Question
The quadratic equation $$p(x) = 0$$ with real co-efficients has purely imaginary roots. Then the equation $$p(p(x)) = 0$$ has
A.
one purely imaginary root
B.
all real roots
C.
two real and two purely imaginary roots
D.
neither real nor purely imaginary roots
Answer :
neither real nor purely imaginary roots
Solution :
Quadratic equation with real co-efficients and purely imaginary roots can be considered as
$$\eqalign{
& p\left( x \right) = {x^2} + a = 0\,\,{\text{where }}a > 0\,\,{\text{and }}a \in R \cr
& {\text{The }}p\left[ {p\left( x \right)} \right] = 0 \cr
& \Rightarrow \,\,{\left( {{x^2} + a} \right)^2} + a = 0 \cr
& \Rightarrow \,\,{x^4} + 2a{x^2} + \left( {{a^2} + a} \right) = 0 \cr
& \Rightarrow \,\,{x^2} = \frac{{ - 2a \pm \sqrt {4{a^2} - 4{a^2} - 4a} }}{2} \cr
& \Rightarrow \,\,{x^2} = - a \pm \sqrt a \,\,i \cr
& \Rightarrow \,\,x = \sqrt { - a \pm \sqrt a \,\,i} \, \cr
& = \,\,\alpha \pm i\,\beta \,\,\,\,{\text{where }}\alpha {\text{,}}\beta \ne {\text{0}} \cr
& \therefore \,\,p\left[ {p\left( x \right)} \right] = 0\,\, \cr} $$
has complex roots which are neither purely real nor purely imaginary.