MCQ
The number $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$
- Aan integer
- Bnot a real number
- ✓an irrational number
- Da rational number
✓
Answer
Correct option: C.
an irrational number
$\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$
$=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times$$\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}$
$=\frac{\big(\sqrt{5}+\sqrt{2}\big)^2}{\big(\sqrt{5}\big)^2-\big(\sqrt{2}\big)^2}$
$=\frac{\big(\sqrt{5}\big)^2+\big(\sqrt{2}\big)^2+2\times\sqrt{5}\times\sqrt{2}}{5-2}$
$=\frac{5+2+2\sqrt{10}}{3}$
$=\frac{7+2\sqrt{10}}{3}$
Here$\sqrt{10}=\sqrt{2}\times\sqrt{5}$
since $\sqrt{2}$ and $\sqrt{5}$ both are an irrational number
Therefore, $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5-\sqrt{2}}}$ is an irrational number.
$=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times$$\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}$
$=\frac{\big(\sqrt{5}+\sqrt{2}\big)^2}{\big(\sqrt{5}\big)^2-\big(\sqrt{2}\big)^2}$
$=\frac{\big(\sqrt{5}\big)^2+\big(\sqrt{2}\big)^2+2\times\sqrt{5}\times\sqrt{2}}{5-2}$
$=\frac{5+2+2\sqrt{10}}{3}$
$=\frac{7+2\sqrt{10}}{3}$
Here$\sqrt{10}=\sqrt{2}\times\sqrt{5}$
since $\sqrt{2}$ and $\sqrt{5}$ both are an irrational number
Therefore, $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5-\sqrt{2}}}$ is an irrational number.
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