1. If the univariate quadratic equation x2 + ax + 1 ≥ 0 holds for all real numbers x, find the value range of real number A. (what is △ at this time?) For all real numbers x, the inequality AX2 + 4x + a ＞ 1-2x2 holds, and the value range of real number a is obtained Ask specific process.. the teacher did not speak carefully And when △ why is constant established? Is constant meaningful?

1. If the univariate quadratic equation x2 + ax + 1 ≥ 0 holds for all real numbers x, find the value range of real number A. (what is △ at this time?) For all real numbers x, the inequality AX2 + 4x + a ＞ 1-2x2 holds, and the value range of real number a is obtained Ask specific process.. the teacher did not speak carefully And when △ why is constant established? Is constant meaningful?

1. The univariate quadratic equation x2 + ax + 1 of X ≥ 0,
If you look at the function y = x & # 178; + ax + 1, its image is open upward, y = x & # 178; + ax + 1 ≥ 0, to be constant, its lowest point is greater than or equal to 0, that is, the function and X-axis can only have one intersection point at most, that is, X & # 178; + ax + 1 = 0 can only have one real root at most (there can be no real root, that is, the image is above the x-axis, there is no intersection point), that is △ = A & # 178; - 4 ≤ 0 solution, - 2 ≤ a ≤ 2
It can also be solved according to "the lowest point is greater than or equal to 0", that is, y = x & # 178; + ax + 1 = (x + A / 2) & # 178; + 1 - (A / 2) & # 178;, the lowest point is at x = - A / 2, y = 1 - (A / 2) & # 178; ≥ 0, and the same solution can be obtained. (the lowest point is greater than or equal to 0, and this function is constant when it is greater than or equal to 0.)
2. Ax & # 178; + 4x + a ＞ 1-2x & # 178;, then (a + 2) x & # 178; + 4x + (A-1) ＞ 0 holds,
When a + 2 ＜ 0, the opening of the image is downward, and (a + 2) x & # 178; + 4x + (A-1) ＞ 0 is not always true. (when a + 2 = 0, the inequality is not always true and can be considered together.)
When a + 2 ＞ 0, that is, when a ＞ - 2, the opening of the image is upward, (a + 2) x & # 178; + 4x + (A-1) ＞ 0 is always true, which is similar to problem 1. It is required that there is no real root with the x-axis, and the image is all above the x-axis, i.e., △ = 4 & # 178; - 4 * (a + 2) (A-1) ＜ 0. The solutions of a ＜ - 3 and a ＞ 2, respectively, intersect with the precondition a ＞ - 2. (note the slight difference between problem 1 ＞ and ≥ 2
That is: the value range of real number a is more than 2
For y = MX & # 178; +,, and so on, (mainly considering how to make the image completely above or below the x-axis. The difference between one real root and two real roots)
If M ＞ 0, the opening of the image is upward, as long as △ ≤ 0, y ≥ 0 can be made constant, and Y < 0 cannot be made constant,
If M ＜ 0, the opening of the image is downward, then it is only possible for y ≤ 0 to be constant. At this time, it is also △ 0. It is impossible for Y > 0 to be constant
Learn to draw inferences from one instance, do a problem to consider comprehensive, pay attention to details