Question

if x= √5+√2/√5-√2 and y=√5-√2/√5+√2,show that 3a^2+4ab -3b^2=4+56/3√10

Answer 1

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Answer 2

Answer:

Given:  $a=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\:\:and\:\:b=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$

To Show: 3a² + 4ab - 3b² = $4+\frac{56\sqrt{10}}{3}$

Consider,

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

$=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}$

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

$=\frac{5+2+2\sqrt{10}}{5-2}$

$=\frac{7+2\sqrt{10}}{3}$

$\implies a^2=(\frac{7+2\sqrt{10}}{3})^2=\frac{7^2+(2\sqrt{10})^2+28\sqrt{10}}{9}=\frac{89+28\sqrt{10}}{9}$

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

$=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}\times\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

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

$=\frac{5+2-2\sqrt{10}}{5-2}$

$=\frac{7-2\sqrt{10}}{3}$

$\implies b^2=(\frac{7-2\sqrt{10}}{3})^2=\frac{7^2+(2\sqrt{10})^2-28\sqrt{10}}{9}=\frac{89-28\sqrt{10}}{9}$

$ab=(\frac{7+2\sqrt{10}}{3})(\frac{7-2\sqrt{10}}{3})=\frac{7^2-(2\sqrt{10})^2}{3^2}=\frac{49-40}{9}=1$

Therefore, $3a^2+4ab-3b^2=3(\frac{89+28\sqrt{10}}{9})+4-3(\frac{89-28\sqrt{10}}{9})=\frac{89}{3}+\frac{28\sqrt{10}}{3}+4-\frac{89}{3}+\frac{28\sqrt{10}}{3}=4+\frac{56\sqrt{10}}{3}$

Hence Proved.