Answer:
0 < m < 4-√12 ≈ 0.535898
Explanation:
You want to know the range of values of m that will give |(x-1)(x-3)| = mx four distinct solutions.
Absolute value
The quadratic function f(x) = (x -1)(x -3) will be negative for values of x between the zeros: 1 < x < 3. Hence the absolute value function will invert the graph in that interval, as shown by the red curve in the attachment.
The line y = mx can only intersect that graph in 4 places in the first quadrant. The value of m must be greater than 0 and less than 1.
Upper limit
The upper limit of the slope will be defined by the value of m that makes the line intersect the inverted quadratic exactly once. That is, the discriminant of mx -(-f(x)) = 0 will be zero.
mx +(x -1)(x -3) = x² +(m -4)x +3 = 0
D = (m -4)² -4(1)(3) = (m -4)² -12 = 0
Solving for m gives ...
(m -4)² = 12
m -4 = ±√12
m = 4 ±√12 ≈ 0.54 or 7.46
We can see from the attached graph that m ≈ 7.46 is an extraneous solution. This means the range of m will be ...
0 < m < 4-√12