SOLUTION: Let r, s, and t be the roots of the equation x^3 - 2x + 1 = 0 in some order. What is the maximal value of r^3 - s- t?

Algebra ->  Inequalities -> SOLUTION: Let r, s, and t be the roots of the equation x^3 - 2x + 1 = 0 in some order. What is the maximal value of r^3 - s- t?      Log On


   

Question 1178729: Let r, s, and t be the roots of the equation x^3 - 2x + 1 = 0 in some order. What is the maximal value of r^3 - s- t?
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E3+-+2x+%2B+1+=+0..........factor
%28x+-+1%29+%28x%5E2+%2B+x+-+1%29+=+0
one root is r=1

use quadratic formula to find other two roots for
x%5E2+%2B+x+-+1=+0

x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x=%28-1%2B-sqrt%281%5E2-4%2A1%2A%28-1%29%29%29%2F%282%2A1%29

x=%28-1%2B-sqrt%281%2B4%29%29%2F2

x=%28-1%2B-sqrt%285%29%29%2F2

=> s=%28-1%2Bsqrt%285%29%29%2F2 =>s=-1%2F2%2Bsqrt%285%29%2F2
and t=%28-1-sqrt%285%29%29%2F2=>t=-1%2F2-sqrt%285%29%2F2

then
r%5E3+-+s-+t=1%5E3+-+%28-1%2F2%2Bsqrt%285%29%2F2%29-+%28-1%2F2-sqrt%285%29%2F2%29
r%5E3+-+s-+t=1%2B1%2F2-sqrt%285%29%2F2%29%2B1%2F2%2Bsqrt%285%29%2F2
r%5E3+-+s-+t=1%2B1%2F2%2B1%2F2
r%5E3+-+s-+t=1%2B1
r%5E3+-+s-+t=2



Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let r, s, and t be the roots of the equation x^3 - 2x + 1 = 0 in some order.
What is the maximal value of r^3 - s- t?
~~~~~~~~~~~~~~~~~~~~~~~~


            The solution in the post by  @MathLover1  is not sufficient.

            I came to bring the correct solution. 


First, you need to read, to interpret and to understand the condition correctly.


    +------------------------------------------------------------------------------------+
    |    The problems asks to find the maximal value of the expression of  r^3 - s - t   |
    |                                                                                    |
    |    over ALL POSSIBLE PERMUTATIONS of the roots r, s and t.                         |
    +------------------------------------------------------------------------------------+



To find the value of  r^3 - s - t  for only one possible permutation, as @MathLover1 does,
IS NOT ENOUGHT to solve the problem.



Now, if r is one of the roots, then  r^3 - 2r + 1 = 0,  which implies


    r^3 = 2r - 1  and further  r^3 - s - t = (2r-1) - s - t = 3r - (r + s + t) - 1.


The sum of the roots  (r + s + t) is the coefficient at x^2 of the original equation, taken with
the opposite sign (the Vieta's theorem).

In our case, this coefficient is zero;  therefore


    r^3 - s - t = 3r - (r + s + t) - 1 = 3r - 1.


THEREFORE,  the expression  r^3 - s - t  is maximal when 3r  is maximal, or, equivalently,  
when the root  "r"  is maximal of the three roots.


The roots of the equation  x^3 - 2x + 1 = 0  are  1,  %28-1%2Bsqrt%285%29%29%2F2  and  -%281%2Bsqrt%285%29%29%2F2,

    as @MathLover1 did find in her post.


Of them, the root  1  has maximum value.

THEREFORE, from what is written above in my post, the maximal value of the expression  r^3 - s - t  is  3*1 - 1 = 2.   


ANSWER.  Under given conditions, the maximal value of the expression  r^3 - s - t  is  2.

Solved (correctly).