Question

find 2 one answer.plz solve this answer with step by step.plz.​

Answer 1

Solution :-

As given :-

$\dfrac{\sqrt{5} + \sqrt{3}}{2\sqrt{5} - 3\sqrt{3}} = a - b\sqrt{15}$

Now via Rationalising Denominator :-

$\dfrac{\sqrt{5} + \sqrt{3}}{2\sqrt{5} - 3\sqrt{3}} \times \dfrac{2\sqrt{5} + 3\sqrt{3}}{2\sqrt{5} + 3\sqrt{3}}$

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

$= \dfrac{ 2(5) + 3\sqrt{15} + 2 \sqrt{15} + 3(3)}{4(5) - 9(3)}$

$= \dfrac{ 10 + 5\sqrt{15} + 9}{20 - 27}$

$= \dfrac{19 + 5\sqrt{15}}{-7}$

$= \dfrac{(-19)}{7} - \dfrac{5\sqrt{15}}{7}$

Now as

$\dfrac{(-19)}{7} - \dfrac{5\sqrt{15}}{7} = a - b\sqrt{15}$

By comparing both sides

$\huge{\boxed{\sf{ a = \dfrac{-19}{7}}}}$

$\huge{\boxed{\sf{ b = \dfrac{5}{7}}}}$

Answer 2

As given :-

$\dfrac{\sqrt{5} + \sqrt{3}}{2\sqrt{5} - 3\sqrt{3}} = a - b\sqrt{15}$

Now via Rationalising

Denominator :-

$\dfrac{\sqrt{5} + \sqrt{3}}{2\sqrt{5} - 3\sqrt{3}} \times \dfrac{2\sqrt{5} + 3\sqrt{3}}{2\sqrt{5} + 3\sqrt{3}}$

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

= $\dfrac{ 2(5) + 3\sqrt{15} + 2 \sqrt{15} + 3(3)}{4(5) - 9(3)}$

= $\dfrac{ 10 + 5\sqrt{15} + 9}{20 - 27}$

=$\dfrac{19 + 5\sqrt{15}}{-7}$

= $\dfrac{(-19)}{7} - \dfrac{5\sqrt{15}}{7}$

Now as

$\dfrac{(-19)}{7} - \dfrac{5\sqrt{15}}{7} = a - b\sqrt{15}$

By comparing both sides

$\huge{\boxed{\sf{ a = \dfrac{-19}{7}}}}$

$\huge{\boxed{\sf{ b = \dfrac{5}{7}}}}$