Finding Mathematics Emperor Junior 3 Equation Problems, Junior 3 Mathematical Equation Problems

1. x²+（m+1)x+m =0

x+1)(x+m)=0

x=-1 or x=-m

i.e. when m<>1, intersect at (-1,0), m,0), and when m=1, intersect at (-1,0).

2. |1-(-m)|=2

m-1|=2

m=3 or m=-1

y=（x+1）（x+m）

When y=0, x=-1, -m two roots.

When m=1 is one intersection point with the x-axis and m≠1, two intersection points.

The length of the line segment is 2, that is, the absolute value of -1-(-m) is equal to 2, m-1 = 2m=3 or -1

It's easy!

1: Since it is the intersection point with the x-axis, then the ordinate is zero, and the x-axis can be solved;

2: Let the equation be y=0, solve the root, and then let the two roots subtract to get the value of m.

Let y=1 use the decidive formula =(m+1) -4m=(m-1) 0, so there are one or two intersections. Discuss m

y=（x+1）（x+m）

The intersection point is (-1,0)(-m,0).

The absolute value of 1-(-m) = 2

m = -1 or 3

1.Analysis: To find the intersection of the parabola and the x-axis like this, it depends on whether it is = or <0 where =b 2 4ac

Answer: It depends on the question.

m+1）²-4m=（m-1）²≥0

When m=1, =0The equation has a root, i.e., the parabola has an intersection with the x-axis.

When m≠1, >0, the equation has two roots, i.e., the parabola has two intersections with the x-axis.

2.Analysis: The title says that the length of the line segment on the x-axis is 2, then the parabola must have two intersections with the x-axis, that is, when m≠1, >0

As the title suggests, x1 x2 = 1 (m) = 2m=3 or m= 1

1. When y=0, x=-m and -1, if m=1, an intersection; m does not = 1, two intersections.

2. Since the length of the line segment intercepted by the parabola on the x-axis is 2, then |m-1|=2, so m=3 or -1

1.Solution: 6x 2-12x+7=6(x 2-2x+1)+1=6(x-1) 2+1>=1>0

2.Solution: This problem only needs to write a few square forms, such as (2x+1) 2 and the like, and add a positive number to the clever shot. The reason for the wide chaos is that it is completely imitating the way of envy and flat, and Evergrande is equal to zero.

6(x 2-2x+1)+1=6(x-1) 2+1x-1) 2 is greater than or equal to zero.

Therefore, the former Evergrande changed to the state of zero boy.

The second question is based on the first question.

Write first (x- or the source + a certain number) 2 + a certain number.

And then take it apart.

For example, (x+2) 2+4=x 2+4x+8

a²-3ab+2b²=(a-b)(a-2b)

(a-b)+(a-2b)=2a-3b

So, the original equation can be made by factoring(x+a-b)(x+a-2b)=0The solution of the equation is:x1=b-ax2=2b-a

This can be calculated using the root finding formula.

(x-1)(x-2)(x-3)(x-4)=0(x-1)=0

or (x-2)=0

or (x-3)=0

or (x-4)=0

This results in x=1 or x=2 or x=3 or x=4

Let's be honest, you know the answer by looking at it without writing the process x=1 or x=2 or x=3 or x=4

First the equation is reduced to a form, and then the classification is discussed when both roots are negative; If one root is negative, it can be solved according to the discriminant formula of the root and the relationship between the root and the coefficient.

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(2) The two sides of the equation are multiplied by (x 2-x-2) at the same time, and the two sides are simplified to obtain (x-1)(x+1)+(x-2)(2-x)=2x+a, that is, 2x=a+5, x is negative, 5+a<0, a<-5

Multiply the left and right sides of the equation by 6'

2（5x-9）- 3（10-x）=6（x-1）10x-18-30+3x-6x+6=0

7x-42=0

So x=6

Multiply both sides of the equation by 6 to get.

2(5x-9)-3(10-x)=6(x-1)

10x-18-30+3x=6x-6

Merge similar items to obtain.

13x-48=6x-6

7x=42x=6

Multiply 6 on both sides to get 10x-18-3(10-x)=6x-6 Simplification: 13x-48=6x-6

So 7x = 42x = 6