Given the equation x ^ 2 + (4x + 1) x + 2m-1 = 0, this equation must have two unequal real roots

Given the equation x ^ 2 + (4x + 1) x + 2m-1 = 0, this equation must have two unequal real roots

Is the equation x ^ 2 + (4m + 1) x + 2m-1 = 0 like this?
△=(4m+1)^2-4(2m-1)
=16m^2+8m+1-8m+4
=16m^2+5
Δ is always greater than zero
So the equation must have two unequal real roots

We know that the sum of two squares of the equation x2 + (M + 9) x + 2m + 6 = 0 is 24, then the value of M is equal to?
A. - 1 or 9 B. - 5 C. - 2 d. - 5 or - 9

D
Suppose that the two roots are a and B, then a + B = - M-9, ab = 2m + 6, and we already know that a ^ 2 + B ^ 2 = 24, that is, (a + b) ^ 2-2ab = 24, we can get - 5 or - 9 by substituting it into the solution

It is known that the sum of two squares of the equation x ^ 2 + (M + 9) x + 2m + 6 = 0 is 24,

x^2+(m+9)x+2m+6=0
Let two roots be X1 and x2
According to Weida's theorem
X1+x2=-(M+9)
X1*X2=2M+6
X1^2+X2^2+2X1X2=24+2X1X2
(X1+X2)^2=24+2X1X2
(m+9)^2=24+12+4m
m^2+14m+45=0
(m+9)(m+5)=0
The solution is m = - 9 or M = - 4
Verification is real root!

The sum of two squares of equation x ^ 2 + (2m + 1) x + m ^ 2-2 = 0 is equal to 11, and m
___ are needed.A.other two B.two else C.two more D.the other two

Let two be X1 and X2, respectively
x1+x2=-(2m+1)
x1x2=m^2 -2
x1^2+x2^2=(x1+x2)^2-2x1x2
=[-(2m+1)]^2-2(m^2-2)
=4m^2+4m+1-2m^2+4
=2m^2+4m+5=11
2m^2+4m-6=0
m^2+2m-3=0
(m+3)(m-1)=0
M = - 3 or M = 1
___ are needed.A.other two B.two else C.two more D.the other two
Choose D

Let the equation of the line l be (m ^ 2-2m-3) x + (2m ^ 2 + m-1) y-2m + 6 = 0, and find the value of m according to the following conditions
(1) The slope of line L is 1
(2) The line L passes through the fixed point P (- 1, - 1)

(1)
(m^2-2m-3)/(2m^2+m-1)=1,
m=2
(2)
m=-2,5/3

When m is a value, the equation 1 / X-1 + m / X-2 = 2m + 2 / (x-1) (x + 2) of X has an increasing root, so we can find the value of M

1 / X-1 + m / X-2 = 2m + 2 / (x-1) (x + 2) [original equation]
(x + 2) (X-2) + m (x + 2) (x-1) = (2m + 2) (X-2) (both sides of the equation are multiplied by the simplest common denominator (x + 2) (X-2) (x-1))
The equation has increasing roots, which may be x = - 2, x = 2, x = 1, because when x is these values, the denominator of the original equation is meaningless
Suppose that the simplified equation with increasing root x = - 2 gets M = - 1, but further consider that when m = - 1, both sides of the original equation are multiplied by the common denominator (X-2) (x-1) at the same time, and finally 3 = 0 is simplified. This equation has no solution. Therefore, there is no increasing root x = - 2
Suppose that the simplified equation with increasing root x = 2 obtains M = 0. Further consider that when m = 0, both sides of the original equation are multiplied by the common denominator (x + 2) (x-1) at the same time, and finally the simplified equation obtains x = 0. Therefore, there is no increasing root x = 2
Suppose that the time of increasing root x = 1 is reduced to obtain M = 1, and further consider that when m = 1, both sides of the original equation are multiplied by the common denominator (x + 2) (X-2) at the same time, and finally x = 0 or 1 is simplified. Therefore, there is increasing root x = 1
In conclusion, only when m = 1, the equation 1 / X-1 + m / X-2 = 2m + 2 / (x-1) (x + 2) has increasing roots

If the equation of a straight line is known to be x + my-2m + 6 = 0, then the straight line passes through a fixed point______ ．

X + my-2m + 6 = 0, can be changed into x + 6 + m (Y-2) = 0, let x + 6 = 0, Y-2 = 0, we can get x = - 6, y = 2, the straight line is always over the fixed point (- 6, 2). So the answer is: (- 6, 2)

Given that the value of solution X and y of equation {MX + y = 2 X-my = - 1 is less than zero, try to find the value of | 2m-1 | divided by 2m-1 - | m + 2 | divided by M + 2
I'm a student in grade one of junior high school. I can write about what we have learned,

mx+y=2
x-my= -1
The results show that x = (2m-1) / 1 + M & sup2;
y=(2+m)/1+m²
According to question 2m-1

It is known that α and β are the two roots of the equation x square + (2m-1) x + 4-2m = 0, and α square + β square is less than 9. The value range of parameter m is obtained

α. β is the two roots of the equation x square + (2m-1) x + 4-2m = 0
α + β = -（2m-1)
α β = 4-2m = 2(2-m)
α squared + β squared < 9
(α+β)^2 - 2α β ＜ 9
{-（2m-1) }^2 - 2* 2(2-m) ＜9
4m^2-7＜9
4m^2＜16
m^2＜4
-2 ＜ m ＜ 2

If a straight line passes through point a (- 3,4), and the sum of intercept on two axes is 12, then the linear equation is______ ．

Let the transverse intercept be a, then the longitudinal intercept be 12-A, and the linear equation be XA + Y12 − a = 1. Substitute a (- 3, 4) to get − 3A + 412 − a = 1, and the solution be a = - 4, a = 9. When a = 9, the linear equation is x9 + Y3 = 1. When a = - 4, the linear equation is x − 4 + Y16 = 1, and the linear equation is y = 4x + 16. To sum up, the linear equation is x + 3y-9 = 0 or y = 4x + 16 6，．