If a, B ∈ R, and | a | + | B | ≤ 1, and at least one absolute value of two X1 and X2 of equation x2 + ax + B = 0 is not less than 1, it is proved that | a | + | B | = 1

If a, B ∈ R, and | a | + | B | ≤ 1, and at least one absolute value of two X1 and X2 of equation x2 + ax + B = 0 is not less than 1, it is proved that | a | + | B | = 1

Let | x1 | ≥ 1,
From the relationship between root and coefficient, we can get: X1 + x2 = - A, X1 + x2 = B,
∴|a|+|b|=|x1+x2|+|x1x2|
≥|x1|-|x2|+|x1||x2|
≥1-|x2|+|x2|≥1.
And | a | + | B | ≤ 1,
So | a | + | B | = 1

Given that one root of the equation x2 + kx-2 = 0 is 1, then the other root is 1______ ．

Let the other root of the equation be x1. According to the meaning of the question, we get x1 × 1 = - 2, so X1 = - 2

If we know that one root of the equation x2 + kx-2 = 0 is 1, then the other root is ()
A. -3B. 3C. -2D. 2

Let the other root of the equation be t. according to the meaning of the problem, we get 1.t = - 2, and the solution t = - 2