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

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(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

$4$

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(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 $4$.

List-$I$	List-$II$
$A$. All the roots are negative	$I$. $(b - 3)^2 = 36 + P^2$ for $P \in R$
$B$. Two roots are complex	$II$. $-3 < b < 9$
$C$. Two roots are positive	$III$. $b \in (-\infty, -3) \cup (9, \infty)$
$D$. All roots are real and distinct	$IV$. $b = 9$
	$V$. $b = -3$

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:

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