Question:
The value of $\sqrt{5+2 \sqrt{6}}$ is
(a) $\sqrt{5}+\sqrt{6}$
(b) $\sqrt{5}-\sqrt{6}$
(c) $\sqrt{3}+\sqrt{2}$
(d) $\sqrt{3}-\sqrt{2}$
Solution:
$5+2 \sqrt{6}=2+3+2 \times \sqrt{3} \times \sqrt{2}$
$=(\sqrt{2})^{2}+(\sqrt{3})^{2}+2 \times \sqrt{3} \times \sqrt{2}$
This is in the form
$a^{2}+b^{2}+2 a b=(a+b)^{2}$
So, we have
$(\sqrt{2})^{2}+(\sqrt{3})^{2}+2 \times \sqrt{3} \times \sqrt{2}=(\sqrt{2}+\sqrt{3})^{2}$
Thus, $\sqrt{5+2 \sqrt{6}}=\sqrt{(\sqrt{2}+\sqrt{3})^{2}}=\sqrt{2}+\sqrt{3}$
Hence, the correct answer is option (c).
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