Question

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

Answer 1

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

$\frac{1}{x}=\frac{7+4\sqrt{3}}{(7+4\sqrt{3})(7-4\sqrt{3})}$ $[\therefore (a+b\left)\right(a-b)=a^2-b^2]$

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

Given, $xy=1$

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

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

$\Rightarrow$$\frac{1}{x^2}+\frac{1}{y^2}=(7-4\sqrt{3}{)}^2+(7+4\sqrt{3}{)}^2$

$\Rightarrow$$\frac{1}{x^2}+\frac{1}{y^2}=(7{)}^2+4(\sqrt{3}{)}^2-2\times 7\times 4\sqrt{3}\times (7{)}^2+(4\sqrt{3}{)}^2+2\times 7\times 4\sqrt{3}$

$\Rightarrow$$\frac{1}{x^2}+\frac{1}{y^2}=49+48+49+48$

$\Rightarrow$$\frac{1}{x^2}+\frac{1}{y^2}=194$