Question 168

$$(1+\sqrt{6})^{2}=?+2\sqrt{6}$$

Solution
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We know that,
$$(a+b)^{2}=a^{2}+b^{2}+2\times{a}\times{b}$$.
Hence,
$$(1+\sqrt{6})^{2}=1+6+2\times{1}\times{\sqrt{6}}$$.
= $$7+2\sqrt{6}$$.
Hence, Option A is correct.

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