# rationalize denominator

• Apr 14th 2010, 05:08 PM
euclid2
rationalize denominator
rationalize the denominator $\displaystyle \frac{\sqrt3 - \sqrt 2}{\sqrt 3 + \sqrt 2}$
• Apr 14th 2010, 05:10 PM
harish21
Quote:

Originally Posted by euclid2
rationalize the denominator $\displaystyle \frac{\sqrt3 - \sqrt 2}{\sqrt 3 + \sqrt 2}$

Multiply the numerator and denominator by $\displaystyle \sqrt3 - \sqrt 2$
• Apr 14th 2010, 05:11 PM
euclid2
Quote:

Originally Posted by harish21
Multiply the numerator and denominator by $\displaystyle \sqrt3 - \sqrt 2$

i know to multiply by the conjugate, i just can't get the right answer
• Apr 14th 2010, 05:14 PM
harish21
Quote:

Originally Posted by euclid2
i know to multiply by the conjugate, i just can't get the right answer

$\displaystyle \frac{\sqrt3 - \sqrt 2}{\sqrt 3 + \sqrt 2}$

$\displaystyle = \frac{\sqrt3 - \sqrt 2}{\sqrt 3 + \sqrt 2} \times \frac{\sqrt3 - \sqrt 2}{\sqrt 3 - \sqrt 2}$

$\displaystyle = \frac{(\sqrt3 - \sqrt2)^2}{(\sqrt3)^2 -(\sqrt2)^2}$

$\displaystyle = \frac{(\sqrt3 - \sqrt2)^2}{3-2}$

Can you finish it now?
• Apr 14th 2010, 05:17 PM
euclid2
Quote:

Originally Posted by harish21
$\displaystyle \frac{\sqrt3 - \sqrt 2}{\sqrt 3 + \sqrt 2}$

$\displaystyle = \frac{\sqrt3 - \sqrt 2}{\sqrt 3 + \sqrt 2} \times \frac{\sqrt3 - \sqrt 2}{\sqrt 3 - \sqrt 2}$

$\displaystyle = \frac{(\sqrt3 - \sqrt2)^2}{(\sqrt3)^2 -(\sqrt2)^2}$

$\displaystyle = \frac{(\sqrt3 - \sqrt2)^2}{3-2}$

Can you finish it now?

i keep getting 5/1 but the answer says 5-2root6
• Apr 14th 2010, 05:19 PM
harish21
Quote:

Originally Posted by euclid2
i keep getting 5/1 but the answer says 5-2root6

Remember: $\displaystyle (a-b)^2 = a^2 -2ab + b^2$

$\displaystyle \therefore (\sqrt3 - \sqrt2)^2 = (\sqrt3)^2 - 2 \sqrt3 \sqrt2 +(\sqrt2)^2$

$\displaystyle = 3 - 2 \sqrt6 +2$