Ok I really need help on this. Can anybody help me Rationalize the denominator of (14)/(3+sqrt(2))?
A angala New member Joined Jun 27, 2010 Messages 2 Jun 27, 2010 #1 Ok I really need help on this. Can anybody help me Rationalize the denominator of (14)/(3+sqrt(2))?
D Deleted member 4993 Guest Jun 27, 2010 #2 angala said: Ok I really need help on this. Can anybody help me Rationalize the denominator of (14)/(3+sqrt(2))? Click to expand... Use \(\displaystyle (a + b)\cdot (a - b) \ = \ a^2 - b^2\) so \(\displaystyle (3 + \sqrt{2})\cdot (3 - \sqrt{2}) \ \ = \ \ 3^2 - (\sqrt{2})^2\) <<< Corrected "sign" \(\displaystyle \frac{14}{3+\sqrt{2}}\) \(\displaystyle = \ \ \frac{14}{3+\sqrt{2}} \cdot \frac{3-\sqrt{2}}{3-\sqrt{2}}\) \(\displaystyle = \ \ \frac{14\cdot (3-\sqrt{2})}{3^2-2}\) \(\displaystyle = \ \ 2\cdot (3-\sqrt{2})\)
angala said: Ok I really need help on this. Can anybody help me Rationalize the denominator of (14)/(3+sqrt(2))? Click to expand... Use \(\displaystyle (a + b)\cdot (a - b) \ = \ a^2 - b^2\) so \(\displaystyle (3 + \sqrt{2})\cdot (3 - \sqrt{2}) \ \ = \ \ 3^2 - (\sqrt{2})^2\) <<< Corrected "sign" \(\displaystyle \frac{14}{3+\sqrt{2}}\) \(\displaystyle = \ \ \frac{14}{3+\sqrt{2}} \cdot \frac{3-\sqrt{2}}{3-\sqrt{2}}\) \(\displaystyle = \ \ \frac{14\cdot (3-\sqrt{2})}{3^2-2}\) \(\displaystyle = \ \ 2\cdot (3-\sqrt{2})\)
