How do you simplify #(3-4sqrt3)/(4sqrt5+3sqrt2)#?

1 Answer
Shwetank Mauria
Jan 24, 2017

#(3-4sqrt3)/(4sqrt5+3sqrt2)=(12sqrt5-9sqrt2-16sqrt15+12sqrt6)/62#

Explanation:

To simplify #(3-4sqrt3)/(4sqrt5+3sqrt2)#, what we need to do is to multiply numerator and denominator by the conjugate of its irrational denominator.

Conjugate of a irrational number #(sqrta+-sqrtb)# is #(sqrta∓sqrtb)#. (If we just have #sqrtp# in denominator, then just multiplying numerator and denominator by #sqrtp# suffices.)

Hence, #(3-4sqrt3)/(4sqrt5+3sqrt2)#

= #((3-4sqrt3)(4sqrt5-3sqrt2))/((4sqrt5+3sqrt2)(4sqrt5-3sqrt2))#

= #(3(4sqrt5-3sqrt2)-4sqrt3(4sqrt5-3sqrt2))/((4sqrt5)^2-(3sqrt2)^2)#

= #(12sqrt5-9sqrt2-16sqrt15+12sqrt6)/(80-18)#

= #(12sqrt5-9sqrt2-16sqrt15+12sqrt6)/62#

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
890 views around the world
You can reuse this answer
Creative Commons License