Question

If for the cubic equation $x^3-4x^2+x+6=0$, the possible no. of positive real roots are $2\text{or}0$ and the possible no. of negative real roots is $1$, then which of the following can be true?

• A. Number of positive roots$:0$
  Number of imaginary roots:$3$
• B. Number of positive roots$:0$
  Number of imaginary roots$:2$
• C. Number of positive roots$:2$
  Number of imaginary roots$:0$
• D. Number of positive roots$:2$
  Number of imaginary roots$:3$

Answer 1

Answer: (B), (C)

Number of positive roots$:2$

Number of imaginary roots$:0$
Given that for the cubic equation $x^3-4x^2+x+6=0$, the possible no. of positive real roots are $2$ or $0$ and the possible no. of negative real roots is $1$.
The total no. of solutions will be $3$ as this is a cubic equation.
Also if there are any imaginary roots, then they will come only in conjugate pairs.

Therefore the different possibilities can be shown as: