How do you evaluate #(5\sqrt { 10} - 4) ^ { 2}#?

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Answer 1 · 2016-12-08T02:44:10

#266 - 40sqrt(10)# or #2(133 - 20sqrt(10))#

This can changed to:

#(5sqrt(10) - 4)(5sqrt(10) - 4)#

Cross multiplying gives:

#(5sqrt(10))^2 - 20sqrt(10) - 20sqrt(10) + 16#

#(5sqrt(10))^2 - 40sqrt(10) + 16#

#25(sqrt(10))^2 - 40sqrt(10) + 16#

#(25*10) - 40sqrt(10) + 16#

#250 - 40sqrt(10) + 16#

#266 - 40sqrt(10)#