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. no. of positive roots:2
  no. of imaginary roots:0
• B. no. of positive roots:2
  no. of imaginary roots:3
• C. no. of positive roots:0
  no. of imaginary roots:3
• D. no. of positive roots:0
  no. of imaginary roots:2

Answer 1

Answer: (A), (D)

Number of positive roots$:0$

Number of imaginary roots$:2$
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: