Question

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Answer 1

Step-by-step explanation:

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Answer 2

Question :

If , $\frac{\sqrt{7}-1}{ \sqrt{7}+1 } - \frac{\sqrt{7}+1}{\sqrt{7}- 1} = a + b\sqrt{7}$, find ${a}^{2} + {b}^2$

$\huge \underline \mathfrak \red {Answer}$

$\frac{ \sqrt{7} - 1 }{ \sqrt{7} + 1 } - \frac{ \sqrt{7} + 1}{ \sqrt{7} - 1}$

$\frac{( \sqrt{7} - 1)( \sqrt{7} - 1) - ( \sqrt{7} + 1)( \sqrt{7} + 1) }{ (\sqrt{7} + 1)( \sqrt{7} - 1) }$

$\frac{7 + 1 - 2 \sqrt{7} - (7 + 1 + 2 \sqrt{7} ) }{7 - 1}$

simplify

$\frac{8 - 2 \sqrt{7} - 8 - 2 \sqrt{7} }{6}$

$\frac{ - 4 \sqrt{7} }{6}$

$\frac{ - 2 \sqrt{7} }{3}$

on comparing,

$a = 0$

$b = \frac{ - 2 }{3}$

so,

${a}^{2} + {b}^{2} = ( { \frac{ -2}{3}) }^{2}$

$\implies \frac{4}{9}$