The equation x / (X-2) + (X-2) / x + (2x + a) / X (X-2) = 0 has only one real root

The equation x / (X-2) + (X-2) / x + (2x + a) / X (X-2) = 0 has only one real root

To the denominator, X & # 178; + (X-2) &# 178; + 2x + a = 0
2x²-2x+4+a=0 （1）
∵ the equation has only one real root
Classification discussion:
The first case is: (1) middle ⊿ = 0
That is: ⊿ = 4-4 * 2 (4 + a) = - 8a-28
-8a-28=0
a=-3.5
In the second case: (1) where ⊿ 0 and one root of the equation is 0,
[(in this case, 0 is the root of (1), but it is not the root of the fractional equation, it is the increasing root of the fractional equation]
That is: ⊿ = 4-4 * 2 (4 + a) = - 8a-28 > 0
a＜-3.5
Substituting x = 0 into (1), we get 4 + a = 0
a=-4
The third case: ⊿ 0 in (1), and the equation has a root of 2,
[(in this case, 2 is the root of (1), but not the root of the fractional equation, it is the increasing root of the fractional equation]
That is: ⊿ = 4-4 * 2 (4 + a) = - 8a-28 > 0
a＜-3.5
Substituting x = 2 into (1), we get 8-4 + 4 + a = 0
a=-8
To sum up, when a = - 3.5, - 4, - 8, the equation x / (X-2) + (X-2) / x + (2x + a) / X (X-2) = 0 has only one real root