1, first judgment △ = B2-4ac, if△

2. If △=0, the original equation has two identical solutions: x =-b/(2a); ?

3. If △ > 0, the solution of the original equation is: x =((-b)√(△)/(2a).

I. Explanation:

A quadratic equation with one variable is a polynomial equation with only one unknown, and the highest degree of the unknown is quadratic. ? The quadratic equation of one variable can be transformed into the general form ax? +bx+c=0(a≠0), where ax? It is called quadratic term, and a is the coefficient of quadratic term; Bx is called a linear term, and b is the coefficient of the linear term; C is called a constant term.

Second, the establishment conditions:

The establishment of a quadratic equation with one variable must meet three conditions at the same time:?

(1) is the whole equation, that is, both sides of the equal sign are algebraic expressions, if there is a denominator in the equation; And the unknown is on the denominator, then this equation is a fractional equation, not a quadratic equation. If there is a root sign in the equation and the unknown is within the root sign, then the equation is not a quadratic equation (it is an irrational number equation). ?

(2) contains only one unknown number; ?

③ The maximum number of unknowns is 2.

Solution:

First, Kaiping method:

1) or the form of a quadratic equation can be solved by direct Kaiping method.

2) If the equation is in the form, then it can be obtained.

3) If the equation can be transformed into a form, then the root of the equation can be found.

Second, the matching method:

Make the quadratic equation of one variable into a table and then solve it by direct Kaiping method. ?

1, the steps of solving a quadratic equation with one variable by matching method:?

(1) transforms the original equation into a general form; ?

② Divide both sides of the equation by the quadratic term coefficient, so that the quadratic term coefficient is 1, and move the constant term to the right of the equation; ?

③ Add half the square of the coefficient of the first term on both sides of the equation; ?

④ The left side is matched into a completely flat mode, and the right side is matched into a constant; ?

⑤ Further use the direct Kaiping method to find the solution of the equation. If the right side is nonnegative, the equation has two real roots. If the right side is negative, then the equation has a pair of imaginary roots of yoke. ?

2. The theoretical basis of the matching method is the complete square formula.

3. The key of the matching method is: first, convert the quadratic coefficient of the quadratic equation of one variable into 1, and then add the square of half the coefficient of the quadratic term on both sides of the equation.