Question

If roots of the equation $f\left( x \right) = {x^6} - 12{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + 64 = 0$ are positive, then which of the following has the greatest absolute value?
A. b
B. c
C. d
D. e

Answer 1

Hint : Given an equation is an equation with degree six which means it will have 6 roots or 6 factors. So first find the sum of all the roots by dividing the negative coefficient of ${x^5}$ with the coefficient of ${x^6}$ , let this sum be M. And also find the product of the roots by dividing the coefficient the constant term with the coefficient of ${x^6}$ , let this product be N. And then develop a relation between M and N and solve further.

Complete step-by-step answer :
We are given that the roots of the equation $f\left( x \right) = {x^6} - 12{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + 64 = 0$ are positive. We have to find the coefficient with the greatest absolute value.
The given equation is a sextic equation or an equation with degree 6. So it will have 6 roots. Let the roots be p, q, r, s, t and u.
Sum of all the roots of the given equation is $\dfrac{{ - \left( {coefficient\\_of\\_{x^5}} \right)}}{{coefficient\\_of\\_{x^6}}}$ . Coefficient of ${x^5}$ is -12 and the coefficient of ${x^6}$ is 1.
Therefore, the sum of the roots is $p + q + r + s + t + u = \dfrac{{ - \left( { - 12} \right)}}{1} = 12$
In the same way, the product of all the roots of the given equation is $\dfrac{{cons\tan t\\_term}}{{coefficient\\_of\\_{x^6}}}$ . The constant term present in the equation is 64 and the coefficient of ${x^6}$ is 1.
Therefore, the product of the roots is $pqrstu = \dfrac{{64}}{1} = 64$
The arithmetic mean of the roots is $\dfrac{{p + q + r + s + t + u}}{6} = \dfrac{{12}}{6} = 2$
The geometric mean of the roots is $\sqrt[6]{{\left( {pqrstu} \right)}} = \sqrt[6]{{64}} = \sqrt[6]{{{2^6}}} = 2$
When the arithmetic and geometric means of a list of positive numbers are equal, then every number in the list is the same, has the same value.
This means that the values of $p = q = r = s = t = u$
$\Rightarrow p + q + r + s + t + u = 12$
$\Rightarrow p + p + p + p + p + p = 12$
$\Rightarrow 6p = 12$
$\therefore p = \dfrac{{12}}{6} = 2$
Therefore, $p = q = r = s = t = u = 2$
The roots of the given equation are equal and they are equal to 2.
The given equation can also be written as ${\left( {x - 2} \right)^6} = 0$
${\left( {x - 2} \right)^6} = {x^6} - 12{x^5} + 60{x^4} - 160{x^3} + 240{x^2} - 192x + 64 = 0$
On comparing the equation ${x^6} - 12{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + 64 = 0$ with ${x^6} - 12{x^5} + 60{x^4} - 160{x^3} + 240{x^2} - 192x + 64 = 0$ , we get $b = 60,c = - 160,d = 240,e = - 192$ .
The absolute values are $\left| b \right| = 60,\left| c \right| = 160,\left| d \right| = 240,\left| e \right| = 192$
As we can see, the absolute value of d is the greatest which is $240$.
So, the correct answer is “Option C”.

Note : The no. of roots of an equation can be determined by the highest degree of the equation. If the degree of the equation is 2, then it will have 2 roots. If the degree of the equation is 3, then it will have 3 roots. Here the degree is 6 so it has 6 roots. The value of ${\left( {x - 2} \right)^6}$ was found by using the binomial theorem formula.