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 (a+b)⋅(a−b) = a2−b2\displaystyle (a + b)\cdot (a - b) \ = \ a^2 - b^2(a+b)⋅(a−b) = a2−b2 so (3+2)⋅(3−2) = 32−(2)2\displaystyle (3 + \sqrt{2})\cdot (3 - \sqrt{2}) \ \ = \ \ 3^2 - (\sqrt{2})^2(3+2)⋅(3−2) = 32−(2)2 <<< Corrected "sign" 143+2\displaystyle \frac{14}{3+\sqrt{2}}3+214 = 143+2⋅3−23−2\displaystyle = \ \ \frac{14}{3+\sqrt{2}} \cdot \frac{3-\sqrt{2}}{3-\sqrt{2}}= 3+214⋅3−23−2 = 14⋅(3−2)32−2\displaystyle = \ \ \frac{14\cdot (3-\sqrt{2})}{3^2-2}= 32−214⋅(3−2) = 2⋅(3−2)\displaystyle = \ \ 2\cdot (3-\sqrt{2})= 2⋅(3−2)
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 (a+b)⋅(a−b) = a2−b2\displaystyle (a + b)\cdot (a - b) \ = \ a^2 - b^2(a+b)⋅(a−b) = a2−b2 so (3+2)⋅(3−2) = 32−(2)2\displaystyle (3 + \sqrt{2})\cdot (3 - \sqrt{2}) \ \ = \ \ 3^2 - (\sqrt{2})^2(3+2)⋅(3−2) = 32−(2)2 <<< Corrected "sign" 143+2\displaystyle \frac{14}{3+\sqrt{2}}3+214 = 143+2⋅3−23−2\displaystyle = \ \ \frac{14}{3+\sqrt{2}} \cdot \frac{3-\sqrt{2}}{3-\sqrt{2}}= 3+214⋅3−23−2 = 14⋅(3−2)32−2\displaystyle = \ \ \frac{14\cdot (3-\sqrt{2})}{3^2-2}= 32−214⋅(3−2) = 2⋅(3−2)\displaystyle = \ \ 2\cdot (3-\sqrt{2})= 2⋅(3−2)
