Root of the equation $x^2+ix+2=0$ 
What are the roots of the equation $x^2+ix+2=0$, where $i=\sqrt{-1}$?

*

*$(-1, 1)$

*$(-2i, i)$

*$( i, 1)$

*no root exists


I don't know the method for finding roots of this type of equation.
I  have tried by the method of $b^2-4ac$ but it doesn't work
 A: Well, we can use the quadratic formula:
$$\text{a}x^2+\text{b}x+\text{c}=0\space\Longleftrightarrow\space x=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{a}\text{c}}}{2\text{a}}\tag1$$
So, for your problem:
$$1x^2+ix+2=0\space\Longleftrightarrow\space x=\frac{-i\pm\sqrt{i^2-4\cdot1\cdot2}}{2\cdot1}=$$
$$\frac{-i\pm\sqrt{-9}}{2}=\frac{i\pm3i}{2}=\frac{i\left(1\pm3\right)}{2}\tag2$$
So, the answers are:
$$x=i\space\vee\space x=-2i\tag3$$
A: To the extent this is a multiple-choice test question, you can solve it quite easily through a process of elimination.  First of all, the Fundamental Theorem of Algebra guarantees that roots exist, which rules out option 4.  Second, it's easy to see that $x=1$ is not a root, since $1^2+i+2=3+i\not=0$, which rules out options 1 and 3.  This leaves only option 2.
You can, if you like, verify that $x=-2i$ and $x=i$ actually are both roots, i.e.,
$$(-2i)^2+i(-2i)+2=-4+2+2=0\qquad\text{and}\qquad i^2+i\cdot i+2=-1-1+2=0$$
but if you're under time pressure, why bother?
In sum, there is no need to solve for the roots of a polynomial when verification suffices.
A: Use the quadratic formula:
\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
&=\frac{-i\pm\sqrt{i^2-4\times 2}}{2}\\
&=\frac{-i\pm\sqrt{(-1)-8}}{2}\\
&=\frac{-i\pm\sqrt{-9}}{2}\\
&=\frac{-i\pm i\sqrt{9}}{2}\\
&=\frac{-i\pm3i}{2}\end{align}
So we have $$x=\frac{-i+3i}{2}=\frac{2i}{2}=i$$ and $$x=\frac{-i-3i}{2}=\frac{-4i}{2}=-2i$$
So our roots are $(-2i,i)$
A: It works ! You can set up $a=1 , b=i, c=2$. 
Then $x1=\frac{-i + \sqrt{i^2 -4*1*2}}{2*1}=i$ and $x2=\frac{-i - \sqrt{i^2 -4*1*2}}{2*1}=-2i$
If you want to verify you can plug them into the equation :
With $x1=i\rightarrow$ $(i)^2 +i*i+2 = -1 -1+2=0$
With $x2=-2i\rightarrow$ $(-2i)^2 +i*(-2i)+2 = -4+2+2=0$
So the roots we found are correct !
