Question

Let \(f(x)=x^2+ax+b\), where \(a,b\in \mathbb{R}\). 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

Hint:

If a quadratic has imaginary roots, its discriminant is negative. Adding derivatives shifts discriminant further negative, preserving imaginary nature.

Answer 1

The Correct Option is B

Step 1: Compute derivatives.
\[ f(x)=x^2+ax+b \] 
\[ f'(x)=2x+a \] 
\[ f''(x)=2 \] 
Step 2: Form new equation. 
\[ f(x)+f'(x)+f''(x)=0 \] 
\[ (x^2+ax+b)+(2x+a)+2=0 \] 
\[ x^2+(a+2)x+(b+a+2)=0 \] 
Step 3: Use condition that roots of \(f(x)\) are imaginary. 
Imaginary roots means discriminant of \(f(x)\) is negative: 
\[ a^2-4b<0 \Rightarrow 4b>a^2 \] 
Step 4: Discriminant of new quadratic. 
\[ \Delta'=(a+2)^2-4(b+a+2) \] 
\[ =a^2+4a+4-4b-4a-8 \] 
\[ =a^2-4b-4 \] 
Step 5: Prove \(\Delta'<0\). 
Given \(a^2-4b<0\). 
So \(a^2-4b-4\) is definitely negative. 
\[ \Delta'<0 \] 
Thus roots are imaginary. 
Final Answer: 
\[ \boxed{\text{imaginary}} \]