Junior high school mathematics has learned four solutions to quadratic equation of one variable: factorization, square root, collocation method and formula method. Please choose one of the following quadratic equation of one variable and choose the appropriate method to solve the equation x²-3x+1=0 2.(x-1)²=3 3.x²-3x=0 4.x²-2x=4 The fastest answer, the right answer,

x²-3x+1=0
x=(3±√(-3)²-4*1)/2
=(3±√5)/2
2.(x-1)²=3
x-1=±√3
x=1±√3
3.x²-3x=0
x(x-3)=0
X = 0 or x = 3
4.x²-2x=4
（x-1)²-1=4
(x-1)²=5
x-1=±√5
x=1±√5
Solve the equation x & sup2; - 2ax-b & sup2; + A & sup2; = 0
One variable quadratic equation
x²-2ax-b²+a²=0
x²-2ax+a²=b²
(x-a)²=b²
X-a = B or x-a = - B
X = a + B or x = a-b
x²-2ax+a²-b²=0
(x-a)²-b²=0
(x-a+b)(x-a-b)=0
X = A-B or x = a + B
x1=a+b
x2=a-b
If x2-x + 3 = 5, then - 4x + 4x2-10=______ .
From the meaning of the question: x2-x = 2-4x + 4x2-10 = 4 (x2-x) - 10 = - 2, so fill in - 2
Several methods of solving quadratic equation of one variable and factorization
The solutions of quadratic equation of one variable are: 1. Direct flattening method; 2. Collocation method; 3. Formula method; 4. Factorization method
There are several methods of factorization: raising common factor, using formula, grouping factorization, cross phase multiplication, splitting and adding term, undetermined coefficient, double cross phase multiplication, rotation symmetry, etc
To solve the equation about X, X & sup2; - 2aX = B & sup2; - A & sup2;
Transformation: X & sup2; - 2aX + A & sup2; = B & sup2;
(x-a)²=b²
X-a = positive and negative B
So x = B + A or x = a-b
(x-a+b)(x-a-b)=0
X-a + B = 0 or x-a-b = 0
X = A-B or x = a + B
X²—2aX+a²=b²
(x-a)^2-b²=0
(x-a+b)(x-a-b)=0
X = A-B or x = a + B
Calculation by formula: 4x & # 178; - root 5x-1 = 0
4x²-√5x-1=0
a=4,b=-√5,c=-1
x=[-b±√(b²-4ac)]/2a
=﹛-(-√5)±√[(-√5)²-4×4×(-1)]﹜/(2×4)
=[√5±√(5+16)]/8
=(√5±√21)/8
x1=(√5+√21)/8
x2=(√5-√21)/8
Factorization (x ^ 2-4x + 2) (x ^ 2-4x + 6) + 6
(x^2-4x+2)(x^2-4x+6)+6
=（x²-4x）²+8(x²-4x)+18
It's not easy to decompose
If you don't understand this question, you can ask,
Calculation: X & # 178; - (1 + 2 √ 3) x + 3 + √ 3 = 0, solved by formula method
Calculation: X & # 178; - (1 + 2 √ 3) x + 3 + √ 3 = 0, solved by formula method
x²-(1+2√3)x+3+√3=0
Discriminant: B ^ 2-4ac = 13 + 4 √ 3-4 * (3 + √ 3) = 1
x=(-b±√b^2-4ac)/2a=(1+2√3±1)/2*1
x1=√3 x2=1+√3
X={（1+2√3）±√[（1+2√3）^2-4(3+√3)]}/2=[（1+2√3）±1]/2，
So X1 = √ 3, X2 = 1 + √ 3.
The arithmetic square root of M + 2 + | n-4 | = 0, factoring (n + MXY) - (x ^ 2 + y ^ 2)
This problem is in the form of 0 + 0. Since the number of square root and absolute value is greater than or equal to 0, and the result is equal to 0, we can imagine that the square root and | n-4 | of M + 2 are both 0, so m + 2 = 0, n-4 = 0, so m = - 2, n = 4 (if you don't understand why you can ask again, I'll answer again)
Substituting (4-2xy) - (x ^ 2 + y ^ 2)
=4-2xy-x^2-y^2
=4-（2xy+x^2+y^2）
=4-（x+y）^2
=（2+x+y）（2-x-y）
How to factorize 9x ^ 5-35x ^ 3-4x
=x(9x^4-35x^2-4)
=x(9x^2+1)(x^2-4)
=x(9x^2+1)(x+2)(x-2)
9x^5-35x^3-4x
=x(9x^4-35x²-4)
=x(9x²+1)(x²-4)
=x(9x²+1)(x-2)(x+2)
9x^5-35x^3-4x=x(9x^4-35x^2-4)=x(x^2-4)(9x^2+1)=x(x-2)(x+2)(9x^2+1).
9x^5-35x^3-4x=x(9x^4-35x^2-4)=x(9x^2+1)(x^2-4)=x(9x^2+1)(x+2)(x-2)

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