For the unary quadratic equation about x: ax? +bx+c=0。 (a≠0)
The method of cross multiplication is to separate the quadratic coefficient A from the constant term C. 。
That is, the quadratic term (x? ) is divided into m and s,
Decomposition of constant term c into n and t,
So that the left multiplication of the split cross is equal to the quadratic coefficient, that is, ms=a,
The multiplication on the right is equal to the constant term, that is, nt=c,
The cross product plus equals the first-order coefficient, that is, mt+ns = b.
(hence the name crossover method)
Common formats are:
After being taken apart.
Constant term of quadratic coefficient
m n
╳ t
How to divide the coefficient depends on the coefficient of the first term. As long as you can get the coefficient of the first term, there is no formula. )
Like x? +2x+ 1=0，
Quadratic equation is not ax? +bx+c=0？
Corresponding to this equation, a= 1, b=2, c= 1.
The so-called crossing is to dismantle a and C.
Here a= 1 is relatively easy to calculate, so it is naturally split into 1× 1.
C is also divided into 1× 1.
Can be written as:
1 1
1 ╳ 1
From/kloc-0 /×1+/kloc-0 /×1= 2.
That is, the cross multiplication and addition (/kloc-0 /×1+/kloc-0 /×1) is equal to the linear term coefficient of 2,
So the original equation can be transformed into (x+ 1)(x+ 1)=0.
Another example is:
Example 1 M? +4m- 12 factorization factor
Analysis: The constant term-12 in this question can be divided into-1× 12, -2×6, -3×4, -6×2,-12× 1.
When-12 is divided by -2×6, this problem is met.
Solution: Because
1 -2
1 ╳ 6
From 1×6+ 1×(-2)=4.
That is, the coefficient that satisfies the cross product and then adds is equal to the first order term.
So m? +4m- 12=(m-2)(m+6)
Example 2 Handle 5x? +6x-8 factorization factor
Analysis: 5 in this question can be divided into 1×5.
-8 can be divided into-1×8, -2×4, -4×2, -8× 1.
When the quadratic term coefficient 5 is divided by 1×5, and the constant term -8 is divided by -4×2, this problem is met.
Solution: Because
1 2
5 ╳ -4
From 1×(-4)+5×2=6.
That is, the coefficient that satisfies the cross product and then adds is equal to the first order term.
So 5x? +6x-8=(x+2)(5x-4)
Example 3 solving equation x? -8x+ 15=0
Analysis: put x? -8x+ 15 is regarded as a quadratic trinomial about x, then1× 15, 3×5, (-1)×(- 15), (-)
When 15 is divided by (-3)×(-5), this problem is met.
Solution: Because
1 -3
1 ╳ -5
From 1×(-5)+ 1×(-3)=-8.
That is, the coefficient that satisfies the cross product and then adds is equal to the first order term.
So the original equation can be transformed into (x-3)(x-5)=0.
So x 1=3 x2=5.
Example 4. Solve equation 6x? -5x-25=0
Analysis: put 6x? -5x-25 is regarded as a quadratic trinomial about x,
Then 6 can be divided into 1×6, 2×3,
-25 can be divided into-1×25,
-5×5,-25× 1。
When the quadratic coefficient 6 is divided by 2×3 and the constant term -25 is divided by -5×5, this problem is met.
Solution: Because
2 -5
3 ╳ 5
From 2×5+3×(-5)=-5.
That is, the coefficient that satisfies the cross product and then adds is equal to the first order term.
So the original equation can be changed to (2x-5)(3x+5)=0.
So x 1=5/2 x2=-5/3.
To put it bluntly, it is to make up four numbers and meet the following requirements: cross left multiplication equals quadratic term coefficient, right multiplication equals constant term, and cross multiplication plus equals linear term coefficient.
2, the use of cross multiplication:
(1) uses cross multiplication to decompose the factor. (2) Solving the quadratic equation with one variable by cross multiplication.
3. Advantages of cross multiplication:
Using cross multiplication to solve problems is faster, can save time, and the calculation is not large, so it is not easy to make mistakes.
4, cross method using conditions:
The coefficient is not complicated and easy to decompose.
(but not all of them can be used, such as 4x? +2x+ 1, however matched, is not equal to 2.
At this time, the root formula should be used for the corresponding factorization. )