# How do you rationalize the denominator and simplify (5sqrt3-3sqrt2)/(3sqrt2-2sqrt3)?

Rewrite this as follows

(5sqrt3-3sqrt2)/(3sqrt2-2sqrt3)=[sqrt3*(5-sqrt3*sqrt2)]/[sqrt3*sqrt2*[sqrt3-sqrt2]]= 1/sqrt2*[5-sqrt3*sqrt2]/[sqrt3-sqrt2]

Multiply denominator and nominator with $\sqrt{3} + \sqrt{2}$ hence

1/sqrt2*[5-sqrt3*sqrt2]/[sqrt3-sqrt2]=1/sqrt2*[(5-sqrt3*sqrt2)*(sqrt3+sqrt2)]/[(sqrt3-sqrt2)*(sqrt3+sqrt2)]= 1/(sqrt2)*[(5-sqrt3*sqrt2)*(sqrt3+sqrt2)]= 1/(sqrt2)*[2sqrt2+3sqrt3]

Mar 18, 2016

Multiply the numerator & denominator by the conjugate of the denominator.

#### Explanation:

Anytime you have "square root plus something" in the denominator, you multiply both num & denom by it's conjugate, i.e. change the sign in the middle.

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

So the new denominator is
( $3 \sqrt{2} - 2 \sqrt{3}$ ) * ($3 \sqrt{2} + 2 \sqrt{3}$) = $9 \left(2\right) - 4 \left(3\right) = 6$, using $\left(A + B\right) \left(A - B\right) = {A}^{2} - {B}^{2}$

The new numerator is $\left(5 \sqrt{3} - 3 \sqrt{2}\right) \left(3 \sqrt{2} + 2 \sqrt{3}\right)$
which (after FOIL) is
$15 \sqrt{6} + 10 \left(3\right) - 9 \left(2\right) - 6 \sqrt{6}$ = $= 9 \sqrt{6} + 12$
So over the new denominator we have
$= \frac{9 \sqrt{6} + 12}{6}$=$\frac{3 \sqrt{6} + 4}{2}$