Question

Equation $x^4+a{x}^3+b{x}^2+cx+1=0$ has real roots ($a,b,c$ are non-negative). Minimum non-negative real value of $b$ is

• A. $12$
• B. $15$
• C. $6$
• D. $10$

Answer 1

Answer: (C)

$6$

Given that a polynomial $x^4+a{x}^3+b{x}^2+cx+1=0$ has real roots.

Let ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4$ be the roots of the polynomial.

Therefore, ${\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4=1$

${\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4=-a$

${\alpha }_1{\alpha }_2+{\alpha }_1{\alpha }_3+{\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_3+{\alpha }_2{\alpha }_4+{\alpha }_3{\alpha }_4=b$

${\alpha }_1{\alpha }_2{\alpha }_3+{\alpha }_1{\alpha }_3{\alpha }_4+{\alpha }_1{\alpha }_2{\alpha }_4+{\alpha }_2{\alpha }_3{\alpha }_4=-c$

We know that $A.M.\ge G.M.$

therefore, $\frac{{\alpha }_1{\alpha }_2+{\alpha }_1{\alpha }_3+{\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_4+{\alpha }_2{\alpha }_3+{\alpha }_3{\alpha }_4}{6}\ge \sqrt[6]{6{\alpha }_1^3{\alpha }_2^3{\alpha }_3^3{\alpha }_4^3}$

$⟹\frac{b}{6}\ge \sqrt[6]{6({\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4{)}^3}$

$⟹\frac{b}{6}\ge \sqrt[6]{6(1{)}^3}$

$⟹\frac{b}{6}\ge 1$

$⟹b\ge 6$

Therefore the minimum value of $b$ is $6$