Question

If $x^2+\frac{1}{x^2}=66$ ,find the value of $x$ .

• A. $\sqrt{17}+3$
• B. $\sqrt{7}+4$
• C. $\sqrt{17}-4$
• D. $\sqrt{17}+4$

Answer 1

The correct option is D $\sqrt{17}+4$

Consider the given quadratic equation

$x^2+\frac{1}{x^2}=66..........\left(1\right)$

On adding 2 in equation (1) and we get,

$x^2+\frac{1}{x^2}+2=66+2$

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

$\Rightarrow x+\frac{1}{x}=\sqrt{68}..........\left(2\right)$

On subtracting 2 in equation (2) and we get,

$x^2+\frac{1}{x^2}-2=66-2$

$\Rightarrow {(x-\frac{1}{x})}^2=64$

$\Rightarrow x-\frac{1}{x}=8.......\left(3\right)$

On adding equation (1) and (2) to,

$x+\frac{1}{x}+x-\frac{1}{x}=\sqrt{68}+8$

$2x=2\sqrt{17}+8$

$x=\sqrt{17}+4$

Hence, this is the answer.