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Solving special higher order equations X^3-3x^2+x+2=0 If the equation x ^ 3-4x ^ 2 + AX = 0 has and only has one real root, what is the value range of a That's the problem
1)x^3-3x^2+x+2=0
(x-2)(x^2-x-1)=0
Then solve X-2 = 0 and x ^ 2-x-1 = 0
2) X ^ 3-4x ^ 2 + AX = 0 has and only has one real root
That is, X (x ^ 2-4x + a) = 0 has and only has one real root, and this root is x = 0
So x ^ 2-4x + a = 0 has no real root
16-4a4
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