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

Answer:

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