How do you evaluate #(2\sqrt { 2} - \sqrt { 7} ) ^ { 2}#?

1 Answer
Dec 7, 2016

Answer:

#(2sqrt2 - sqrt7)^2# simplifies to #-4sqrt(14) + 15#.

Explanation:

For the square of any binomial, it follows this general rule: #(x - y)^2 = x^2 - 2xy + y^2#. We can apply this here where #x# is #2sqrt2# and #y# is #sqrt7#

#(2sqrt(2) - sqrt(7))^2#
#= (2sqrt2)^2 - 2(2sqrt2)(sqrt7) + (sqrt7)^2#
#= 8 - 4sqrt(7*2) + 7#
#= -4sqrt(14) + 15#

So, our final simplified answer is #-4sqrt(14) + 15#.