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}.
The given conditions generate four simultaneous equations:
b2^2 +b1* b2 +c1 = 0
c2^2 + b1 *c2 + c1 = 0
b1^2 + b1*b2 + c2 = 0
c1^2 + b2 * c1 + c2 = 0
This generates four cases:
CASE 1: b2=c2 and b1=c1
CASE 2: b2=c2 and b1+c1+b2= 0
CASE 3: b2+c2+b1 =0 and, b1=c1
CASE 4: b2+c2+b1=0 and, b1+c1+b2=0
From here on it gets extremely lengthy and messy.
To cut a long story short, solving the cases one by one, we obtain:
(b1, b2, c1, c2)
= (-1/2, -1/2, -1/2, -1/2)
or, (1, 1, -2, -2)
or, (m, -m, 0, 0) where m is any real number including zero.
Checking the first validity of the first quadruplet, we observe that
x^2-x/2-1/2 =0
or, x=1, -1/2, which is a contradiction.
For the second quadruplet, we observe that:
x^2+x-2=0, gives: x=1, -2
-> This checks out
For the last one:
x^2-mx =0 has roots (m, 0)
x^2+mx=0 has roots (-m, 0)
Consequently, (b1, b2, c1, c2) = (1, 1, -2, -2) and, (m, -m, 0, 0) gives all possible solutions to the given equation.
Note: (b1, b2, c1, c2) =(0, 0, 0, 0) is included in the 2nd quadruplet when m=0
Edited on June 16, 2022, 7:43 am
Puzzle Solution: Part 2
