How do you perform the operation and write the result in standard form given #(sqrt14+sqrt10i)(sqrt14-sqrt10i)#?

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Answer 1 · 2018-03-15T21:46:36

It is well known that the product yields the difference of two squares.

#(sqrt14+sqrt10i)(sqrt14-sqrt10i) = (sqrt14)^2-(sqrt10i)^2#

The square is the inverse of the square root, therefore, only the argument remains:

#(sqrt14+sqrt10i)(sqrt14-sqrt10i) = 14-10i^2#

Substitute #i^2 = -1#:

#(sqrt14+sqrt10i)(sqrt14-sqrt10i) = 14-10(-1)#

Multiplication by -1 changes the sign:

#(sqrt14+sqrt10i)(sqrt14-sqrt10i) = 14+10#

#(sqrt14+sqrt10i)(sqrt14-sqrt10i) = 24#