How do you simplify $\frac{- 2 + \sqrt{8}}{- 3 - \sqrt{2}}$ $(-2+sqrt8)/(-3-sqrt2)$ ?

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Answer 1 · 2015-06-09T21:11:38

Multiply numerator and denominator by $(3 - \sqrt{2})$ $(3-sqrt(2))$ , expand, group, combine and eliminate common factors to get:

$\frac{- 2 + \sqrt{8}}{- 3 - \sqrt{2}} = 1 - \sqrt{2}$ $(-2+sqrt(8))/(-3-sqrt(2)) = 1-sqrt(2)$

$\frac{- 1 + \sqrt{8}}{- 3 - \sqrt{2}}$ $(-1+sqrt(8))/(-3-sqrt(2))$

$= \frac{1 - \sqrt{8}}{3 + \sqrt{2}}$ $=(1-sqrt(8))/(3+sqrt(2))$

$= \frac{1 - \sqrt{8}}{3 + \sqrt{2}} \cdot \frac{3 - \sqrt{2}}{3 - \sqrt{2}}$ $=(1-sqrt(8))/(3+sqrt(2))*(3-sqrt(2))/(3-sqrt(2))$

$= \frac{(1 - 2 \sqrt{2}) \cdot (3 - \sqrt{2})}{3^{2} - {(\sqrt{2})}^{2}}$ $=((1-2sqrt(2))*(3-sqrt(2)))/(3^2-(sqrt(2))^2)$

[using the difference of squares identity: $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2 = (a+b)(a-b)$ ]

$= \frac{3 - \sqrt{2} - 6 \sqrt{2} + 2 {(\sqrt{2})}^{2}}{9 - 2}$ $=(3-sqrt(2)-6sqrt(2)+2(sqrt(2))^2)/(9-2)$

$= \frac{3 - 7 \sqrt{2} + 4}{7}$ $=(3-7sqrt(2)+4)/7$

$= \frac{7 - 7 \sqrt{2}}{7}$ $=(7-7sqrt(2))/7$

$= 1 - \sqrt{2}$ $=1-sqrt(2)$

How do you simplify (-2+sqrt8)/(-3-sqrt2)−2+√8−3−√2(-2+sqrt8)/(-3-sqrt2)?