The equation MX2 + 2 (M + 3) x + 2m + 14 = 0 of X has two real roots, one of which is greater than 4 and the other is less than 4

The equation MX2 + 2 (M + 3) x + 2m + 14 = 0 of X has two real roots, one of which is greater than 4 and the other is less than 4

The constructor f (x) = MX2 + 2 (M + 3) x + 2m + 14 ∵ one is greater than 4, the other is less than 4, ∵ MF (4) < 0 ∵ m (26m + 38) < 0 ∵ 1913 < m < 0

It is known that for X-2 (m-1) x + m-2m = 0, if one of the two real roots of the equation is less than 1 and the other is greater than 2, then the value range of M is 0___

Because the equation has two real roots, if the square of B - 4ac > 0 is 4 > 0, there will always be two roots
If you draw a picture again, you can see f (1)

If a focus of hyperbola 2x2-y2 = m is (0, 3), then the value of M is______ ．

Hyperbola 2x2-y2 = m, i.e. 2x2m − y2m = 1. From the meaning of the title, we know that m < 0, its focus is (0, ± − 3M2), {− 3M2 = 3,} M = - 2, so the answer is: - 2

If the distance from the point on the hyperbola X & # 178; / M-Y & # 178; = 1 to the left quasilinear is 1 / 3 of the distance to the left focus, then M =
:

According to the second definition of hyperbola, eccentricity e = 3,
a^2=m,b^2=1,e=c/a=3,c=3a,c^2=9a^2
b^2=c^2-a^2=9m-m=8m,
8m=1,
∴m=1/8.