Question

(√3 +1)(1-√12)+9/(√3+√12) Simplify.

Answer 1

Identity used :

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

First, Simplifying √12 = √2 × 2 × 3

i.e. √2² × 3 = 2√3

Now,

$\frac{( \sqrt{3} + 1)(1 - 2\sqrt{3}) + 9 }{ \sqrt{3} + 2\sqrt{3} } \\ \\ = \frac{ \sqrt{3}(1 - 2 \sqrt{3}) + 1(1 - 2 \sqrt{3}) + 9 }{ \sqrt{3} + 2 \sqrt{3} } \\ \\ = \frac{ \sqrt{3} - 2 \sqrt{ {3}^{2} } + 1 - 2 \sqrt{3} + 9}{ \sqrt{3} + 2 \sqrt{3} } \\ \\ = \frac{ \sqrt{3} - 2 \times 3 + 1 + 9 - 2 \sqrt{3} }{ \sqrt{3} + 2 \sqrt{3} } \\ \\ = \frac{ \sqrt{3} - 2 \sqrt{3} - 6 + 1 + 9 }{ \sqrt{3} + 2 \sqrt{3} } \\ \\ = \frac{ - \sqrt{3} + 4 }{ \sqrt{3} + 2 \sqrt{3} } \\ \\ = \frac{4 - \sqrt{3} }{ \sqrt{3} + 2 \sqrt{3} }$

Rationalizing the denominator we get,

$= \frac{4 - \sqrt{3} }{ \sqrt{3} + 2 \sqrt{3} } \times \frac{ \sqrt{3} - 2 \sqrt{3} }{ \sqrt{3} - 2 \sqrt{3} } \\ \\ = \frac{4( \sqrt{3} - 2 \sqrt{3} ) - \sqrt{3} ( \sqrt{3} - 2 \sqrt{3}) }{ {( \sqrt{3} )}^{2} - {(2 \sqrt{3}) }^{2} } \\ \\ = \frac{4 \sqrt{3} - 8 \sqrt{3} - 3 + 2 \times 3 }{3 - 4 \times 3} \\ \\ = \frac{ - 4 \sqrt{3} - 3 + 6}{3 - 12} \\ \\ = \frac{ - 4 \sqrt{3} + 3 }{ - 9} \\ \\ = \frac{4 \sqrt{3} - 3}{9}$

Hope it helps! ;)