Question

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

Answer 1

$\huge\bf{\underline{\underline{Solution:}}}$

→ $\frac{1}{x} = 7 - 4 \sqrt{3} \\ \frac{1}{y} = 7+ 4 \sqrt{3} \\ \frac{1}{ {x}^{2} } + \frac{1}{ {y}^{2} } = \binom{1}{x}^{2} + \binom{1}{y} ^{2} \\ = (7 - 4 \sqrt{3})^{2} + (7 + 4 \sqrt{3} )^{2} = \\ 2(49 + 48) = 194$

✨${Hope \: it \: helps}$✨

$\huge{Be \: Brainly}$

Answer 2

$\rule{200}{2}$

$\huge\boxed{\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\pink{R:-}}}}}$

$\qx = 7 + 4 \sqrt{3}$........eq 1

$\wy = 7 - 4 \sqrt{3}$........eq 2

eq 1 + eq 2

$7 + 4 \sqrt{3}$ + $7 - 4 \sqrt{3}$

= 2 (48+49)

= $\boxed{\boxed{194}}$