Question

If roots of the equation $x^4-8x^3+b{x}^2+cx+16=0$ are positive, then

• A. $b=c=8$
• B. $b=-24,c=-32$
• C. $b=24,c=-32$
• D. $b=24,c=32$

Answer 1

The correct option is C $b=24,c=-32$

given that a polynomial $x^4-8x^3+b{x}^2+cx+16=0$ has real roots.

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

Therefore, ${\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4=(-1{)}^{\ast }16=16$

${\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4=8$

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

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

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

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

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

$⟹\frac{b}{6}\ge \sqrt[6]{6(16{)}^3}$

$⟹\frac{b}{6}\ge {16}^{\frac{3}{6}}$

$⟹\frac{b}{6}\ge {16}^{\frac{1}{2}}$

$⟹\frac{b}{6}\ge 4$

$⟹b\ge 24$

therefore the value of $b$ is $24$

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

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

$⟹\frac{-c}{4}\ge \sqrt[4]{4(16{)}^3}$

$⟹\frac{-c}{4}\ge {16}^{\frac{3}{4}}$

$⟹\frac{-c}{4}\ge 2^{4\ast \frac{3}{4}}$

$⟹\frac{-c}{4}\ge 2^3$

$⟹\frac{-c}{4}\ge 8$

$⟹-c\ge 32$

$⟹c\le -32$

therefore the value of $c$ is $-32$