( 1)x 1+x2； (2)x 1 * x2； (3) 1/x 1+ 1/x2； (4)x 1^2+x2^2.
Classroom assignments-basic standards
1. if the two roots of the equation aX 2+bx+c = 0 (a =/0) are x 1, x 2, then x1+x2 = _ _ _ _ _ _, x1* x2 = _. x 1 * x2 = _ _ _ _ _ _ _； 1/x 1+ 1/x2 = _ _ _ _ _ _ _ _； x 1^2+x2^2=________； (x 1+ 1)(x2+ 1)= _ _ _ _ _ _ _ _。 3. It is known that the two roots of quadratic equation 2x 2-3x- 1 = 0 are x 1, x 2, x2. Then x12+x2 2 = _ _ _ _ _. 5. It is known that x 1, x2 is two real roots of the equation x 2+MX+m = 0, and x 1+x2= 1/3.
A.3x^2-2x+3=0
B.3x^2+2x-3=0
C.3x^2-6x-9=0
D 3x 2+6x-9 = 0 7。 Let x 1 and x2 be two roots of the equation 2x 2-2x- 1 = 0, and use the relationship between roots and coefficients to find the following values:
( 1)(2x 1+ 1)(2x 2+ 1)；
⑵(x 1^2+2)(x2^2+2)；
(3) x 1-x2。
Homework-basic extension 1. (Problem solving skills) Given α 2+α- 1 = 0, β 2+β- 1 = 0, and α is not equal to β, the value of α β+α+β is ().
A.2
B.-2
C.- 1
D.02 (Error-prone) Given that two sides of a triangle are 2 and 9 respectively, and the third side is a root of the unary quadratic equation x 2-14x+48 = 0, then the circumference of this triangle is ().
A. 1 1
B 17
C. 17 or 19
D. 19 3。 If the two real roots of the unary quadratic equation x2+KX+4K 2-3 = 0 about x are x 1 and x2 satisfies x 1+x2=x 1*x2, then the value of k is ().
A.- 1 or 3/4
B.- 1
C.3/4
D. no 4. It is known that one root of equation 2x 2+MX-4 = 0 is -2, so find the value of the other root. (It can be solved in two ways)
Answer: 1. -P Q
2.5 3 The third formula is combined with (X 1+X2)/X 1*X2=5/3 The fourth formula = (x1+x2) 2-2x1* x2 =19.
Classroom assignments-basic standards
1.-B/A C/A
2.-3/2 -2 3/4 25/4
3.3/2
4.3
5.- 1/3
6.C
7. Let x 1 and x2 be two roots of the equation 2x 2-2x- 1 = 0, and use the relationship between the roots and the coefficient to find the following values:
( 1)(2x 1+ 1)(2x 2+ 1)； Extension =2
Because x1+x2 =1x1x2 =-1/2.
⑵(x 1^2+2)(x2^2+2)； Expansion =29/4
(3) x 1-x2。 = (x 1-x2) 2 squared = x12+x2 2-2x1x2 =
= (x 1+x2) 2-4x1x2 = 3 square root operation-basic extension1. (Problem solving skills) It is known that α 2+α- 1 = 0, β 2+β- 1 = 0.
A.2
B.-2
C.- 1
D.02 (Error-prone) It is known that the two sides of a triangle are 2 and 9 respectively, and the third side is a root of the unary quadratic equation x 2-14x+48 = 0, so the circumference of this triangle is (d). Note that the sum of two sides is greater than the difference between the third side and the third side, so it can only be 8.
A. 1 1
B 17
C. 17 or 19
D. 19 3。 If the two real roots of the unary quadratic equation x2+KX+4K 2-3 = 0 about X are x 1 and X2 respectively, and x 1+x2=x 1*x2 is satisfied, then the value of K is (c) Note.
A.- 1 or 3/4
B.- 1
C.3/4
D. no 4. It is known that one root of equation 2x 2+MX-4 = 0 is -2, so find the value of the other root. (It can be solved in two ways)
1. sum of two roots =-M/2 =-2+ X2 product of two roots =-2.
So X2= 1 M=2.
2. (b 2-4ac under -b+ or-radical sign) /2a =-2 Solve the following equation.
1.(2x- 1)^2- 1=0 1
2.—(x+3)^2=2
2
3.x^2+2x-8=0
4.3x^2=4x- 1
5.x(3x-2)-6x^2=0
6.(2x-3)2 = x ^ 2 1。 Match the completely flat way (write the answer directly)
1.x2-4x+_ _ _ _ _ = (x-_ _ _ _ _) 22. x2+MX+9 is completely flat, so m = _ _ _ _ _ _ 2.
X 2-8x-9 = 0 basic standard
1 If the equation x 2-6x-5 = 0 is solved by the matching method, the formula will get ().
A.(x-6)^2= 14
B.(x-3)^=8
C.(x-3)^= 14
d(x-6)2 = 4 1 ^ 2。 Put the quadratic trinomial 2x-3x+5 formula, and the correct formula is ().
