1 Answer

Fat Shogun · Answer 1 · 2021-12-29T09:55:51+0000

The trick is to exploit the difference of squares formula,

a^2-b^2=(a-b)(a+b)

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

$(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2$

Whatever you do to the denominator, you have to do to the numerator too. So

$(\sqrt2+\sqrt6)/(\sqrt8+\sqrt6)=((\sqrt2+\sqrt6)(\sqrt8-\sqrt6))/((\sqrt8+\sqrt6)(\sqrt8-\sqrt6))=\frac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2$

Expand the numerator:

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=√(2\cdot8)+√(6\cdot8)-√(2\cdot6)-(\sqrt6)^2$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=√(16)+√(48)-√(12)-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+√(48)-√(12)-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+√(12)(\sqrt4-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+√(12)(2-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-√(12)$

So we have

$(\sqrt2+\sqrt6)/(\sqrt8+\sqrt6)=-\frac{2+√(12)}2$

But √12 = √(3•4) = 2√3, so

$(\sqrt2+\sqrt6)/(\sqrt8+\sqrt6)=-\frac{2+2\sqrt3}2=\boxed{-1-\sqrt3}$

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