Equations for solving several factorizations urgent!! Ask a professional to help me answer, just sa

Observing these four problems, they all use the square difference formula to defactor the factor to solve the one-dimensional quadratic equation drop, mainly using a -b = (a+b)(a-b).

1，x²-2x=0

x(x-2)=0

x1=0 x2=2

2，4x²=（x-1)²

4x²-(x-1)²=0

2x）²-x-1)²=0

2x+x-1)(2x-x+1)=0

3x-1)(x+1)=0

x1=1/3 . x2=-1

3,49x²-25=0

7x)²-5²=0

7x+5)(7x-5)=0

x1=5/7 x2=-5/7

4. The question is not written completely.

5，（3x-2）²=(1-5x)²

3x-2)²-1-5x)²=0

3x-2+1-5x)(3x-2-1+5x)=0-2x-1）(8x-3)=0

x1=-1/2 x2=3/8

1, x squared - 2x = 0

x(x-2)=0

x=0 or x=2

2, 4x squared = (x-1) squared.

4x²-(x-1)²=0

2x+x-1)(2x-x+1)=0

x=1 3 or x=-1

3, 49x squared - 25 = 0

7x+5)(7x-5)=0

x=±5/7

4, 4x squared =

5, (3x-2) squared = (1-5x) squared.

3x-2)²-1-5x)²=0

3x-2+1-5x)(3x-2-1+5x)=0x=-1 2 or x=3 8

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

Solution: 3x+2=0 or 2x-1=0

x1=-2/3,x2=1/2

2x-1) 2=3 (1-2x).

Solution: 4x 2-4x+1=3-6x

2x^2+x-1=0

x+1)(2x-1)=0

x1=-1, x2=1 Infiltration and Lack of Praise 2

Solution: (x +6x+9)-17=0

x+3+root17)(x+3-root17)=0x1=-3-root17 x2=root17-3(2x-1) =0

x(x-3)=0

x1=0，x2=3

x²-2x-4=0

It is not possible to cross multiply, it is better to use the root finding formula).

1: The original test is decomposed into x(x-3)=0;So x=0 or x=3;(The main problem in this question is that 0 and any number is equal to 0).

2: The original test is decomposed into (x-1)(x-1)-1=4;（x-1）（x-1）=5；x-1=√5；x=√5+1；

x（x-2）=0

x1=0x2=2

x+10）（x-10）=0

x1=-10

x2=10 friends, correct answer, you only ask questions, incorrect answers, no energy!!

Friend, please [answer], yours is the motivation for me to answer the question, if you don't understand, please ask. Thank you.

(1) x(3x+2)=0 x=0 or x=-3 2

2) x 2+8x-6x=0 x 2+2x=0 x(x+2)=0 x=0 or -2

3) x 2+6x+9=2x+6 x 2+4x+3=0 (x+1)(x+3)=0 x=-1 or -3

4）2（x-3）^2=(x+3)(x-3) 2(x-3)=x+3 2x-6-x-3=0 x=9

5)-(root number 2-x) = 5x (root number 2-x) Approximate (root number 2-x) to find no solution, so x = root number 2

6）4（x^2-6x+9）=3(x^2-8x+16)

x 2 + 36 = 48 x 2 = 12 x = +-2 root number 3

Solution: (1).

3x²+2x=0

x(3x+2)=0

x=0 or x=-2 3

2）x(x+8)-6x=0

x(x+8-6)=0

x=0 or x+2=0

x=0 or x=2

3）(x+3)²-2(x+3)=0

x+3)(x+3-2)=0

x+3=0 or x+1=0

x=-3 or x=-1

4）2(x-3)²-x+3)(x-3)=0(x-3)(2x-6-x-3)=0

x-3 = 0 or x-9 = 0

x=3 or x=9

5）(x-√2)+5x(x-√2)=0

x-√2)(1+5x)=0

x-2=0 or 5x+1=0

x=2 or x=-1 5

6）4(x²-6x+9)-3x²+24x-48=04x²-24x+36-3x²+24x-48=0x²-12=0

x²-(2√3)²=0

x+2√3)(x-2√3)=0

x=-2 3 or x=2 3

(1) to (5) can be used to extract the common factor.

Methods and skills of factorization in junior high school 1Extract the common factor.

This is the most basic. It's just that if there is a common factor, it will be brought up, and everyone will know this, so I won't say much.

2.Perfectly squared.

a^2+2ab+b^2=(a+b)^2

a^2-2ab+b^2=(a-b)^2

If you see that there are two numbers squared in the formula, you should pay attention to it, find out if there is twice the product of the two numbers, and if so, follow the above formula.

3.Square Difference Formula.

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

This should be memorized, because it is possible to add terms when matching perfect squares, and if the front is perfectly squared, and then subtract a number, you can use the square difference formula to break it down.

4.Cross multiplication.

x^2+(a+b)x+ab=(x+a)(x+b)

This one is very practical, but it is not easy to use.

When the above method cannot be used to decompose, the lower cross multiplication method can be used.

Example: x 2 + 5 x + 6

First of all, it is observed that there are quadratic terms, primary terms, and constant terms, which can be multiplied by crosses.

The coefficient of the primary term is 1So it can be written as 1*1

The constant term is 6It can be written as 1*6, 2*3, -1*-6, -2*-3 (decimals are not recommended).

Then arrange it like this.

The positions of the following columns can be reversed, as long as the product of these two numbers is a constant term).

Then multiply diagonally, 1*2=2, 1*3=3Add the product again. 2+3=5, which is the same as the coefficient of the primary term (it may not be equal, so you should try another time at this time), so it can be written as (x+2) (x+3) (at this time, it will be done horizontally).

In fact, the most important thing is to use it yourself, the above methods can actually be used together, and practice is always better than teaching others.

By the way. If the b 2-4ac of an equation is less than 0, the formula cannot be decomposed in any way (in the range of real numbers, b is the coefficient of the first term, a is the coefficient of the quadratic term, and c is the constant term).

These methods are generally applicable when the highest order is secondary!

After reading it, you'll be able to know.,I won't ask me.。。。