The quadratic equation p ( x ) = 0 with real coefficients has purely imaginary roots. Then the equation p ( p ( x ) ) = 0 has [2014]

Question

The quadratic equation $p (x) = 0$ with real coefficients has purely imaginary roots. Then the equation $p (p (x)) = 0$ has [2014]

Smart Achievers · Accepted Answer

(4)

Quadratic equation with real coefficients and purely imaginary roots can be considered as

$p (x) = x^{2} + a = 0 where a > 0 and a \in ℝ$

$Then p [p (x)] = 0 \Rightarrow {(x^{2} + a)}^{2} + a = 0$

$\Rightarrow x^{4} + 2 a x^{2} + (a^{2} + a) = 0$

$\Rightarrow x^{2} = \frac{- 2 a \pm \sqrt{4 a^{2} - 4 (a^{2} + a)}}{2}$ $= \frac{- 2 a \pm \sqrt{4 a^{2} - 4 a^{2} - 4 a}}{2}$

$= \frac{- 2 a \pm \sqrt{- 4 a}}{2}$ $\Rightarrow x^{2} = - a \pm \sqrt{a} i$

$\Rightarrow x = \sqrt{- a \pm \sqrt{a} i} = α \pm i β, where α, β \neq 0$

$∴$ $p [p (x)] = 0 has complex roots which are neither purely real nor purely imaginary.$

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