Purely imaginary equation $p(x)=0$ with real coefficient

Question

The quadratic equation $p(x)=0$ with real coefficient has purely imaginary roots. Then the equation $p(p(x))=0$ has

(A) only purely imaginary roots

(B) all real roots

(C) two real and two purely imaginary roots

(D) neither real nor purely imaginary roots

The official answer is (D)

My approach us as follow

Though the question has been solved in this website but dont consider it as a duplicate as I would like to put forward my method for perusal

$p\left( {p\left( x \right)} \right) = 0$

$p\left( x \right) = a{x^2} + bx + c = 0$

${b^2} - 4ac < 0$

$T = p\left( x \right)\& T > 0$

$a{x^2} + bx + c > 0$

$p\left( T \right) = 0 = a{T^2} + bT + c$

${b^2} - 4ac < 0$ but $T > 0$ contradiction so the equation is neither real nor imaginary

Is my approach of solving matches with the standard procedure

Answer 1

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.

Answer 2

As it has real coefficients. We can WLOG assume the roots are $li,-li$ where $l$ is some real $l\neq 0$.Thus $$p(x)=a(x-li)(x+li)=a(x^2+l^2)$$ Here $a$ is some real $a\neq 0$

Now $$p(p(x))=0 \Rightarrow p(x)=li,-li$$

It is easy to see $$a(x^2+l^2)=li,a(x^2+l^2)=-li$$ has neither real nor purely imaginary roots

Answer 3

Let $r$ be a purely imaginary root of $p(x)$, then the other root is $\bar r=-r$ so we can write $p(x)=(x-r)(x-(-r))=x^2-r^2$.

Let $T=p(x)$, then we want to find roots of $p(p(x))=p(T)=0$. We know the roots to be $T=r,-r$. Thus, $x^2-r^2=\pm r\implies x^2=r^2\pm r$. Since $r^2\in\Bbb R-\{0\}$ and $r$ is purely imaginary non-zero complex number, $r^2\pm r$ is not real. So $x$ can't be purely imaginary or real.

Answer 4

We could also look at the quadratic polynomial $ \ P(x) \ $ in terms of how it "maps" numbers. Since it has real coefficients, the zeroes are $ \ z = \zeta·i \ , \ \overline{z} = -\zeta·i \ \ , $ so we have $$ \ P(x) \ \ = \ \ (x + \zeta·i)·(x - \zeta·i) \ \ = \ \ x^2 + \zeta^2 \ \ = \ \ x^2 + |z|^2 \ \ . $$ As this is a sum of two "squared" terms, the second being a positive real number, $ \ P(x) \ $ "maps"

• all real numbers to positive real numbers;

• all imaginary numbers other than the two zeroes $ \ \pm \zeta·i \ $ to non-zero real numbers; and

• all complex numbers to complex numbers which are not purely imaginary [since $ \ (a + bi)^2 + \zeta^2 \ = \ (a^2 - b^2 + \zeta^2) + 2ab·i \ ] \ $ , with the exception of those with $ \ b \ = \ \pm \sqrt{a^2 + \zeta^2} \ \ , $

Hence, $ \ P( \ P(x) \ ) \ = \ 0 \ $ requires that $ \ P(x) \ = \ \pm \zeta·i \ \ , $ which means that $ \ x \ $ must be a number of the form $ \ \pm a \ \pm \ i(\sqrt{a^2 + \zeta^2}) \ \ $ (a member of two complex-conjugate pairs), which is consistent with choice $ \ \mathbf{(D)} \ \ . $