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Common root in quadratic combinations (Posted on 2025-03-08) Difficulty: 3 of 5
Let P(x) = x2 - 3x - 7, and let Q(x) and R(x) be two quadratic polynomials also with the coefficient of x2 equal to 1. David computes each of the three sums P + Q, P + R, and Q + R and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If Q(0) = 2, then find R(0).

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There are 3 quadratics: PQ, QR and RP (the sum results, named for 
their components). Each has 2 roots that are symmetric about its 
axis of symmetry, x=-b/2. In order to have roots match in pairs, 
their arrangement on the x-axis number line, where "o" is x=-b/2 and 
the X's are the roots, must overlap schematically like this:
(The X's align)


X----------o----------X
X---o---X           
           X------o-----X

One of the 3 functions is the widest. For each of selection of the 
widest, there are two possibilities for which one of the remaining 
two is the lower function, and which one is the upper. The six possible 
arrangements are:

1) PQ spanning the contiguous pair QR, RP  
2) PQ spanning the contiguous pair RP, QR  
3) QR spanning the contiguous pair RP, PQ  
4) QR spanning the contiguous pair PQ, RP  
5) RP spanning the contiguous pair PQ, QR  
6) RP spanning the contiguous pair QR, PQ  

We have:
P = x^2 - 3x -7  with coefficients: 
(ap,bp,cp) = (1,-3,-7)
Q = x^2 + bq x + 2
R = x^2 + br x + cr   Note: cr = R(0)
So, the summed coefficients are: 
PQ = (2, -3+bq, -5)
QR = (2, bq+br, 2+cr) 
RP = (2, br+3,  cr-7) 

In each case above we have 3 equations 
("-", "+" indicate lower and upper roots): 
For example case 1) gives:
root PQ- = root QR-     
root QR+ = root RP-    
root PQ+ = root RP+   

and 3 unknowns: bq, br, cr -> I will call these x,y, and z.
(Not to be confused with the x and y or the quadratics!) 
The roots for each function are: 
 
lower PQ
(3-x) - sqrt( (3-x)^2 + 40)
upper PQ
(3-x) + sqrt( (3-x)^2 + 40)
lower QR
 -(x+y) - sqrt((x+y)^2 - 8(2+z) )
upper QR
-(x+y) + sqrt ((x+y)^2 - 8(2+z))
lower RP
-(y+3) - sqrt((y-3)^2-8(z-7))
upper RP 
-(y+3) + sqrt((y-3)^2-8(z-7))

Here are the 6 cases
1) PQ spanning the contiguous pair QR, RP  (we arbitrarily illustrated this)
(3-x) - sqrt( (3-x)^2 + 40)    =  -(x+y)-sqrt((x+y)^2 -8(2+z)),
 -(x+y) + sqrt((x+y)^2 - 8(2+z)) = -(y-3) - sqrt((y-3)^2-8(z-7)) ,
 (3-x) + sqrt( (3-x)^2 + 40) =  -(y-3) + sqrt((y-3)^2-8(z-7))

2) PQ spanning the contiguous pair RP, QR 
(3-x) - sqrt( (3-x)^2 + 40)    = -(y-3) - sqrt((y-3)^2-8(z-7)), 
-(y-3) +   sqrt((y-3)^2-8(z-7)) = -(x+y) - sqrt ((x+y)^2 - 8(2+z)),
 (3-x) + sqrt( (3-x)^2 + 40)   = -(x+y) + sqrt ((x+y)^2 - 8(2+z))

3) QR spanning the contiguous pair RP, PQ  
-(x+y) - sqrt((x+y)^2 -8(2+z)) = -(y-3) - sqrt((y-3)^2-8(z-7)),
 -(y-3) + sqrt((y-3)^2-8(z-7))   = (3-x)  - sqrt( (3-x)^2 + 40)  ,
-(x+y) + sqrt ((x+y)^2 -8(2+z))= (3-x) + sqrt( (3-x)^2 + 40)

4) QR spanning the contiguous pair PQ, RP  
-(x+y) -sqrt((x+y)^2 - 8(2+z)) =  (3-x) - sqrt( (3-x)^2 - 40) ,
 (3-x) + sqrt( (3-x)^2 - 40)     = -(y+3) - sqrt((y+3)^2-8(z-7)) ,
-(x+y) + sqrt ((x+y)^2 -8(2+z))= -(y+3) + sqrt((y+3)^2-8(z-7))

5) RP spanning the contiguous pair PQ, QR
-(y-3) - sqrt((y-3)^2-8(z-7)) =   (3-x) -   sqrt( (3-x)^2 +40),
 (3-x) + sqrt( (3-x)^2 + 40)     = -(x+y) - sqrt((x+y)^2-8(2+z)) ,
  -(y-3) + sqrt((y-3)^2-8(z-7))= -(x+y) + sqrt((x+y)^2 - 8(2+z) )
  
6) RP spanning the contiguous pair QR, PQ 
-(y-3) - sqrt((y-3)^2-8(z-7)) =  -(x+y) - sqrt((x+y)^2-8(2+z)),
 -(x+y)+ sqrt((x+y)^2-8(2+z))    = (3-x) - sqrt( (3-x)^2 + 40)   ,
 -(y-3) + sqrt((y-3)^2-8(z-7)) = (3-x) + sqrt( (3-x)^2 + 40)

With a typo corrected (3/11) this give the right answer (Case 1):
x=10/3 (b_Q), y=54/19 (b_R) and z= 52/19 R(0)

Edited on March 11, 2025, 8:33 am
  Posted by Steven Lord on 2025-03-10 00:46:05

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