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If the Quadratic equation $p(x)=0$ with real coefficient has purely Imaginary roots.Then the

equation $p(p(x)) = 0$ has

$\bf{OPTIONS::}$

$(a)\;\; $ Only purely Inaginary Roots. $\;\;\;\;\;\;(b)$ all real roots.

$(c)$ Two real and Two purely imaginary roots.$\;\;\;\;\;\;(d)$ neither real nor purely imaginary roots.

$\bf{My\; Try}::$ Let $p(x) = x^2+1.$ Then $p(p(x)) = (x^2+1)^2+1 = x^4+2x^2+2$.

Now Roots of $\displaystyle p(p(x))=0\Rightarrow x^4+2x^2+2=0\Rightarrow x^2 = \frac{-2\pm\sqrt{4-8}}{2} = \frac{-2\pm 2i}{2} = -1\pm i$

So $x^2=-1+i\;\;,-1-i$. Means $x$ must be in the form of $a+ib\;,$ where $a,b\neq 0$ and $i=\sqrt{-1}$

So option $(d)$ mut be Correct.

But my question is how can we solve the given question in Analytical method. means not by guessing.

Help me

Thanks

juantheron
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4 Answers4

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If $p(r) = 0$ it must be that $p(p(x)) = 0$ when $p(x) = r$ or when $p(x)$ equals one of it's roots.

The only (quadratic) polynomial with purely imaginary roots is $p(x) = ax^2 + b$.

If we let the roots $r = \pm iq$ then the solution to $p(x) = ax^2 + b =\pm iq$ cannot be pure real or pure imaginary.

If $x$ were pure real or pure imaginary the left hand side would be real so there would be no way to solve the equation.

Brad
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If a quadratic equation has purely imaginary roots, then it can be written in the form $p(x)=k(x-ai)(x+ai)=k(x^2+a^2)$ for some $k,a\ne0$.

Hence $p(p(x))=k[k^2(x^2+a^2)^2+a^2]=k^3x^4+2k^3a^2x^2+k^3a^4+ka^2$. By the quadratic equation, this gives us $$x^2=\frac{-2k^3a^2\pm\sqrt{4k^6a^4-4k^3(k^3a^4+ka^2)}}{2k^3}=\frac{-2k^3a^2\pm\sqrt{-4k^4a^2}}{2k^3}=\frac{-ka^2\pm ia}{k}$$

This means that $x$ can be neither purely real or purely imaginary, since if it were, then $x^2$ would be real.

Peter Woolfitt
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The general form for p(x) is:

$p(x)=k(x-ai)(x+ai)=\\=kx^{2}+ka^2$

where k and a are real numbers.

poolpt
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Another version on @Brad's answer :

$p(p(x)) = 0$

$a(ax^2+b)^2 + b = 0$

$ax^2 + b = ki$ (where $k=\sqrt\frac{b}{a}$ )

But, since $a,b\in R$,

Therefore

LHS will always be real (if $x\in R$) or Complex (if $x \in C$) However RHS is purely imaginary.

Therefore the only solutions of the equation $p(p(x)=0)$ will be when x is a complex number.

Jdeep
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