How do you use FOIL to multiply #(sqrt7-sqrt8)(sqrt7-sqrt8)#?

1 Answer
George C.
Jun 9, 2015

Using FOIL:

#(sqrt(7)-sqrt(8))(sqrt(7)-sqrt(8))#

#= 7-sqrt(7)sqrt(8)-sqrt(8)sqrt(7)+8#

#=15-4sqrt(2)sqrt(7) = 15-4sqrt(14)#

Explanation:

FOIL helps us pick pairs of terms from each binomial to multiply, before adding together to get the result:

First: #sqrt(7) * sqrt(7) = 7#
Outside: #sqrt(7) * (-sqrt(8)) = -sqrt(7) * sqrt(8)#
Inside: #(-sqrt(8)) * sqrt(7) = -sqrt(7) * sqrt(8)#
Last: #(-sqrt(8)) * (-sqrt(8)) = 8#

Added together:

#7 - 2*sqrt(7)*sqrt(8) + 8#

Now #sqrt(8) = sqrt(2^2*2) = 2*sqrt(2)#

So our expression can be simplified to:

#15-4*sqrt(7)sqrt(2) = 15-4sqrt(14)#

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