Question
If $\text{a}=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ and $\text{b}=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}},$ show that $3\text{a}^2+4\text{ab}-3\text{b}^2=4+\frac{56}{3}\sqrt{10}.$
✓
Answer
| $\text{a}=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ | $\text{b}=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ |
| $=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ | $=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}\times\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ |
| $=\frac{\big(\sqrt{5}+\sqrt{2}\big)^2}{\big(\sqrt{5}\big)^2-\big(\sqrt{2}\big)^2}$ | $=\frac{\big(\sqrt{5}-\sqrt{2}\big)^2}{\big(\sqrt{5}\big)^2-\big(\sqrt{2}\big)^2}$ |
| $=\frac{5+2\sqrt{10}+2}{5-2}$ | $=\frac{5-2\sqrt{10}+2}{5-2}$ |
| $=\frac{7+2\sqrt{10}}{3}$ | $=\frac{7-2\sqrt{10}}{3}$ |
| Now, | |
| $\text{a}^2=\Big(\frac{7+2\sqrt{10}}{3}\Big)^2$ | $\text{b}^2=\Big(\frac{7-2\sqrt{10}}{3}\Big)^2$ |
| $=\frac{49+28\sqrt{10}+40}{9}$ | $=\frac{49-28\sqrt{10}+40}{9}$ |
| $=\frac{89+28\sqrt{10}}{9}$ | $=\frac{89-28\sqrt{10}}{9}$ |
$\text{L.H.S.}=3\text{a}^2+4\text{ab}-3\text{b}^2$
$=3\times\frac{89+28\sqrt{10}}{9}+4\times\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}-3\times\frac{89-28\sqrt{10}}{9}$
$=\frac{89+28\sqrt{10}}{3}+4-\frac{89-28\sqrt{10}}{3}$
$=4-\frac{89+28\sqrt{10}-89+28\sqrt{10}}{3}$
$=4+\frac{56}{3}\sqrt{10}$
$=\text{R.H.S.}$
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