Given two real roots of the equation x2 + 2px + 1 = 0 about X, one is less than 1 and the other is greater than 1, then the value range of real number P is______ ．

Given two real roots of the equation x2 + 2px + 1 = 0 about X, one is less than 1 and the other is greater than 1, then the value range of real number P is______ ．

Let f (x) = x2 + 2px + 1, ∵ the equation x2 + 2px + 1 = 0 with respect to X has two real roots, ∵ f (1) = 1 + 2p + 1 = 2p + 2 ＜ 0, ∵ P ＜ - 1, ∵ the equation x2 + 2px + 1 = 0 with respect to X has an opening upward, ∵ one of the two real roots is greater than 1, the other is less than 1 (as shown in the sketch), ∵ f (1) = 1 + 2p + 1 = 2p + 2 ＜ 0, ∵ P ＜ - 1, ∵ the range of P ＜ - 1

The equation x ^ 2 + 2 (M + 3) x + 2m + 14 = 0 of X has two real roots, one is less than 1, the other is more than 3. Find the range of M

This problem is that the distribution of solutions of quadratic equation with one variable is often found in the college entrance examination and the final examination, which is an important knowledge point in compulsory 1. Generally, the function method is used
To solve the problem, first draw a schematic diagram according to the meaning of the problem, master the three elements, the scope of the axis of symmetry (you can not write if you are uncertain), △ and the sign corresponding to the left and right values, so the solution is as follows
Discriminant > 0
1 ＜ axis of symmetry ＜ 3
f(1)＜0
f（3）＜0
We can solve the M range