asked 137k views
8 votes
Find the zeros of f(x)= 2x^2 + 32
the zeros of f are x=_ & x=_

asked
User Bertine
by
7.7k points

1 Answer

9 votes

Answer:


2x^2+32=0\\2x^2=-32\\x^2=-16\\x=16i \\x=-16i

are you completely sure you don't mean 2x^2-32? that would have real roots

answered
User Jarek Kulikowski
by
9.0k points

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