SOLUTION: Find an equation of the form sqrt(ax+b) = cx+d where a, b, c, and d are nonzero integers and 1 is a solution, but -5 is an extraneous solution.

The question is:              ____  
Find an equation of the form Öax+b = cx+d where
a, b, c, and d are nonzero integers and 1 is a solution, but -5 is an
extraneous solution.
----------------------
Set each side of the equation equal to y:
      ____ 
y = -Öax+b = cx+b

Consider the system of equations
      ____ 
y = -Öax+b 
y = cx+b

The first is the equation of the top half of
a parabola whose vertex is on the x-axis.  Something like this:

 

whereas the right side y = cx+d is the equation of the line, and
its point of intersection with the half parabola is is (1,c+d)

The extraneous solution comes from the point where the line intersects
the lower half of the parabola, whose equation is 
      ____ 
y = -Öax+b

 


The complete parabola has the equation y = ±Öax+b which amounts
to the parabola whose equation is

y² = ax+b

The line y = cx+d must intersect this parabola at two points, 
1. the point of intersection above the x-axis must have x-coordinate 1
2. the point of intersection below the x-axis must have x-coordinate -5

Substituting x = 1 and x = -5 into 

y² = ax+b = (cx+d)²
      a+b = (c+d)²
  -5a + b = (-5c+d)² 

Subtract the equations:

       6a = (c+d)² - (-5c+d)² 

Factor the right side:

       6a = [(c+d) - (-5c+d)][(c+d) + (-5c+d)] 

       6a = [c+d+5c-d][c+d-5c+d] 

       6a = 6c(2d-4c) 

        a = c(2d-4c) 

The y-coordinate c+d must be above the x-axis so 

c+d > 0, or d > -c

The y coordinate -5c+d must be below the x-axis, so

-5c+d < 0, or d < 5c

Putting those together we have

-c < d < 5c
 
So we have this set of rules for making equations

1. pick a positive value of c.
2. pick a value of d, such that -c < d < 5c.
3. Calculate a = c(2d-4c). 
4. Calculate b = (c+d)²-a 

OK, let's make one:

1. pick a positive value of c.  Say we pick c = 1
2. pick a value of d, such that -c < d < 5c, say we pick d = 1
3. Calculate a = c(2d-4c). That's a = 1(2·1-4·1) = -2
4. Calculate b = (c+d)²-a = (1+1)²-(-2) = 2²+2 = 4+2 = 6
 ____ 
Öax+b = cx+d 
 _____
Ö-2x+6 = x+1
  
Solving it, we get, squaring both sides:

-2x+6 = x²+2x+1
    0 = x²+4x-5
    0 = (x-1)(x+5)

solutions, 1, -5

1 is a solution, -5 is extraneous.     

Here are some other ones:
 _____
Ö-6x+6 = x-1
 _____ 
Ö2x+14 = x+3
 _______
Ö-24x+49 = 3x+2
 _____
Ö4x+45 = 2x+5
 
Use the above rules to make all you want.

Edwin