Question:
Find the value
$2 a^{2}+2 \sqrt{6} a b+3 b^{2}$
Solution:
$=(\sqrt{2} a)^{2}+2 \times \sqrt{2} a \times \sqrt{3} b+(\sqrt{3} b)^{2}$
Using the identity $(p+q)^{2}=p^{2}+q^{2}+2 p q$
$=(\sqrt{2} a+\sqrt{3} b)^{2}$
$=(\sqrt{2} a+\sqrt{3} b)(\sqrt{2} a+\sqrt{3} b)$
$\therefore 2 a^{2}+2 \sqrt{6} a b+3 b^{2}$
$=(\sqrt{2} a+\sqrt{3} b)(\sqrt{2} a+\sqrt{3} b)$
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