How do you rationalize the denominator and simplify $\frac{8}{3 - \sqrt{2}}$ $8/(3-sqrt2)$ ?

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Answer 1 · 2018-03-02T01:57:43

The answer is $\frac{24 + 8 \sqrt{2}}{7}$ $(24+8sqrt2)/7$ .

To rationalize the denominator of a fraction, multiply both the numerator and the denominator by the denominator's conjugate.

The conjugate of $3 - \sqrt{2}$ $3-sqrt2$ is $3 + \sqrt{2}$ $3+sqrt2$

$= \frac{8}{3 - \sqrt{2}}$ $color(white)=8/(3-sqrt2)$

$= \frac{8}{(3 - \sqrt{2})} \cdot \frac{(3 + \sqrt{2})}{(3 + \sqrt{2})}$ $=8/((3-sqrt2))color(red)(*((3+sqrt2))/((3+sqrt2)))$

$= \frac{8 (3 + \sqrt{2})}{(3 - \sqrt{2}) (3 + \sqrt{2})}$ $=(8(3+sqrt2))/((3-sqrt2)(3+sqrt2))$

$=(24+8sqrt2)/(3^2+color(red)cancel(color(black)(3sqrt2-3sqrt2))-(sqrt2)^2)$

$=(24+8sqrt2)/(9-2)$

$=(24+8sqrt2)/7$

This answer is as simplified as possible.

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