Let's say you want a quadratic that has 2i has a root.  When your quadratic has real coefficients, but imaginary roots, it means the roots must come in complex conjugate pairs.  For example, if 3+4i is a root, then 3-4i is also a root.  Or, if -2-5i is a root, then -2+5i is a root as well.  For us, that means if 0+2i is a root, then so is 0-2i.

Now that we know our two roots, it means our quadratic must be of the form:

y=a(x-2i)(x+2i)

We can choose a=1 to make things simple.  (Or you can pick a to be any real number except 0).

Then:

y=(x-2i)(x+2i) , so:

`y=x^2 + 2ix - 2ix -4i^2`

`y=x^2+4`

Thus, y=x^2+4 has two imaginary roots, at 2i and -2i

We can plug these into our equation to check:

`y=(2i)^2+4 = 4i^2 + 4 = -4+4 = 0`

`y=(-2i)^2+4=4i^2+4=-4+4 = 0`

We can do the same procedure to get some other quadratics with purely imaginary roots.

Some other examples are:

`y=x^2+9`

has the roots 3i and -3i, and

`y=x^2+16`

has the roots 4i and -4i