Question

If $x=\frac{5-\sqrt{21}}{2}$, then $(x^2+\frac{1}{x^2})-(x+\frac{1}{x})=$?

• A. $21$
• B. $18$
• C. $4$
• D. $5$

Answer 1

The correct option is B $18$

Given, $x=\frac{5-\sqrt{21}}{2}$.
Then, $\frac{1}{x}=\frac{2}{5-\sqrt{21}}\times \frac{5+\sqrt{21}}{5+\sqrt{21}}=\frac{2(5+\sqrt{21})}{25-21}=\frac{2(5+\sqrt{21})}{4}$.

$\therefore {(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2x\frac{1}{x}$
$⟹$ ${(\frac{5-\sqrt{21}}{2}+\frac{5+\sqrt{21}}{2})}^2=x^2+\frac{1}{x^2}+2$

$⟹$ ${\left(\frac{5-\sqrt{21}+5+\sqrt{21}}{2}\right)}^2=x^2+\frac{1}{x^2}+2$
$⟹$ ${\left(\frac{10}{2}\right)}^2-2=x^2+\frac{1}{x^2}$
$⟹$ $x^2+\frac{1}{x^2}=25-2=23$.
and $x+\frac{1}{x}=\frac{10}{2}=5$.
$\therefore (x^2+\frac{1}{x^2})-(x+\frac{1}{x})=23-5=18$.

Therefore, option $B$ is correct.