4(x2+3)=0
x2=-3
x=sqrt(-3)
=±(sqrt(3)*i)
where i2=-1
The roots are therefore complex.
4x^2+12 = 0
I factored this out, but how do I show the final answer? Thanks!
4(x^2+3)
4 answers
Would iroot3 and -iroot3 be acceptable answers? Thanks!
Yes, they are acceptable answers.
When in doubt (as in an exam), the best way to check an answer is to square your proposed roots and see if you get back -3.
When in doubt (as in an exam), the best way to check an answer is to square your proposed roots and see if you get back -3.
Thanks for the help!