Question

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

• A. $-10$
• B. $-9$
• C. $-6$
• D. $-4$

Answer 1

Answer: (D)

$-4$

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

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

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

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

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

Therefore, $\frac{{\alpha }_1{\alpha }_2{\alpha }_3+{\alpha }_1{\alpha }_3{\alpha }_4+{\alpha }_1{\alpha }_2{\alpha }_4+{\alpha }_2{\alpha }_3{\alpha }_4}{4}\ge \sqrt[4]{4{\alpha }_1^3{\alpha }_2^3{\alpha }_3^3{\alpha }_4^3}$

$⟹\frac{-c}{4}\ge \sqrt[4]{4({\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4{)}^3}$

$⟹\frac{-c}{4}\ge \sqrt[4]{4(1{)}^3}$

$⟹\frac{-c}{4}\ge 1$

$⟹-c\ge 4$

$⟹c\le -4$