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.