Question

If $x=\frac{1}{7-3\sqrt{3}},y=\frac{1}{7+3\sqrt{5}}$ then $x^2+y^2$ is

• A. $\frac{47}{2}$
• B. $\frac{7}{4}$
• C. $\frac{11}{4}$
• D. $\frac{47}{4}$

Answer 1

The correct option is D $\frac{47}{4}$

$x=\frac{1}{7-3\sqrt{5}}=\frac{7+3\sqrt{5}}{(7-3\sqrt{5})(7+3\sqrt{5})}=\frac{7+3\sqrt{5}}{4}$
$y=\frac{1}{7+3\sqrt{5}}=\frac{7-3\sqrt{5}}{4}$
$x^2+y^2={(x+y)}^2-2xy$
$={(\frac{7+3\sqrt{5}}{4}+\frac{7-3\sqrt{5}}{4})}^2-2\left(\frac{7+3\sqrt{5}}{4}\right)\left(\frac{7-3\sqrt{5}}{4}\right)$
$={\left(\frac{14}{4}\right)}^2-2\times \frac{4}{4\times 4}$
$={\left(\frac{7}{2}\right)}^2-\frac{1}{2}$
$=\frac{49}{4}-\frac{1}{2}=\frac{47}{4}$