Part 1:
Find any and all quadratic functions
f(x)=x
2+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}.
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
Puzzle Solution: Part 1
