Consider the cubic equation x^3+ax^2+bx+c=0 w | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $f(x) = x^3+ax^2+bx+c$. Then $f'(x) = 3x^2+2ax+b$. The discriminant of the quadratic $f'(x)$ is $D = (2a)^2 - 4(3)(b) = 4a^2 - 12b = 4(a^2-3b)

Vedclass · Accepted Answer

If $a^2-2b < 0$,then the equation has one real and two imaginary roots.

Vedclass · Answer

(A) Let $f(x) = x^3+ax^2+bx+c$. Then $f'(x) = 3x^2+2ax+b$. The discriminant of the quadratic $f'(x)$ is $D = (2a)^2 - 4(3)(b) = 4a^2 - 12b = 4(a^2-3b)$. If $a^2-2b Since $b$ must be positive for $a^2 Thus,$D  0$ for all $x$. Since $f'(x)$ is always positive,$f(x)$ is a strictly increasing function. $A$ strictly increasing cubic polynomial has exactly one real root and two complex conjugate roots. Therefore,the statement in option $A$ is correct.

Consider the cubic equation $x^3+ax^2+bx+c=0$ where $a, b, c$ are real numbers. Which of the following statements is correct?

Similar Questions

The least integer $k$ which makes the roots of the equation $x^2 + 5x + k = 0$ imaginary is

Let $a, b, c$ be three real numbers such that $a + 2b + 4c = 0$. Then the equation $ax^2 + bx + c = 0$

What is the nature of the roots of the equation $x^2 + x = 2(x - 1)$?

If $f(x) \in \mathbb{Q}[x]$ is a non-zero polynomial such that all its roots are irrational,then the degree of $f(x)$ is

If $-1+i$ is a root of the equation $x^4+4x^3+5x^2+2x-2=0$,then the real roots of this equation are

Vedclass Test Series

Exam Paper Generator

Online Exam Module