tenistasenorita
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- Mar 10, 2006
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Hi. Complex numbers are really confusing to me. Please help me figure out how to do this:
Find the zeros of t(x)=2x^2+ix+3.
Thanks!
Find the zeros of t(x)=2x^2+ix+3.
Thanks!
Finding complex zeros.
- Thread starter tenistasenorita
- Start date
Hello, tenistasenorita!
We have: \(\displaystyle \,a\,=\,2,\;b\,=\,i,\;c\,=\,3\)
Hence: \(\displaystyle \L\,x\;=\;\frac{-i\,\pm\,\sqrt{i^2\,-\,4(2)(3)}}{2(2)} \;= \;\frac{-i\,\pm\,\sqrt{-25}}{4}\;=\;\frac{-i\,\pm\.5i}{4}\)
Therefore: \(\displaystyle \L\,x\:=\:\frac{-i\,+\,5i}{4}\:=\:\frac{4i}{4}\:=\:i\)
. . . . and: \(\displaystyle \L\,x\:=\:\frac{-i\,-\,5i}{4}\:=\:\frac{-6i}{4}\:=\:-\frac{3}{2}i\)
How about the Quadratic Formula? \(\displaystyle \;x\;=\;\frac{-b\,\pm\,\sqrt{b^2\,-\,4ac}}{2a}\)
We have: \(\displaystyle \,a\,=\,2,\;b\,=\,i,\;c\,=\,3\)
Hence: \(\displaystyle \L\,x\;=\;\frac{-i\,\pm\,\sqrt{i^2\,-\,4(2)(3)}}{2(2)} \;= \;\frac{-i\,\pm\,\sqrt{-25}}{4}\;=\;\frac{-i\,\pm\.5i}{4}\)
Therefore: \(\displaystyle \L\,x\:=\:\frac{-i\,+\,5i}{4}\:=\:\frac{4i}{4}\:=\:i\)
. . . . and: \(\displaystyle \L\,x\:=\:\frac{-i\,-\,5i}{4}\:=\:\frac{-6i}{4}\:=\:-\frac{3}{2}i\)
