Question

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

• A. Real and distinct
• B. Imaginary
• C. Equal
• D. Rational and equal

Answer 1

Answer: (B)

Imaginary

Answer 2

Given, $f(x) = x^2 + ax + b$ has imaginary roots.
$\therefore$ Discriminant, $D < 0 \, \Rightarrow \, a^2 - 4b < 0$
Now, $f'(x) = 2x + a$
$f'(x) = 2$
Also , $f(x) + f (x) + f" (x) = 0$
$\Rightarrow \, \, \, x^2 + ax + b + 2x + a + 2 = 0$
$\Rightarrow \, \, \, x^2 + (a + 2)x + b + a + 2 = 0$
$\therefore \, \, x = \frac{ - a + 2 \pm \overline{ a+2^2 - 4 \, a + b + 2}}{2}$
$= \frac{- a + 2 \, \pm \, \overline{a^2 - 4 b - 4}}{2}$
Since, $a^2 - 4b < 0$
$a^2 - 4b - 4 < 0$
Hence, E (i) has imaginary roots