Question

Please answer fast..​

Answer 1

Solution :-

Given :-

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

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

First we have to rationalise the denominators :-

For a :-

$a = \dfrac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \times \dfrac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}}$

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

$a = \dfrac{ 5 + 2\sqrt{10} + 2}{5 - 2}$

$a = \dfrac{7 + 2\sqrt{10}}{3}$

For b :-

$b = \dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} \times \dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}}$

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

$b= \dfrac{ 5 - 2\sqrt{10} + 2}{5 - 2}$

$b = \dfrac{7 -2\sqrt{10}}{3}$

Now

3a² + 4ab - 3b²

$\small{= 3.\left( \dfrac{7 + 2\sqrt{10}}{3}\right)^2 + 4.\left( \dfrac{7 + 2\sqrt{10}}{3}\right)\left( \dfrac{7 - 2\sqrt{10}}{3}\right) - 3.\left( \dfrac{7 - 2\sqrt{10}}{3}\right)^2}$

$\small{ = \left(\dfrac{49 + 28\sqrt{10} + 40}{3}\right) + 4.\left( \dfrac{(7)^2 - (2\sqrt{10})^2}{3^2}\right) - \left(\dfrac{49 - 28\sqrt{10} + 40}{3}\right)}$

$\small{ = \dfrac{89 + 28\sqrt{10} - 89 + 28\sqrt{10}}{3} + 4.\left(\dfrac{49 - 40}{9}\right)}$

$= \dfrac{56\sqrt{10}}{3} + \dfrac{4 \times 9 }{9}$

$= 4 + \dfrac{56\sqrt{10}}{3}$