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

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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?

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