3 3 1
A.(x- —)^2+ —
4 16
3 34
B.(x- —)^2- —
4 16
3 3 1
C.2(x- —)^2+ —
4 16
3 3 1
D.2(x- —)^2+ —
4 8 1/ uc， 1: 14:06
3. Fill in the blanks:
1.x^2+8x+______=(x+______)^2
2.2x 2- 12x+_ _ _ _ = 2(x-_ _ _ _)24。 Solve the following equation (important process) 1.x+5x+3 = 0 2.2x 2-x-3 =
A.-4
B.2
C.- 1 or 4
D.2 or 4
2. (Complex number) Solving the problem about x 2+2mx-n 2 = 0 (requires writing process) 3. (innovative question) Xiaoli and Xiaoqing are good friends, but Xiaoli is addicted to the internet recently. Xiaoqing decided not to make this friend, so she gave her a quadratic equation and said, "Solve this equation, this is our result!" " Xiaoli was surprised when she solved this equation. It turns out that putting these two together is "886" (network term "goodbye"). Classmate, can you design such a quadratic equation with one variable? 4. (Open inquiry) Let the algebraic expression 2x 2+4x-3 = m, and use the matching method to explain that no matter what value X takes, M is always not less than a certain value, and find this value (the whole process is necessary).
Answer: Solve the following equation.
1、(2X)^2- 1=0
If you move items, you will get: (2x) 2 = 1.
The square root gives 2X=+- 1.
Divide both sides of the equation by 2 to get: X=+- 1/2 2, 1/2 (x+3) 2 = 2.
Multiply the two sides of the equation to get: (x+3) 2 = 4.
Square root, X+3=+-2.
Subtract 3 from both sides of the equation to get: X=- 1 or -5 3, x 2+2x-8 = 0.
The factorization on the left gives: (X+2)(X-4)=0.
X+2=0 or X-4=0.
X=-2 or X=4 4, 3x 2 = 4x- 1.
Move the item and get: 3x 2-4x+ 1 = 0.
Factorizing on the left, we get: (3X- 1)(X- 1)=0.
3X- 1=0 or X- 1=0.
X= 1/3 or X= 1 5, X (3x-2)-6x 2 = 0.
3X^2-2X-6X^2=0
Finishing, get: -3x 2-2x = 0.
Divide both sides of the equation by-1 to get 3x 2+2x = 0.
Left factorization, X(3X+2)=0.
X=0 or 3X+2=0
X=0 or X=-2/3 6, (2x-3) 2 = x 2.
4X^2- 12X+9=X^2
Subtract X 2 from both sides of the equation to get 3x 2- 12x+9 = 0.
Divide both sides of the equation by 3 to get: x 2-4x+3 = 0.
Factorization on the left, we get: (X- 1)(X-3)=0.
X- 1=0 or X-3=0.
X= 1 or X=3. First, use the full plane mode.
1、x^2-4x+4=(x-2)^2
2.X 2+MX+9 is completely flat, so m=6. Second, the collocation method is used to solve the quadratic equation of one variable.
X^2-8X-9=0
X^2-8X=9
X^2-8X+ 16=9+ 16
(X-4)^2=25
(X-4)^2=5^2
X-4=+-5
X=9 or-1 foundation meets the standard.
1、C
2、D
Step 3 fill in the blanks
① x^2+8x+ 16=(x+4)^2
②2x^2- 12x+ 18=2(x-3)^2
4. Solve the following equation (important process) by matching method.
①X^+5X+3=0
X^+5X=-3
x^+5X+(5/2)^2=(5/2)^2-3
(X+5/2)^2= 13/4
X+5/2=+-√ 13/2
X=(√ 13-5)/2 or -(√ 13+5)/222x 2-x-3 = 0.
X^2- 1/2X=3/2
x^2- 1/2x+( 1/4)^2=3/2+( 1/4)^2
(X- 1/4)^2=25/ 16
X- 1/4=+-5/4
X=3/2 or X=- 1 basic extension
1、B 2、X^2+2mX-n^2=0
X^2+2mX=n^2
X^2+2mX+m^2=n^2+m^2
(X+m)^2=n^2+m^2
X+m=+-√(n^2+m^2)
X =-m+-√ (n 2+m 2) 3。 I'm not quite sure what this question means. Two roots together are 886. How to add them up or combine them? If it's 8 and 6, it's easy. (X-8)(X-6)=0, and the expansion is X 2-6544.
If the two roots are 88 and 6, (X-88)(X-6)=0, the expansion is x 2-94x+528 = 0 4, 2x2+4x-3 = m.
M=2X^2+4X-3
=2(X^2+2X)-3
=2(X^2+2X+ 1- 1)-3
=2(X^2+2X+ 1)-5
= 2 (x+ 1) 2 -5 No matter what the value of x is, 2 (x+ 1) 2 is always greater than 0, so m is always greater than -5.
The specific process of the matching method is as follows
1. Transform this unary quadratic equation into the form of AX 2+BX+C = 0 (this unary quadratic equation has real roots).
2. Convert the quadratic coefficient into 1.
3. Move the constant term to the right of the equal sign.
4. Add the square of half of the first coefficient to the left and right sides of the equal sign at the same time.
5. Write the algebraic expression on the left side of the equal sign as a complete square.
6. Square the left and right sides at the same time.
7. The root example of the original equation can be obtained by sorting: solving the equation 2x 2+4 = 6x.
1、2x^2-6x+4=0
2、x^2-3x+2=0
3、x^2-3x=-2
4、x^2-3x+2.25=0.25
5、(x- 1.5)^2=0.25
6、x- 1.5= 0.5
7、x 1=2
x2= 1