If a and B are fixed values, the equation 2kx + a3 = 2 + X − bk6 about X, no matter what the value of K is, its root is always 1. Find the value of a and B
Both sides of the equation are multiplied by 6 at the same time to get: 4kx + 2A = 12 + x-bk, (4k-1) x + 2A + bk-12 = 0 ①, ∵ no matter what the value of K is, its root is always 1, ∵ substitute x = 1 into ①, 4k-1 + 2A + bk-12 = 0, then when k = 0, k = 1, we can get the equations: − 1 + 2A − 12 = 04 − 1 + 2A + B − 12 = 0, the solution is a = 132, B = - 4
If a and B are fixed values, the equation 2kx + a3 = 2 + X − bk6 about X, no matter what the value of K is, its root is always 1. Find the value of a and B
Both sides of the equation are multiplied by 6 at the same time to get: 4kx + 2A = 12 + x-bk, (4k-1) x + 2A + bk-12 = 0 ①, ∵ no matter what the value of K is, its root is always 1, ∵ substitute x = 1 into ①, 4k-1 + 2A + bk-12 = 0, then when k = 0, k = 1, we can get the equations: − 1 + 2A − 12 = 04 − 1 + 2A + B − 12 = 0, the solution is a = 132, B = - 4
If the equation x ^ 2 + 2 (K + 1) x + K ^ 2-1 = 0 has two roots x 1 and x 2, find the minimum value of x 1 ^ 2 + x 2 ^ 2
∵ X1 and X2 are the two roots of the equation ∵ X1 + x2 = - 2 (K + 1) X1 * x2 = k ^ 2-1 ∵ X1 ^ 2 + x2 ^ 2 = (x1 + x2) ^ 2-2x1 * x2 = [- 2 (K + 1)] ^ 2-2 (k ^ 2-1) = 2K ^ 2 + 8K + 6 ∵ let y = 2K ^ 2 + 8K + 6, the quadratic function has the minimum value ∵ y = - 2, that is, the minimum value of X1 ^ 2 + x2 ^ 2 = - 2
Let proposition p: for the equation x ^ 2 + KX + 1 = 0, there is no real root, and the function f (x) = x ^ 3-kx ^ 2 + 3kx + 1 is on (negative infinity, positive infinity)
If the proposition "P" or "Q" is true and "P" and "Q" are false, the value range of real number k can be obtained
Q must be false
According to the meaning of the topic, the proposition "P" or "Q" is true, and "P and Q" are false,
P is true
k^2-4
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