Question

If $x=\sqrt{7+4\sqrt{3}}$, then the value of $(x+\frac{1}{x})$ is equal to

• A. $4$
• B. $6$
• C. $3$
• D. $2$

Answer 1

The correct option is A $4$

Given, $x=\sqrt{7+4\sqrt{3}}$

$\therefore \frac{1}{x}$ $=$ $\frac{1}{\sqrt{7+4\sqrt{3}}}$

Now, $x+\frac{1}{x}$ $=\sqrt{7+4\sqrt{3}}$ $+$ $\frac{1}{\sqrt{7+4\sqrt{3}}}$

After squaring both sides we get

${(x+\frac{1}{x})}^2$ $={\left(\sqrt{7+4\sqrt{3}}\right)}^2$ + ${\left(\frac{1}{\sqrt{7+4\sqrt{3}}}\right)}^2+2$

$=$ $7+4\sqrt{3}+\frac{1}{7+4\sqrt{3}}+2$

$=$ $\frac{(9+4\sqrt{3})(7+4\sqrt{3})+1}{7+4\sqrt{3}}$

$=$ $\frac{112+64\sqrt{3}}{7+4\sqrt{3}}=16$

Taking square roots on both sides, we get

$\therefore (x+\frac{1}{x})$ = + $4$

Hence, option A is correct.