rationalize denominator

**euclid2** · April 14th 2010, 05:08 PM

rationalize the denominator $\frac{\sqrt3 - \sqrt 2}{\sqrt 3 + \sqrt 2}$

**harish21** · April 14th 2010, 05:10 PM

Originally Posted by euclid2

rationalize the denominator $\frac{\sqrt3 - \sqrt 2}{\sqrt 3 + \sqrt 2}$

Multiply the numerator and denominator by $\sqrt3 - \sqrt 2$

**euclid2** · April 14th 2010, 05:11 PM

Originally Posted by harish21

Multiply the numerator and denominator by $\sqrt3 - \sqrt 2$

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

**harish21** · April 14th 2010, 05:14 PM

Originally Posted by euclid2

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

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

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

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

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

Can you finish it now?

**euclid2** · April 14th 2010, 05:17 PM

Originally Posted by harish21

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

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

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

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

Can you finish it now?

i keep getting 5/1 but the answer says 5-2root6

**harish21** · April 14th 2010, 05:19 PM

Originally Posted by euclid2

i keep getting 5/1 but the answer says 5-2root6

Remember: $(a-b)^2 = a^2 -2ab + b^2<br />$

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

$= 3 - 2 \sqrt6 +2$

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