Question

If $x=\frac{5-\sqrt{21}}{2}$, find the value of $x^2+\frac{1}{x^2}$

• A. $23$
• B. $5$
• C. $4$
• D. $6$

Answer 1

Answer: (A)

$23$

Given, $x=\frac{5-\sqrt{21}}{2}$

Thus $\frac{1}{x}=\frac{2}{5-\sqrt{21}}$

On rationalizing, we get

$\frac{1}{x}=\frac{2}{5-\sqrt{21}}\times \frac{5+\sqrt{21}}{5+\sqrt{21}}=\frac{10+2\sqrt{21}}{4}$

$\Rightarrow \frac{1}{x}=\frac{5+\sqrt{21}}{2}$

Hence, $x+\frac{1}{x}=5$

We know $x^2+\frac{1}{x^2}=(x+\frac{1}{x}{)}^2-2$

$=5^2-2$

$=25-2=23$