Question

Please solve and give steps too!

Answer 1

❄️ Question :-

$\bf \frac{4+3 \sqrt{5} }{4-3 \sqrt{5} } + \frac{4-3 \sqrt{5} }{4+3 \sqrt{5} }$

Simplify this by rationalizing the denominator.

❄️ Solution :-

Rationalizing both -

$\mapsto \bf \frac{4+3 \sqrt{5} }{4-3 \sqrt{5} } \times \frac{4 + 3 \sqrt{5} }{4 + 3 \sqrt{5} }$

Using the identities, (a+b)²=a²+b²+2ab and (a-b)(a+b)=a²-b².

$\implies \small \bf \dfrac{(4)²+(3 \sqrt{5}) ²+2(4)(3 \sqrt{5} )}{(4)²-(3 \sqrt{5)²} }$

$\implies \bf \frac{16 + 45 + 24 \sqrt{5} }{16 - 45}$

$\implies \bf \frac{61 + 24\sqrt{5} }{ - 29}$

$\mapsto\bf \frac{4 - 3 \sqrt{5} }{4 + 3 \sqrt{5} } \times \frac{4 - 3 \sqrt{5} }{4-3 \sqrt{5} }$

$\implies \bf \frac{4 - 3 \sqrt{5} }{4 + 3 \sqrt{5} } \times \frac{4 - 3 \sqrt{5} }{4 - 3 \sqrt{5} }$

$\implies \Large \bf \frac{(4)²+(3 \sqrt{5})²-2(4)(3 \sqrt{5}) }{(4)²-(3 \sqrt{5})² }$

$\implies \bf \frac{16+45 - 24 \sqrt{5} }{(4)² - (3 \sqrt{5})² }$

$\implies \bf \frac{61-24 \sqrt{5} }{16-45}$

$\implies \bf \frac{61 - 24 \sqrt{5} }{ - 29}$

Solving them :-

$\mapsto \bf \frac{61 + 24 \sqrt{5} }{ - 29} + \frac{61 - 24 \sqrt{5} }{ - 29}$

$\implies \bf \frac{61 + 24 \sqrt{5} + 61 - 24 \sqrt{5} }{ - 29 + ( - 29)}$

$\implies \bf \frac{61 + 61\cancel{ - 24 \sqrt{5}}\cancel{+24 \sqrt{5}} }{-29-29}$

$\implies \bf \frac{\cancel{122}}{\cancel{ - 58}}$

$\implies \bf \frac{61}{ - 29}$

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Hope it helps you :)