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

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Answer 1 · 2015-06-02T22:36:53

Multiply both numerator (top) and denominator (bottom) by the conjugate $(2 + \sqrt{2})$ $(2+sqrt(2))$ ...

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

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

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

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

$= \frac{6 + 3 \sqrt{2}}{4 - 2}$ $=(6+3sqrt(2))/(4-2)$

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

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

How do you rationalize the denominator 3/(2-sqrt2)32−√23/(2-sqrt2)?