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Cycles of roots as coefficients (Posted on 2022-06-16) Difficulty: 3 of 5
Part 1:
Find any and all quadratic functions
f(x)=x2+bx+c with roots {b,c}.

Part 2:
Find any and all pairs of quadratic functions
f1(x)=x2+b1x+c1 with roots {b2,c2} and
f2(x)=x2+b2x+c2 with roots {b1,c1}.

Part 3:
Find any and all trios of quadratic functions
f1(x)=x2+b1x+c1 with roots {b2,c2},
f2(x)=x2+b2x+c2 with roots {b3,c3}, and
f3(x)=x2+b3x+c3 with roots {b1,c1}.

No Solution Yet Submitted by Jer    
Rating: 5.0000 (1 votes)

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Solution Puzzle Solution: Part 1 | Comment 1 of 4
Since {b, c} are the roots of the equation:
x^2+bx+c=0
we must have:
2b^2+c=0, and c^2+bc+c=0
2b^2+c gives c= - 2b^2....(#) 
Substituting this in the other equation, we must have:
(- 2b^2)^2 +(b)*(-2b^2) +(-2b^2l =0
Simplifying this , we obtain:
2b^2 (2b^2 -b -1) = 0
So, b=c=0 is a solution.
2b^2 -b-1=0, gives:
(b-1)(2b+1)=0
=> b=1, -1/2
b=1 gives c=-2
b= -1/2 gives c = -1/2  from (#)

Now, checking for the validity of the foregoing, we observe that:
If (b,c) = (0, 0 ), then we have:
x^2=0 so that: x=0, 0

If (b, c) = (1, -2) 
Then we have:
x^2+x-2=0
or, x= 1, -2

If (b, c) = (-1/2, -1/2)
or, x^2-x-1/2 = 0
or, (x-1)(x+1/2) =0
or, x= 1, -1/2, which is a contradiction. 

Consequently,  (b, c) = (0, 0), (1, -2) constitutes all possible solutions to the given problem.

Edited on June 16, 2022, 7:21 am
  Posted by K Sengupta on 2022-06-16 07:13:17

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