Given that X1 = - 2 is a root of the equation x2 + mx-6 = 0, find the value of M and the other root of the equation x2

Given that X1 = - 2 is a root of the equation x2 + mx-6 = 0, find the value of M and the other root of the equation x2

From the meaning of the question: (- 2) 2 + (- 2) × M-6 = 0, the solution is m = - 1, when m = - 1, the equation is x2-x-6 = 0, the solution is X1 = - 2 & nbsp; & nbsp; & nbsp; x2 = 3, so the other root of the equation is x2 = 3

Given that X1 = - 1 is a root of the equation x ^ 2 + MX-5 = 0, find the root of M and the other root of the equation x2

It is known that X1 = - 1 is a root of the equation x ^ 2 + MX-5 = 0
M = - 4
x^2-4x-5=0
(x-5)(x+1)=0
So the other root x = 5

If one of the equations 2x ^ 2 + 3x + 5m = 0 is greater than 1 and the other is less than 1, the range of M is obtained

I think:
First of all: B ^ 2-4ac > 0, there is a value range of M, M

It is known that the quadratic equation MX & sup2; + (2m-1) x + (M + 1) = 0 has two unequal real roots. The range of M is expressed by interval

The equation has two unequal real roots, so m ≠ 0,
△=(2m-1)^2-4m(m+1)＞0
The solution is m ≠ 0 and m ＜ 1 / 8
m∈(-∞,0)∪(0,1/8)

If the quadratic equation (m-1) x2 MX + 1 = 0 of X has two unequal real roots, then the value range of M is______ ．

∵ equation is a quadratic equation with one variable, ∵ M-1 ≠ 0, that is, m ≠ 1, ∵ equation has two unequal real roots, ∵ Δ= (- M) 2-4 (m-1) = (m-2) 2 ＞ 0, ∵ m ≠ 2, m ≠ 1 and m ≠ 2

The image of the quadratic function y = two-thirds X & sup2; is shown in Fig. 6, and the point A0 is at the origin of the coordinate
A1, A2,..., A2010 are on the positive half axis of y-axis, B1, B2,..., b2010 are on the quadratic function y = two-thirds X

This problem has been done, equilateral triangle side length = 2010
a1=1
a2=2
a3=3
.
a2010=2010

Given the quadratic function y = 3x & sup2; + 12x + 13, then the minimum value of function value y is?
Why is the answer one,

If we use the formula method
When x = - 12 / 6 = - 2, the function gets the minimum value, which is (4x3x13-12x12) / 12 = 1

Y = 2x2 + 5x + 18, please convert the quadratic function of y = AX2 + BX + C into the form of Y + a (x + H) + K,

y=2(x+5/4)^2+119/8

According to the corresponding values of the independent variable X of the quadratic function y = AX2 + BX + C and the function value Y in the following table, the symmetry axis of the function image can be determined as
According to the corresponding values of the independent variable X of the quadratic function y = AX2 + BX + C and the function value Y in the table below, X is: - 3 - 201 5 and Y is: - 29 - 151 3 - 29 respectively, the symmetry axis of the function image can be determined as a.x = - 1, b.x = 0, c.x = 1, D.X = - 29

X are: - 3 - 201 5, respectively,
Y are: - 29 - 15 13 - 29,
When x = - 3 or 5, y = - 29
So the axis of symmetry x = (- 3 + 5) / 2 = 1
Choose C

It is known that the ordinate of the intersection of the quadratic function y = AX2 + BX + C and the Y axis is - 2, and when x = 1, this function can obtain the maximum value of 2, and the analytic expression of this function is obtained

"It is known that the ordinate of the intersection of quadratic function y = AX2 + BX + C and Y axis is - 2" and C = - 2 (let x = 0, y = - 2) is obtained
The maximum value is x = 1 = - B / 2A (axis of symmetry) and the maximum value is 2 = (4ac-b2) / 4A
The solution is a = - 4, B = 8, C = - 2