设集合A=﹛χ|χ²+4χ=0﹜,B=﹛χ|χ²+2(α+1﹚χ+α²-1=0﹜,如果B包含于A,求实数α的取值范围
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![设集合A=﹛χ|χ²+4χ=0﹜,B=﹛χ|χ²+2(α+1﹚χ+α²-1=0﹜,如果B包含于A,求实数α的取值范围](/uploads/image/z/12627833-41-3.jpg?t=%E8%AE%BE%E9%9B%86%E5%90%88A%3D%EF%B9%9B%CF%87%EF%BD%9C%CF%87%26%23178%3B%EF%BC%8B4%CF%87%EF%BC%9D0%EF%B9%9C%2CB%3D%EF%B9%9B%CF%87%EF%BD%9C%CF%87%26%23178%3B%EF%BC%8B2%EF%BC%88%CE%B1%EF%BC%8B1%EF%B9%9A%CF%87%EF%BC%8B%CE%B1%26%23178%3B%EF%BC%8D1%EF%BC%9D0%EF%B9%9C%2C%E5%A6%82%E6%9E%9CB%E5%8C%85%E5%90%AB%E4%BA%8EA%2C%E6%B1%82%E5%AE%9E%E6%95%B0%CE%B1%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
设集合A=﹛χ|χ²+4χ=0﹜,B=﹛χ|χ²+2(α+1﹚χ+α²-1=0﹜,如果B包含于A,求实数α的取值范围
设集合A=﹛χ|χ²+4χ=0﹜,B=﹛χ|χ²+2(α+1﹚χ+α²-1=0﹜,如果B包含于A,求实数α的取值范围
设集合A=﹛χ|χ²+4χ=0﹜,B=﹛χ|χ²+2(α+1﹚χ+α²-1=0﹜,如果B包含于A,求实数α的取值范围
B是A的子集
若B是空集
则方程无解
判别式小于0
4(a²+2a+1)-4(a²-1)