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

2 Answers
Jun 8, 2017

Given: #3/(4+sqrt5)#

The factor #(4-sqrt5)# is the conjugate of #(4+sqrt5)#.

Please understand that, to make the conjugate, one merely replaces the sign between the two terms with the opposite sign. i.e.-- If the sign is a plus, then replace it with a minus or, if the sign is a minus, then replace is with plus.

Multiply by 1 in the form of #(4-sqrt5)/(4-sqrt5)#:

#3/(4+sqrt5)(4-sqrt5)/(4-sqrt5)#

Multiplying the denominator by its conjugate makes it become the difference of two squares, in accordance with the pattern #(a+b)(a-b) = a^2-b^2#

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

# (12-3sqrt5)/(16-5) =#

# (12-3sqrt5)/11#

Done.

Jun 8, 2017

Answer:

#(12-3sqrt(5))/(11)#

Explanation:

Simplifying rational numbers with square roots in the bottom involves a process called Rationalizing the Denominator. We can rewrite the quotient so that the denominator contains no square roots.

This is accomplished by multiplying both the numerator and the denominator by it's conjugate, a term which just means the middle sign is different. Like this,

#3/(4+sqrt(5))*(4-sqrt(5))/(4-sqrt(5))#

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

You can use the FOIL method to multiply the denominator.

#=(12-3sqrt(5))/(16-5)#

#=(12-3sqrt(5))/(11)#

That is pretty much it. If you want, you can actually break it down one more step by dividing both sides the numerator by #11#.

#=12/11-3/11 sqrt(5)#