Question

Express 21 by 3√5+√3 with Rationalising denomination

Answer 1

Answer:

$\frac{21}{3 \sqrt{5} + \sqrt{3} } = \frac{21}{3 \sqrt{5} + \sqrt{3} } \times \frac{3 \sqrt{5} - \sqrt{3} }{3 \sqrt{5} - \sqrt{3} } \\ \: \: \: = \frac{21(3 \sqrt{5} - \sqrt{3} )}{(3 \sqrt{5}) {}^{2} - ( \sqrt{3}) {}^{2} } \\ \: \: \: = \frac{63 \sqrt{5} - 21 \sqrt{3} }{45 - 3} \\ \: \: \: = \frac{63 \sqrt{5} - 21 \sqrt{3} }{42}$

Answer 2

Answer :

$\sf \dfrac{2\sqrt{5}-\sqrt{3}}{2}$

Solution :

$\sf \dfrac{21}{3\sqrt{5} + \sqrt{3}}$

$\sf = \dfrac{21}{3\sqrt{5}+\sqrt{3}} \times \dfrac{3\sqrt{5}-\sqrt{3}}{3\sqrt{5}-\sqrt{3}}$

∵ a² - b² = ( a + b ) ( a - b )

$\sf = \dfrac{21(3\sqrt{5}-\sqrt{3})}{(3\sqrt{5})^2 - (\sqrt{3})^2}$

$\sf = \dfrac{63\sqrt{5}-21\sqrt{3}}{45-3}$

$\sf = \dfrac{21(2\sqrt{5}-\sqrt{3})}{42}$

$\sf = \dfrac{7(2\sqrt{5}-\sqrt{3})}{14}$

Answer : $\sf = \dfrac{2\sqrt{5}-\sqrt{3}}{2}$

$\rule{200}2$