A problem about quadratic equation of one variable If the equation x & sup2; + MX + 1 = 0 and the equation x & sup2; - x-m = 0 have only one real root, the value of M is obtained [note]: & sup2; means square; it's better to do it by formula method. I don't know the right answer-|||

A problem about quadratic equation of one variable If the equation x & sup2; + MX + 1 = 0 and the equation x & sup2; - x-m = 0 have only one real root, the value of M is obtained [note]: & sup2; means square; it's better to do it by formula method. I don't know the right answer-|||

Let the same root be a, then there is:
a^2+ma+1=0
a^2-a-m=0
By subtracting the two formulas, we get: (M + 1) a + 1 + M = 0
(m+1)a=-(m+1)
Because m is not = - 1, then M + 1 is not = 0
So, a = - 1. That is to say, the common solution is - 1
Substituting into the equation: 1-m + 1 = 0, M = 2

It is known that the univariate quadratic equation (M + 1) x2-x + m2-3m-3 = 0 has a root of 1, then the value of M is 1______ ．

∵ the univariate quadratic equation (M + 1) x2-x + m2-3m-3 = 0 of X has a root of 1, ∵ m + 1-1 + m2-3m-3 = 0, and M + 1 ≠ 0, the solution is m = 3

The problem of solving a quadratic equation of one variable
Given that C is a constant and the opposite number of a root of the equation x ^ 2-3x + C = 0 is a root of the equation x ^ 2 + 3x-c = 0, find the root of the equation x ^ 2-3x + C and the value of C

Let a root of x ^ 2-3x + C = 0 be m
Then - M is the root of the equation x ^ 2 + 3x-c = 0
Separately substituting
X=M,M^2-3M+C=0
X=-M,M^2-3M-C=0
Subtraction, 2C = 0
C=0
X^2-3X+C=X^2-3X=0
X(X-3)=0
X=0,X=3
So the root of x ^ 2-3x + C = 0 is x = 0, x = 3
C=0

The minimum value of the function y = x ^ 3-3x + 9 is

y'=3x²-3=0
x=±1
X1, y '> 0, increasing function
-1

The number of intersections between the image of function y = f (x) and the line x = a in the same coordinate system is
A. 1 B.2 c.0 d.0 or 1 is possible

Since it is a function domain, every x value corresponds to a f (x)
So choose D. if x = a is defined, it will be one. If x = a is not defined, it will be zero

In the same coordinate system, it is better to make the image of function y = 2 / X and function y = X-1, and use the image to find their intersection coordinates

But I really don't understand your so-called two methods, which are as simple as solving equations

In the same coordinate system, make the image of function y = X-1 and function y = 2 / x, and use the image to find their intersection coordinates. How to list and draw?

List: x y = X-1 y = 2 / X
-2 -3 -1
-1 -2 -2
0 - 1 is meaningless
1 0 2
2 1 1
Draw: do it yourself

When m takes what value, the image of the function y / (m ^ 2-5M) = 1 / ([x ^ (2m ^ 2-33)] is a straight line passing through the origin and the first and third quadrants
When m takes what value, the image of the function y / (m ^ 2-5M) = 1 / ([x ^ (2m ^ 2-33)] is a straight line passing through the origin and the first and third quadrants

y/(m^2-5m)=1/([x^(2m^2-33)]
y=(m^2-5m)x^(-2m^2+33)
∵ an image is a straight line passing through the origin and the first and third quadrants
∴m^2-5m>0,-2m^2+33=1
m5;m=±4
∴m=-4
y=36x

Given that the image of function y = (2k + 1) x + 1 / 2 does not pass through the third quadrant, then the value range of K is

Without passing through the third quadrant, 2K + 1

Please write out a quadratic function expression which satisfies the following conditions: (1) the opening is downward; (2) the function image passes through the point (- 2,0) (5,0)
I know the answer y = - x2 + 3x + 10, the answer is not unique.. but I don't know how to calculate it ~ please advise!

So let y = a (x + 2) (X-5),
The opening is downward, as long as a is negative. There are countless such functions
y=-(x+2)(x-5)=-x^2+3x+10
y=-2(x+2)(x-5)=-2x^2+6x+20
.