If all roots of the equation x^3 - 2ax^2 + 3b | vedclass

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

Question

(A) Let the roots of the equation be $x_1, x_2, x_3 > 0$.From Vieta's formulas:$x_1 + x_2 + x_3 = 2a$$x_1x_2 + x_2x_3 + x_3x_1 = 3b$$x_1x_2x_3 = 8$Usi

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Let the roots of the equation be $x_1, x_2, x_3 > 0$.From Vieta's formulas:$x_1 + x_2 + x_3 = 2a$$x_1x_2 + x_2x_3 + x_3x_1 = 3b$$x_1x_2x_3 = 8$Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality for the terms $\frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3}$:$\frac{\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3}}{3} \geq \sqrt[3]{\frac{1}{x_1} \cdot \frac{1}{x_2} \cdot \frac{1}{x_3}}$$\frac{\frac{x_1x_2 + x_2x_3 + x_3x_1}{x_1x_2x_3}}{3} \geq \left(\frac{1}{8}\right)^{1/3}$$\frac{\frac{3b}{8}}{3} \geq \frac{1}{2}$$\frac{b}{8} \geq \frac{1}{2}$$b \geq 4$Thus,the minimum value of $b$ is $4$.

If all roots of the equation $x^3 - 2ax^2 + 3bx - 8 = 0$ are positive,where $a, b \in R$,then the minimum value of $b$ is:

Similar Questions

If the difference between the roots of the equations $x^2 - bx + c = 0$ and $x^2 - cx + b = 0$ is the same,then $b + c = \dots$

The equation $\sqrt{3x^2 + x + 5} = x - 3$,where $x$ is real,has

If one root of the equation $x^2 + px + q = 0$ is $2 + \sqrt{3}$,then the values of $p$ and $q$ are:

Let $f(x) = x^2 + ax + b$,where $a, b \in R$. If $f(x) = 0$ has all its roots imaginary,then the roots of $f(x) + f'(x) + f''(x) = 0$ are

The number of positive real roots of the equation $3^{x+1}+3^{-x+1}=10$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module