How do you rationalize #(2sqrt5-8)/ (2sqrt5+3)#?

2 Answers
Mar 4, 2018

Answer:

#2(2-sqrt5)#

Explanation:

#(2 sqrt5-8)/(2sqrt5+3)#. Multiplying by #(2sqrt5-3)# on

both numerator and denominator we get,

#=((2 sqrt5-8)(2sqrt5-3))/((2sqrt5+3)(2sqrt5-3))#

#=(20-2sqrt5(8+3)+24)/((2sqrt5)^2-3^2)#

#=(44-22sqrt5)/(20-9)=(22(2-sqrt5))/11#

#=2(2-sqrt5)# [Ans]

Mar 4, 2018

Answer:

#(2sqrt5-8)/(2sqrt5+3)=4-2sqrt5#

Explanation:

To rationalize the denominator, we multiply by the conjugate and use the difference of squares rule. In this case, the conjugate is #2sqrt5-3#, so we multiply by it on both top and bottom:

#(2sqrt5-8)/(2sqrt5+3)=((2sqrt5-8)(2sqrt5-3))/((2sqrt5+3)(2sqrt5-3))#

The difference of squares rule says:
#(a+b)(a-b)=a^2-b^2#

Applying this to the denominator, we get:
#((2sqrt5-8)(2sqrt5-3))/(4*5-3)#

Then we multiply out the top:
#(20-6sqrt5-16sqrt5+24)/11=(44-22sqrt5)/11=4-2sqrt5#