How do you simplify #4/(sqrt2-5sqrt3)#?

1 Answer
George C.
Aug 3, 2017

#4/(sqrt(2)-3sqrt(5)) = -4/43(sqrt(2)+3sqrt(5)) = -4/43sqrt(2)-12/43sqrt(5)#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Using this with #a=sqrt(2)# and #b=5sqrt(3)#, we can rationalise the denominator of the given expression by multiplying both numerator and denominator by #sqrt(2)+5sqrt(5)#...

#4/(sqrt(2)-3sqrt(5)) = (4(sqrt(2)+3sqrt(5)))/((sqrt(2)-3sqrt(5))(sqrt(2)+3sqrt(5)))#

#color(white)(4/(sqrt(2)-3sqrt(5))) = (4(sqrt(2)+3sqrt(5)))/((sqrt(2))^2-(3sqrt(5))^2)#

#color(white)(4/(sqrt(2)-3sqrt(5))) = (4(sqrt(2)+3sqrt(5)))/(2-45)#

#color(white)(4/(sqrt(2)-3sqrt(5))) = -4/43(sqrt(2)+3sqrt(5))#

#color(white)(4/(sqrt(2)-3sqrt(5))) = -4/43sqrt(2)-12/43sqrt(5)#

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