Let a = {the square of X / x-3x + 2 = 0}, B = {the square of X / x + 2 (a + 1) x + (the square of a-5) = 0. If AUB = a, find the value range of real number a? Complete solution steps, at least I can understand! The answer is given, where △ is the square of 4 (a + 1) - 4 (the square of a - 5) = 8 (a + 3); how did it come from? Why do we have to use it to judge? This △ = 4 (a + 1) square-4 (A's square-5) = 8 (a + 3); how can we get it? Or is there a dead formula for such a similar problem, as long as it's hard? Sorry, it's a bit stupid @!

Let a = {the square of X / x-3x + 2 = 0}, B = {the square of X / x + 2 (a + 1) x + (the square of a-5) = 0. If AUB = a, find the value range of real number a? Complete solution steps, at least I can understand! The answer is given, where △ is the square of 4 (a + 1) - 4 (the square of a - 5) = 8 (a + 3); how did it come from? Why do we have to use it to judge? This △ = 4 (a + 1) square-4 (A's square-5) = 8 (a + 3); how can we get it? Or is there a dead formula for such a similar problem, as long as it's hard? Sorry, it's a bit stupid @!

AUB = a, so there are three possibilities for B sets
The first one has two equal roots, which are 1 or 2. Suppose that △ = 0, calculate the value of a and verify it again
The second kind of B is an empty set, as long as △ 0, calculate the value of ah
Third, B is equal to A. use the relationship between root and coefficient to find a, and then check
The size of △ is zero and equal to zero is the key to whether the quadratic equation has a real number solution