Given that the zeros of the function f (x) = x ^ 2 + 2mx + 2m + 3 are x1, X2, find the minimum value of X1 ^ 2 + x2 ^ 2 It's a process!

x1+x2=m,x1x2=2m+3,
X1^2+X2^2=(x1+x2)^2-2x1x2=m^2-4m-6=(m-2)^2-10,
The minimum is - 10

The function f (x) = x2-x-m has zero point at (- 1,1) to find the value range of M

F (x) = x ^ 2-x-m has zero point on (- 1,1), x ^ 2-x-m = 0 △ = 1 + 4m ≥ 0, m ≥ - 1 / 4, X1 = [1 + radical (1 + 4m)] / 2x2 = [1-radical (1 + 4m)] / 2 - 1 ＜ [1 + radical (1 + 4m)] / 2 ＜ 1, m ＜ 0 or - 1 ＜ [1-radical (1 + 4m)] / 2 ＜ 1, m ＜ 2 is obtained

If the function f (x) = x2 + MX-1 has zeros in the interval (0,1), then the value range of real number m

f(0)0
m>0

The minimum value of function y = (SiNx) ^ 2 + (3 / (SiNx) ^ 2) (x is not equal to K π, K belongs to Z)

Let t = (SiNx) ^ 2, then 0

Given the quadratic function y = AX2 ax (a is constant, and a is not equal to 0), the vertex of the image is a, and the quadratic function y = x2-2x + 1, the vertex of the image is B
Given the quadratic function y = AX2 ax (a is constant, and a is not equal to 0), the vertex of the image is a, and the quadratic function y = x2-2x + 1, the vertex of the image is B
(1) Judge whether point B is on the image of y = AX2 ax, why
(2) If the image of quadratic function y = x2-2x + 1 passes through point a, find the value of A
0 0

(1) Because the vertex of the quadratic function y = x2-2x + 1 image is B (1,0), substituting x = 1 into y = AX2 ax to get y = 0
(2) If the image of quadratic function y = x2-2x + 1 passes through point a, a is (0.5, - 0.25A), substitute x = 0. Into y = x2-2x + 1, y = 0.25, because a is (0.5, - 0.25A), so a = - 1

Given that the zeros of the function f (x) = x ^ 2 + 2mx + 2m + 3 are x1, X2, find the minimum value of X1 ^ 2 + x2 ^ 2 It's a process!