If $$P(z)=\sum_{i=0}^na_iz^i$$
is a polynomial having real coefficients with $z_0$ as a root, then taking conjugate on both sides we see that $\overline{z_0}$ is also a root.
Now, if $a+ib$ is a root of a quartic with real coefficients, then by this reasoning, $a-ib$ is also a root. Hence, given polynomial is divisible by the quadratic $(x-a-bi)(x-a+bi)=(x-a)^2+b^2$. Divide it by this quadratic, get a quadratic and apply the same procedure.
Note
The above also shows that:
If a non-real complex number $z$ is a root of a polynomial with real coefficients with multiplicity $k$, then its conjugate is also a root with same multiplicity (irrespective of the degree of the polynomial)
This answer shows that $(b\overline{c}+c\overline{b})(a\overline{b}+\overline{a}b)+(c\overline{a}-a\overline{c})^2=0$ is a necessary condition.
$iz^2-i=i(z+1)(z-1)=0$ shows that it is not a sufficient condition.
From the following part :
Equations $(1)$ and $(2)$ must have both roots common because we want only one value of $k$.
After applying condition for both roots common I obtained $\dfrac{a{_1}}{a_{2}}=\dfrac{-b_2}{b_1}=\dfrac{c_1}{c_2} \qquad(3)$
it seems that you think "equation has one purely imaginary root" means that "equation has a purely imaginary double root".
I think that you have to deal with the case $a_2b_1c_2=0$ separately.
Another way is to start with $a(z-ki)(z-ki)=0$. Since $b=-2kia$ and $c=-ak^2$, we have $(a\bar b+\bar a b)(b\bar c+ \bar b c)+(c \bar a -\bar c a)^2=0$.
Using your idea, we can show that if $az^2+bz+c=0$ has at least one purely imaginary root, then $(b\overline{c}+c\overline{b})(a\overline{b}+\overline{a}b)+(c\overline{a}-a\overline{c})^2=0$.
You have
$$a_1k^2+b_2k-c_1=0\tag1$$
$$a_2k^2-b_1k-c_2=0\tag2$$
Now, $(1)\times a_2-(2)\times a_1$ gives
$$(a_2b_2+a_1b_1)k=a_2c_1-a_1c_2$$
which can be written as
$$\frac{a\bar b+\bar a b}{2}k=-\frac{c \bar a -\bar c a}{2i}$$
i.e.
$$i(a\bar b+\bar a b)k=-(c \bar a -\bar c a)\tag3$$
Also, $(1)\times c_2-(2)\times c_1$ gives
$$(a_2c_1-a_1c_2)k^2=(b_1c_1+b_2c_2)k\tag4$$
If $k=0$, then we have $c=0$ for which $(a\bar b+\bar a b)(b\bar c+ \bar b c)+(c \bar a -\bar c a)^2=0$ holds.
If $k\not=0$, then dividing the both sides of $(4)$ by $k$ gives
$$(a_2c_1-a_1c_c)k=b_1c_1+b_2c_2$$
which can be written as
$$-\frac{c \bar a -\bar c a}{2i}k=\frac{b\bar c+ \bar b c}{2}$$
i.e.
$$(c \bar a -\bar c a)k=-i(b\bar c+ \bar b c)\tag5$$
Multiplying the both sides of $(3)$ by $(c \bar a -\bar c a)$ gives
$$i(a\bar b+\bar a b)\color{red}{(c \bar a -\bar c a)k}=-(c \bar a -\bar c a)^2$$
Finally, using $(5)$ to eliminate $k$, we get
$$i(a\bar b+\bar a b)\color{red}{\bigg(-i(b\bar c+ \bar b c)\bigg)}=-(c \bar a -\bar c a)^2$$
i.e.
$$(a\bar b+\bar a b)(b\bar c+ \bar b c)+(c \bar a -\bar c a)^2=0$$
Best Answer
If it has purely imaginary roots, that basically means that $$p(x)=ax^2+b\,,$$ where $a,b$ are nonzero real numbers of the same sign. It follows that $$p(p(x))=a(ax^2+b)^2+b\,.$$ The solution to this is $$ax^2+b=\pm\sqrt{\frac{b}{a}}i\,,$$ which clearly shows that $x$ can neither be real nor purely imaginary because in that case the left-hand side would be a real number whereas the right-hand side is purely imaginary.