How do you simplify # (sqrt2 + sqrt6)(sqrt2 - sqrt6)#?

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Answer 1 · 2016-06-02T22:11:20

The result is -4.

When you multiply two numbers in the form #(A+B)(A-B)# you obtain as result #A^2-B^2#.

In your case #A=sqrt(2)# and #B=sqrt(6)# then

#(sqrt(2)+sqrt(6))(sqrt(2)-sqrt(6))=sqrt(2)^2-sqrt(6)^2=2-6=-4#.

Answer 2 · 2016-06-02T22:15:34

#2 - 6 = -4#

This is in the same format as #(x + y)(x - y)# which are the factors of #x^2 - y^2, " known as the difference between two squares"#

So if #(x + y)(x - y) = x^2 - y^2#, then

#(sqrt2 + sqrt6)(sqrt2 - sqrt6) = (sqrt2)^2 - (sqrt6)^2#

This simplifies to #2 - 6 = -4#

This could also be simplified by multiplying out the brackets to get 4 terms, but the answer will be the same - it will just take longer.