Let $f(x)=x^2+a x+b$, where $a, b \in R$. If $f(x)=0$ has all its roots imaginary, then the roots of $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=0$ are

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Aicharya · Accepted Answer

Given, $f(x)=x^2+a x+b$ has imaginary roots. $\therefore$ Discriminant, $D < 0 \Rightarrow a^2-4 b < 0$ Now, $\begin{aligned} f^{\prime}(x) & =2 x+a \\ f^{\prime \prime}(x) & =2 \end{aligned}$

$\begin{array}{ll}\Rightarrow & x^2+a x+b+2 x+a+2=0 \\ \Rightarrow & x^2+(a+2) x+b+a+2=0\end{array}$ $\begin{aligned} & \therefore \quad x=\frac{-(a+2) \pm \sqrt{(a+2)^2-4(a+b+2)}}{2} \\ & =\frac{-(a+2) \pm \sqrt{a^2-4 b-4}}{2} \\ & \text { Since, } \quad a^2-4 b < 0 \\ & \therefore \quad a^2-4 b-4 < 0 \\ & \end{aligned}$ Hence, Eq. (i) has imaginary roots.

Let $f(x)=x^2+a x+b$, where $a, b \in R$. If $f(x)=0$ has all its roots imaginary, then the roots of…

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