Question

If $x=4+\sqrt{3},$ then find the value of $x^2+\frac{1}{x^2}$.

• A. $\frac{-3230+1344\sqrt{3}}{169}$
• B. $\frac{-3230-1344\sqrt{3}}{169}$
• C. $\frac{3230-1344\sqrt{3}}{169}$
• D. $\frac{3230+1344\sqrt{3}}{169}$

Answer 1

The correct option is D $\frac{3230+1344\sqrt{3}}{169}$

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

$x^2=(4+\sqrt{3}{)}^2=4^2+(\sqrt{3}{)}^2+$$2\times 4\times \sqrt{3}$ (Using $(a+b{)}^2=a^2+b^2+2ab$)

$x^2=16+3+8\sqrt{3}=19+8\sqrt{3}$

$\frac{1}{x^2}=\frac{1}{(4+\sqrt{3}{)}^2}=(\frac{1}{4+\sqrt{3}}{)}^2$
On rationalising, we get,
$\frac{1}{x^2}=(\frac{4-\sqrt{3}}{16-3}{)}^2=(\frac{4-\sqrt{3}}{13}{)}^2$
$=\left(\frac{19-8\sqrt{3}}{169}\right)$

$x^2+\frac{1}{x^2}=19+8\sqrt{3}+\frac{19-8\sqrt{3}}{169}$

$=\frac{3211+1352\sqrt{3}+19-8\sqrt{3}}{169}$

$=\frac{3230+1344\sqrt{3}}{169}$