Question

If $a=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\mathrm{and}b=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$, show that $3a^2+4ab-3b^2=4+\frac{56}{3}\sqrt{10}$.

Answer 1

According to question,
$a=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\mathrm{and}b=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$

$$\begin{gathered} a=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\ =\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times \frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}} \\ =\frac{{\left(\sqrt{5}+\sqrt{2}\right)}^2}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{{\left(\sqrt{5}\right)}^2+{\left(\sqrt{2}\right)}^2+2\sqrt{5}\sqrt{2}}{5-2} \\ =\frac{5+2+2\sqrt{10}}{3} \\ =\frac{7+2\sqrt{10}}{3}...\left(1\right) \\ b=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}} \\ =\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\ =\frac{{\left(\sqrt{5}-\sqrt{2}\right)}^2}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{{\left(\sqrt{5}\right)}^2+{\left(\sqrt{2}\right)}^2-2\sqrt{5}\sqrt{2}}{5-2} \\ =\frac{5+2-2\sqrt{10}}{3} \\ =\frac{7-2\sqrt{10}}{3}...\left(2\right) \end{gathered}$$
Now,
$$\begin{gathered} 3a^2+4ab-3b^2 \\ =3\left(a^2-b^2\right)+4ab \\ =3\left(a+b\right)\left(a-b\right)+4ab \\ =3\left(\frac{7+2\sqrt{10}}{3}+\frac{7-2\sqrt{10}}{3}\right)\left(\frac{7+2\sqrt{10}}{3}-\frac{7-2\sqrt{10}}{3}\right)+4\left(\frac{7+2\sqrt{10}}{3}\times \frac{7-2\sqrt{10}}{3}\right) \\ =3\left(\frac{14}{3}\right)\left(\frac{4\sqrt{10}}{3}\right)+4\left(\frac{{\left(7\right)}^2-{\left(2\sqrt{10}\right)}^2}{9}\right) \\ =\frac{56}{3}\sqrt{10}+4\left(\frac{49-40}{9}\right) \\ =\frac{56}{3}\sqrt{10}+4 \end{gathered}$$
Hence, $3a^2+4ab-3b^2=4+\frac{56\sqrt{10}}{3}$.