Question

If $x=7+4\sqrt{3}$ and xy = 1, then $\frac{1}{x^2}+\frac{1}{y^2}=$

• A. 64
• B. 134
• C. 194
• D. $\frac{1}{49}$

Answer 1

The correct option is C

194

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

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

$\therefore y=\frac{1}{x}=\frac{1(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})}$

(Rationalising the denominator)

$=\frac{7-4\sqrt{3}}{(7{)}^2-(4\sqrt{3}{)}^2}=\frac{7-4\sqrt{3}}{49-48}=\frac{7-4\sqrt{3}}{1}$

$=7-4\sqrt{3}$

$\because y=\frac{1}{x}\Rightarrow \frac{1}{y}=x$

$\therefore \frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{x^2}+x^2$

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

=$\left[\right(7+4\sqrt{3})+7-4\sqrt{3}{]}^2-2$

$=(14{)}^2-2=196-2=194$