Question

If $x=7+4\sqrt{3},y=7-4\sqrt{3}$ find the value of $\frac{1}{x^2}+\frac{1}{y^2}$

• A. 194
• B. 198
• C. 212
• D. 318

Answer 1

The correct option is A 194

We have, $x=7+4\sqrt{3}$ So,

$x^{-2}=\frac{1}{{(7+4\sqrt{3})}^2}=\frac{1}{49+48+56\sqrt{3}}=\frac{(97-56\sqrt{3})}{(97+56\sqrt{3})(97-56\sqrt{3})}=\frac{97-56\sqrt{3}}{9409-9408}=97-56\sqrt{3}$

Now, $y=7-4\sqrt{3}$ So,

$y^{-2}=\frac{1}{{(7-4\sqrt{3})}^2}=\frac{1}{49+48-56\sqrt{3}}=\frac{(97+56\sqrt{3})}{(97+56\sqrt{3})(97-56\sqrt{3})}=\frac{97+56\sqrt{3}}{9409-9408}=97+56\sqrt{3}$

So, $x^{-2}+y^{-2}=97-56\sqrt{3}+97+56\sqrt{3}=194$