If the equation 4 ^ x-m * 2 ^ (x + 1) + 5 = 0 has two unequal real roots, then the value range of real number m

If the equation 4 ^ x-m * 2 ^ (x + 1) + 5 = 0 has two unequal real roots, then the value range of real number m

Because m * 2 ^ (x + 1) = 2m * 2 ^ X4 ^ x = (2 ^ x) & sup2; let t = 2 ^ x > 0, then the above formula becomes T & sup2; - 2Mt + 5 = 0 equation 4 ^ x-m * 2 ^ (x + 1) + 5 = 0 has two unequal real roots, then T & sup2; - 2Mt + 5 = 0 has two unequal real roots

It is known that the quadratic equation AX ^ 2-4x + 1 = 0 with respect to X has no real root, then the value range of a is_____

⊿=（-4）²-4a＜0
16-4a＜0
4a＞16
a＞4

a. The equation | x2 + ax + B | = 2 about X has three unequal real roots. (1) prove that a2-4b-8 = 0; (2) if the three unequal real roots of the equation are exactly the degrees of the three inner angles of a triangle, prove that the triangle must have an inner angle of 60 °; (3) if the three unequal real roots of the equation are exactly the three sides of a right triangle, find the values of a and B

It is proved that: (1) from the original equation: x2 + ax + B-2 = 0 (1), X2 + ax + B + 2 = 0 (2), the discriminant of the two equations are respectively: △ 1 = a2-4b + 8, △ 2 = a2-4b-8, ∵ the original equation has three roots, ∵ one equation has two unequal real roots, the other equation has two equal real roots, namely △ 1, ∵ 2

If the equation x2 + ax + A2-1 = 0 about X has a positive root and a negative root, then the value range of a is______ ．

Let f (x) = x2 + ax + A2-1, the opening of quadratic function be upward, if the equation has a positive root and a negative root, then only f (0) < 0 is needed, that is, A2-1 < 0, ∧ - 1 < a < 1