Consider the cubic equation x^3+c x^2+b x+c=0 | vedclass

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Question

(a)

We have, $x^3+a x^2+b x+c=0$

Let $f(x)=x^3+a x^2+b x+c$

$P^{\prime}(x) =3 x^2+2 a x+b$

$D =(2 a)^2-4(3)(b)$

$D =4 a^2-12 b$

$D =

Vedclass · Accepted Answer

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

Vedclass · Answer

(a)

We have, $x^3+a x^2+b x+c=0$

Let $f(x)=x^3+a x^2+b x+c$

$P^{\prime}(x) =3 x^2+2 a x+b$

$D =(2 a)^2-4(3)(b)$

$D =4 a^2-12 b$

$D =4\left(a^2-3 bight)$

For three roots $a^2-3 b > 0$

but given $a^2-2 b < 0$

$	herefore \quad D < 0$

Hence, $f(x)$ has one real roots and two imaginary roots.

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

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