Question

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 $f(x)+f'(x)+f''(x)=0$ are

• A. real and distinct
• B. imaginary
• C. equal
• D. rational and equal

Answer 1

Answer: (B)

imaginary

Answer 2

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