Question

Rationalize the denominator of 4/2+root3+root7

Answer 1

Heya !!

Here's your answer.. ⬇⬇

______________________________

$= \frac{4}{2 + \sqrt{3} + \sqrt{7} } \\ \\ = \frac{4}{(2 + \sqrt{3)} + \sqrt{7} } \\ \\ = \frac{4}{(2 + \sqrt{3)} + \sqrt{7} } \times \frac{(2 + \sqrt{3)} - \sqrt{7} }{(2 + \sqrt{3)} - \sqrt{7}} \\ \\= = > \: {a}^{2} - {b}^{2} = (a + b)(a - b) \\ \\ = \frac{4(2 + \sqrt{3} - \sqrt{7}) }{ {(2 + \sqrt{3}) }^{2} - {( \sqrt{7} )}^{2} } \\ \\ = \frac{4(2 + \sqrt{3} - \sqrt{7})}{4 + 4 \sqrt{3} + 3 - 7} \\ \\ = \frac{4(2 + \sqrt{3} - \sqrt{7})}{4 \sqrt{3} + 7 - 7 } \\ \\ = \frac{4(2 + \sqrt{3} - \sqrt{7})}{4 \sqrt{3} } \\ \\ = \frac{(2 + \sqrt{3} - \sqrt{7})}{ \sqrt{3} } \\ \\ = \frac{(2 + \sqrt{3} - \sqrt{7})}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ \\ = \frac{2 \sqrt{3} + 3 - \sqrt{21} }{3}$

________________________________
Hope it helps

Thanks :))

Answer 2

Answer: Our rationalized form is

$\frac{1}{12}(8\sqrt{3}+12-16\sqrt{21})$

Step-by-step explanation:

Since we have given that

$\frac{4}{2+\sqrt{3}+\sqrt{7}}$

We will rationalizing the denominator:

$\frac{4}{2+\sqrt{3}+\sqrt{7}}\\\\=\frac{4}{(2+\sqrt{3})+\sqrt{7}}\times \frac{(2+\sqrt{3}-\sqrt{7})}{(2+\sqrt{3}-\sqrt{7})}\\\\=\frac{4(2+\sqrt{3}-\sqrt{7})}{(2+\sqrt{3})^2-7}\\\\=\frac{8+4\sqrt{3}-16\sqrt{7}}{4+3+4\sqrt{3}-7}\\\\=\frac{(8+4\sqrt{3}-16\sqrt{7})\times \sqrt{3}}{4\sqrt{3}\times \sqrt{3}}\\\\=\frac{8\sqrt{3}+12-16\sqrt{21}}{12}\\\\=\frac{1}{12}(8\sqrt{3}+12-16\sqrt{21})$

Hence, our rationalized form is

$\frac{1}{12}(8\sqrt{3}+12-16\sqrt{21})$