Question

If $a,b$ are real numbers such that $x^3-a{x}^2+bx-6=0$ has its roots real and positive then the minimum value of $b$ is:

• A. $1$
• B. $2$
• C. $3(36{)}^{1/3}$
• D. None of these

Answer 1

The correct option is C $3(36{)}^{1/3}$

Let $x_1,x_2,x_3$ be the roots of the equation

$⟹x_1{x}_2+x_1{x}_3+x_2{x}_3=b$

$x_1{x}_2{x}_3=6$

Use $A.M\ge G.M$

$⟹\frac{x_1{x}_2+x_1{x}_3+x_2{x}_3}{3}\ge \sqrt[3]{3x_1^2{x}_2^2{x}_3^2}$

$⟹\frac{b}{3}\ge \sqrt[3]{336}$

$⟹b\ge 3\sqrt[3]{336}$

So, the answer is option (C).