I need to rationalize the denominator and simplify 6 over the sq root of 2 minus the sq root of 3.
Simplify the following:
6/(sqrt(2)-sqrt(3))
Multiply numerator and denominator of 6/(sqrt(2)-sqrt(3)) by -1:
-(6)/(sqrt(3)-sqrt(2))
Multiply numerator and denominator of (-6)/(sqrt(3)-sqrt(2)) by sqrt(2)+sqrt(3):
(-6 (sqrt(2)+sqrt(3)))/((sqrt(3)-sqrt(2)) (sqrt(2)+sqrt(3)))
(sqrt(3)-sqrt(2)) (sqrt(2)+sqrt(3)) = -sqrt(2) sqrt(2)-sqrt(2) sqrt(3)+sqrt(3) sqrt(2)+sqrt(3) sqrt(3) = -2-sqrt(6)+sqrt(6)+3 = 1:
(-6 (sqrt(2)+sqrt(3)))/(1)
(-6 (sqrt(2)+sqrt(3)))/(1) = -6 (sqrt(2)+sqrt(3)):
Answer: | -6 (sqrt(2)+sqrt(3)) / 1
