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 negative, then the number of positive roots is and the number of imaginary roots is

• A. $1$
• B. $2$
• C. $0$
• D. $3$

Answer 1

Answer: (A), (C)

$0$

Given: $f\left(x\right)=-a{x}^3+b{x}^2+cx+d\forall a,b,c,d\in R^{+}$Now, to find the number of negative roots of $f\left(x\right)=0$:
$f(-x)=a{x}^3+b{x}^2-cx+d$
Now, this equation has 2 sign changes, thus the equation will have either 2 or 0 negative roots.
Given $f\left(x\right)$ has one negative root.
Thus, it will have one more negative roots.
As we know that imaginary roots of a polynomial equation with real coefficients, if exist, occurs in conjugate pair.
Thus the third root will be positive.