct option is C $\sqrt{20}$

$x^2+\frac{1}{x^2}=18$

$\Rightarrow$ $x^2+\frac{1}{x^2}=20-2$ [ Since, $20-2=18$ ]

$\Rightarrow$ $x^2+\frac{1}{x^2}+2=20$

$\Rightarrow$ $x^2+\frac{1}{x^2}+2\times x\times \frac{1}{x}=20$

$\Rightarrow$ ${(x+\frac{1}{x})}^2=20$ [ Since, $a^2+b^2+2ab=(a+b{)}^2$ ]

$\Rightarrow$ $(x+\frac{1}{x})=\sqrt{20}$