The square of (2Y + 1) - 8 (2Y + 1) + 15 = 0 is used to solve the equation

The square of (2Y + 1) - 8 (2Y + 1) + 15 = 0 is used to solve the equation

The square of (2Y + 1) - 8 (2Y + 1) + 15 = 0
[(2y+1)-3][(2y+1)-5]=0
(2y-2)(2y-4)=0
(y-1)(y-2)=0
So y = 1 or y = 2
[(2y+1)-3][(2y+1)-5]=0
(2y-2)(2y-4)=0
4(y-1)(y-2)=0
y=1,y=2
Break 15 into 16-1, and then there's 16-1
[(2Y + 1) - 4] Square-1 = 0,,
Here is the square difference formula..
You should understand.
The square of (2Y + 1) - 8x + 15 = 0
(2Y+1-3)(2Y+1-5)=0
The curve represented by equation x2 + xy = x is ()
A. One point B. one line C. two lines D. one point and one line
The equation x2 + xy = x, that is, X (x + Y-1) = 0, can be simplified to x = 0 or x + Y-1 = 0. X = 0 represents a straight line, and X + Y-1 = 0 also represents a straight line, so the curve of equation x2 + xy = x is two straight lines, so select C
The difference between quadratic equation with one variable and quadratic equation with two variables and linear equation with one variable
One variable once: an integral equation with only one unknown number and one degree of unknown number is called one variable once equation. The usual form is KX + B = 0 (k, B is constant and K ≠ 0). One variable once equation belongs to integral equation, that is, both sides of the equation are integral. One variable means that the equation has only one unknown number and the degree of one time is 1
Simplification: 3x (2x + y) - 2x (X-Y) x is a letter X, not a multiplication sign
3x（2x+y)-2x(x-y)
=6x^2+3xy-2x^2+2xy
=4x^2+5xy
=x(4x+5y)
Remove brackets 6x ^ 2 + 3xy-2x ^ 2 + 2XY
Sum and union like terms 4x ^ 2 + 5xy
The common factor (4x + 5Y) x
3x（2x+y)-2x(x-y) =3x²+3xy-2x²+2xy=x²+5xy
The number of solutions of linear equation with one variable, linear equation with two variables and quadratic equation with one variable?
There is only one solution for the first-order equation of one variable and the second-order equation of two variables,
The solution of quadratic equation of one variable should be based on the discriminant. If the discriminant is greater than 0, there are two solutions, equal to 0, one solution, less than 0, and no solution
1. (4 / 5-1 / 2) x = 9 / 5 2. 1 / 2 + 3 / 10 x = 3 / 4 x is unknown x is not a multiplier
1. (4 / 5-1 / 2) x = 9 / 5
（4/5-1/2)x=9/5
(8/10-5/10)x=9/5
3/10x=9/5
Multiply both sides by 10
3x=18
X=6
2. 1 / 2 + 3 / 10 x = 3 / 4
1/2+3/10x=3/4
3/10x=3/4-2/4
3/10x=1/4
Multiply both sides by 20
6x=5
x=5/6
When did one variable quadratic equation, one variable linear equation and two variable linear equation learn?
I'm a student living in Zhuhai, Guangdong Province. In our primary school, I learned the one yuan one time equation in grade five,
Bivariate quadratic equation was learned in junior one, and bivariate quadratic equation was learned in junior two. It may be different in other provinces and cities
3x²=4x-1
Let's go to the following equation
3x²=4x-1
3x²-4x+1=0
(3x-1)(x-1)=0
x1=1/3,x2=1
Why should the right side of quadratic equation of one variable be changed to 0 by factorization?
That's a good question. If the right is not zero, it's any non-zero number, then you can decompose the factor on the left and get two formulas. The product of the two formulas should be equal to the non-zero number on the right. In this way, there are n kinds of values of the two formulas, because a non-zero number can be divided into two numbers. There are infinite kinds. How can you solve the unknown number
Because this is the necessary step of factorization?
3x²+4x+1=0
（3x+1）（x+1）=0
X = - 1 / 3 or x = - 1