First, expand (K \cdot (X+2) = 1):
[ KX + 2K = 1 ]
Next, we imagine (KX+2K = 1) as a linear equation about (x). A linear equation has only one variable, so (k) must be regarded as a constant.
We can rewrite (KX+2K = 1) into the standard form of (KX+2K- 1 = 0). The root of this equation is the value of (x). But requiring different solutions means that it has two different solutions. This shows that the discriminant (\ delta = b 2-4ac) of equation (KX+2K- 1 = 0) must be positive.
Where (a = K), (b = 2K) and (c =-1).
[\ delta =(2k)^2-4 \ cdot k \ cdot(- 1)]
[ \Delta = 4K^2 + 4K ]
Discriminate (\δ> 0) for different solutions. Therefore, we have:
[4k^2+4k & gt； 0 ]
[ 4K(K+ 1)>0 ]
This means that (k) and (K+ 1) must have the same sign. Because we are solving the value of (k), we can determine the possible solution by analyzing the positive and negative of (k).
If (K > 0), then (k+1>; 0 )。 At this time, the signs of (k) and (K+ 1) are the same, and the conditions are met.
If (k
So we come to the conclusion that the range of (k) is a real number. Any positive number or negative number can be taken as the value of (k), so the equation (KX+2K = 1) has two different solutions.