The quadratic equation p(x) = 0 with real coe | vedclass

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(D) Let $p(x) = ax^2 + c$ where $a, c \in \mathbb{R}$. Since the roots are purely imaginary,let them be $\pm i k$ $(k 
eq 0)$.Then $p(ik) = a(ik)^2 +

Vedclass · Accepted Answer

neither real nor purely imaginary roots

Vedclass · Answer

(D) Let $p(x) = ax^2 + c$ where $a, c \in \mathbb{R}$. Since the roots are purely imaginary,let them be $\pm i k$ $(k 
eq 0)$.Then $p(ik) = a(ik)^2 + c = -ak^2 + c = 0$,which implies $c = ak^2$.Thus,$p(x) = a(x^2 + k^2)$.Now,consider $p(p(x)) = 0$,which implies $p(x) = \pm ik$.$a(x^2 + k^2) = ik$ or $a(x^2 + k^2) = -ik$.$x^2 + k^2 = \pm \frac{ik}{a}$.$x^2 = -k^2 \pm \frac{ik}{a}$.Since $k^2$ is real and $\pm \frac{ik}{a}$ is purely imaginary,$x^2$ is a complex number with a non-zero imaginary part.Therefore,$x$ cannot be purely real (as $x^2$ would be real) and $x$ cannot be purely imaginary (as $x^2$ would be real).Thus,the roots are neither real nor purely imaginary. The correct option is $(D)$.

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

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