How do you rationalize the denominator and simplify #sqrt(3)/(sqrt(7)-2)#?

1 Answer
Mar 2, 2018

Answer:

#(sqrt21+2sqrt3)/3#

Explanation:

To rationalize the denominator, the best move is to multiply the numerator and denominator by the reciprocal of the denominator (which is the denominator but with the middle sign reversed).

#sqrt3/(sqrt7-2) * color(blue)((sqrt7+2)/(sqrt7+2))#

#(sqrt(3)(sqrt(7)+2))/((sqrt(7)-2)(sqrt(7)+2)#

From here, we can simplify the numerator and denominator.

#sqrt(alpha)sqrt(beta)=sqrt(alphabeta) -> sqrt3sqrt7=sqrt21#

#sqrt3(sqrt7+2)=color(red)(sqrt21+2sqrt3)#

#(sqrt7-2)(sqrt7+2) = 7+cancel(2sqrt7-2sqrt7)-4 = color(red)3#

#sqrt3/(sqrt7-2)=color(red)((sqrt21+2sqrt3)/3)#