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Question: Answered & Verified by Expert
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
MathematicsQuadratic EquationAP EAMCETAP EAMCET 2009
Options:
  • A real and distinct
  • B imaginary
  • C equal
  • D rational and equal
Solution:
2440 Upvotes Verified Answer
The correct answer is: imaginary
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.

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