Solve:
$ \frac{6-2 \sqrt{20}+3 \sqrt{5}-\sqrt{100}}{3^{2}-(\sqrt{20})^{2}} $

Given:

\( \frac{6-2 \sqrt{20}+3 \sqrt{5}-\sqrt{100}}{3^{2}-(\sqrt{20})^{2}} \)

To do:

We have to evaluate \( \frac{6-2 \sqrt{20}+3 \sqrt{5}-\sqrt{100}}{3^{2}-(\sqrt{20})^{2}} \).

Solution:

We know that,

$(a-b)^2=a^2-2ab+b^2$

Therefore,

$\frac{6-2 \sqrt{20}+3 \sqrt{5}-\sqrt{100}}{3^{2}-(\sqrt{20})^{2}}$

$=\frac{2 \times 3-2 \times \sqrt{20}+\sqrt{5} \times 3-\sqrt{5} \times \sqrt{20}}{(3)^{2}-(\sqrt{20})^{2}}$

$=\frac{6-2\times\sqrt{4\times5}+3\sqrt5-\sqrt{5}\times\sqrt{4\times5}}{9-20}$

$=\frac{6-2\times\sqrt{2^2\times5}+3\sqrt5-\sqrt{5}\times\sqrt{2^2\times5}}{-11}$

$=\frac{6-2\times2\sqrt{5}+3\sqrt5-2\times\sqrt{5}\times\sqrt{5}}{-11}$

$=\frac{6-4\sqrt{5}+3\sqrt5-2\times5}{-11}$

$=\frac{6-10+\sqrt{5}(-4+3)}{-11}$

$=\frac{-4-\sqrt5}{-11}$

$=\frac{-(4+\sqrt5)}{-(11)}$

$=\frac{4+\sqrt{5}}{11}$

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Updated on: 10-Oct-2022

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