Finding complex zeros.

[tenistasenorita]

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!

 

[Gene]

It factors nicely.
2x^2+ix+3 =
2x^2+ix-3i^2 =
(2x+3i)(x-i)

 

[soroban]

Hello, tenistasenorita!

Find the zeros of: $\displaystyle t(x)\:=\:2x^2\,+\,ix\,+\,3$

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: \(\displaystyle \L\,x\:=\:\frac{-i\,-\,5i}{4}\:=\:\frac{-6i}{4}\:=\:-\frac{3}{2}i\)

 

[Gene]

How about completing the square?
2x^2+ix+3 =
x^2+ix/2+3/2 =
x^2+ix/2+(i/4)^2+3/2-(i/4)^2 =
(x+i/4)^2+(24-i^2)/16 =
(x+i/4)^2+(25)/16 = 0
x+i/4=+5i/4
x=4i/4 = i or
x=-6i/4 = -3i/2
Your turn soroban :evil:
---------------
Gene

 

[pka]

Gene,
I think that your first solution is brilliant.