Question

If one of the roots of the equation $a{x}^3-b{x}^2+cx+d=0\forall a,b,c,d\in R^{+}$ is positive, then the number of negative roots is and the number of imaginary roots is

• A. $2$
• B. $1$
• C. $0$
• D. $3$

Answer 1

Answer: (B), (C)

$0$

$\mathrm{let}f\left(x\right)=a{x}^3-b{x}^2+cx+d\forall a,b,c,d\in R^{+}$
The number of sign change in $f\left(x\right)$ is $2,$ this means the equation can have $2$ or no positive roots.

It is given that the equation has a positive root, thus one more root will be positive.
Now, complex roots of a polynomial equation with real coefficients, if exist, occurs in conjugate pair.
As the given equation is cubic, so it must have 3 roots. Out of which two are positive.
Thus, the last root should be negative.