How do you rationalize the denominator 3/(2-sqrt2)?

Jun 2, 2015

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

$\frac{3}{2 - \sqrt{2}}$

$= \frac{3}{2 - \sqrt{2}} \cdot \frac{2 + \sqrt{2}}{2 + \sqrt{2}}$

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

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

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

$= \frac{6 + 3 \sqrt{2}}{2}$

$= 3 + \frac{3}{2} \sqrt{2}$