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

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

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?

Similar Questions

The number of real roots of the equation $\mathrm{e}^{4 \mathrm{x}}-\mathrm{e}^{3 \mathrm{x}}-4 \mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{\mathrm{x}}+1=0$ is equal to $.....$

If the sum of all the roots of the equation $e^{2 x}-11 e^{x}-45 e^{-x}+\frac{81}{2}=0$ is $\log _{ e } P$, then $p$ is equal to

Complete solution set of the inequality $\left( {{{\sec }^{ - 1}}\,x - 4} \right)\left( {{{\sec }^{ 1}}\,x - 1} \right)\left( {{{\sec }^{ - 1}}\,x - 2} \right) \ge 0$ is

The number of solutions for the equation ${x^2} - 5|x| + \,6 = 0$ is

If $a+b+c=1, a b+b c+c a=2$ and $a b c=3$, then the value of $a^{4}+b^{4}+c^{4}$ is equal to $....$