Question

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

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

Answer 1

The correct option is B Imaginary

Given, $f\left(x\right)=x^2+ax+b$ has imaginary roots.

$\therefore$ Discriminant, $D<0\Rightarrow a^2-4b<0$

Now, $f^{\prime }\left(x\right)=2x+a$

$f^{\prime }\left(x\right)=2$

Also, $f\left(x\right)+f^{\prime }\left(x\right)+f"\left(x\right)=0$ ..........(i)

$\Rightarrow x^2+ax+b+2x+a+2=0$

$\Rightarrow x^2+(a+2)x+b+a+2=0$

$\therefore x=\frac{-(a+2)\pm \sqrt{(a+2{)}^2-4(a+b+2)}}{2}$

$x=\frac{-(a+2)\pm \sqrt{a^2-4b-4}}{2}$

Since, $a^2-4b<0$

$a^2-4b-4<0$

Hence, eqn$\left(i\right)$ has imaginary roots.