Question

Roots of the equation $f\left(x\right)=x^6-12x^5+b{x}^4+c{x}^3+d{x}^2+ex+64=0$ are positive. Which of the following has the greatest absolute value?

• A. $b$
• B. $c$
• C. $d$
• D. $e$

Answer 1

The correct option is C $d$

Given

$x^6-12x^5+b{x}^5+c{x}^3+d{x}^2+ex+64=0$

Since ${\alpha }_{1,}{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5,{\alpha }_6$ are the roots of the equation

Let us consider the sum of the roots

${\alpha }_1,{\alpha }_1,+.......+{\alpha }_6=12$ ..... $\left(1\right)$

The product of roots would be

${\alpha }_1,{\alpha }_2,.......{\alpha }_6=64$..... $\left(2\right)$

If we apply arithmetic and geometric mean to equation roots

$A.M>G.M$

${\alpha }_1+{\alpha }_2+.......+{\alpha }_6\ge ({\alpha }_1,{\alpha }_2.......+{\alpha }_6{)}^{{1/6}^{}}$

$\frac{12}{6}\ge ({\alpha }_1,{\alpha }_2,.......{\alpha )}^{1/6}$

$2^6\ge \frac{12}{6}\ge ({\alpha }_1,{\alpha }_2,.......{\alpha }_6)$

${\alpha }_1,{\alpha }_2,.......{\alpha }_6)\le 64$

$({\alpha }_1,{\alpha }_2,.......{\alpha }_6)=64$

The roots product to equality if all the roots are equal

$\Rightarrow ({\alpha }_1,{\alpha }_2,.......{\alpha }_6)=2$

Hence the polynomial

$x^6-12x^5+b{x}^4+c{x}^3+d{x}^2+ex+64=0$

Since all the root are $2$

Hence the maximum of the $d$ is greatest absolute value to cancel out the polynomial to zero.