How do you multiply #(sqrt5- sqrt2)(sqrt5+sqrt2)#?

2 Answers
Jun 16, 2015

Answer:

#(sqrt(5)-sqrt(2))(sqrt(5)+sqrt(2))# is of the form

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

So #(sqrt(5)-sqrt(2))(sqrt(5)+sqrt(2)) = 5-2 = 3#

Explanation:

#(sqrt(5)-sqrt(2))(sqrt(5)+sqrt(2))#

#=(sqrt(5))^2 - (sqrt(2))^2 = 5 - 2 = 3#

using the difference of squares identity:

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

Alternatively use FOIL to pick pairs of terms to multiply and add together:

First: #sqrt(5) * sqrt(5) = 5#
Outside: #sqrt(5)*sqrt(2)#
Inside: #-sqrt(2)*sqrt(5)# = #-sqrt(5)*sqrt(2)#
Last: #-sqrt(2)*sqrt(2) = -2#

F + O + I + L = #5 + cancel(sqrt(5)sqrt(2)) - cancel(sqrt(5)sqrt(2)) - 2 = 3#

Jun 16, 2015

Answer:

The answer is #3#.

Explanation:

You can use "FOIL" (First-Outside-Inside-Last) as follows:

#(sqrt(5)-sqrt(2))*(sqrt(5)+sqrt(2))=sqrt(5)*sqrt(5)+sqrt(5)sqrt(2)-sqrt(5)sqrt(2)-sqrt(2)sqrt(2)#

After cancellation and use of the facts that #sqrt(5)^2=5# and #sqrt(2)^2=2#, this becomes #5-2=3#.

The reason "FOIL" works is because of abstract properties of numbers systems, such as the Distributive Property, which says that #a*(b+c)=a*b+a*c# for all numbers #a, b#, and #c#. You can use it twice when you expand #(a+b)*(c+d)# as follows:

#(a+b)*(c+d)=(a+b)*c+(a+b)*d#

#=a*c+b*c+a*d+b*d#

This can be rearranged, by the Commutative Property, to #=a*c+a*d+b*c+b*d#. (FOIL)

It's also worthwhile to memorize the difference of two squares factoring formula: #a^2-b^2=(a-b)(a+b)#. This can be applied to the problem at hand "in reverse" by letting #a=sqrt(5)# and #b=sqrt(2)#.