How do you solve #(x +3)(x^2-2)=0#?

1 Answer
Gió
Jan 28, 2017

I got three possible values satisfying it:
#x_1=-3#
#x_(2,3)=+-sqrt(2)#

Explanation:

You need that the product of the two brackets must be zero; to get this either the first or the second must be zero or:
either:
#x+3=0#
so:
#x=-3#
or
#x^2-2=0#
so
#x^2=2#
#x=+-sqrt(2)#

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