Find six problems to solve a quadratic equation using the matching method

1. x²-2x-3=0

2. 2x²+12x+10=0

3. x²-4x+3=0

4. x²/4 +x-3=0

5. 9x²-6x-8=0

6. x²+12x-15=0

7. 2x²+1=3x

8. 3x²-6x+4=0

9. 3x²+6x-4=0

10. 4x²-6x-3=0

11. x²+4x-9=2x-11

12. x(x+4)=8x+12

13.（x+1）²=4x

14. x²+x/3-2=0

15. x²/3+x/4-1/2=0

16.x -x 5-2 5=x Actually, you can find it all on the Internet, and you can do it yourself.

1. x +x+1=0 x has no solution.

2. x -18x+9=0 x=6 2+9 or x=-6 2+9x 5x 3=0 x=1 2 or x= 3

4, x +6x+5=-3 x=-2 or x=-45, x-2x+8=9 x= 2+1 or x=- 2+16, x +5x-20=4 x=3 or x=-87, x +13x+36=0 x=-4 or x=-9

1 Solve the equation: x 6x 4=0, x=3 132 Solve the equation: x +4x 1=0, x= 2 53 Solve the equation:

x 6x+5=0,x1=5,x2=14 Solve the equation: x 2x=4, x=1 5

5 Solve the equation: 2x 3x 3=0, x=(3 33) 46 Solve the equation: x +2x 5=0, x= 1 67 Solve the equation 2x 4x 3=0, x=1 10 28 Solve the equation:

x 2x 2=0,x=1 39 Solve the equation: x 2x 4=0,x=1 510 Solve the equation: 2x 4x+1=0,x=1 2 2

1. x +x+1=0 x has no solution.

2. x -18x+9=0 x=6 2+9 or x=-6 2+9

x 5 x 3 = 0 x = 1 2 or x = 3

4. x +6x+5=-3 x=-2 or x=-4

5. x -2x+8=9 x= 2+1 or x=- 2+1

6. x +5x-20 = 4 x = 3 or x = -8

7. x +13x+36=0 x=-4 or x=-9

8. x -3x+4=2 x=1 or x=2

x -40x+25=1 x=(13+5)2 or x=(5-13)2

10、x²-|x|-2=0 Remove the absolute value (need to be discussed) and then solve the quadratic equations separately.

Answer: When x is greater than 0, x=2 or x=-1

When x is equal to 0, x has no solution.

When x is less than 0, x=-2 or x=1

An integral equation that contains only one unknown (unary) and the highest order of the unknown term is 2 (quadratic) is called a quadratic equation. The unary quadratic equation can be formed into a general form ax2+bx+c=0(a≠0).

where ax2 is called the quadratic term, and a is the coefficient of the quadratic term; bx is called the primary term, and b is the coefficient of the primary term; c is called a constant term.

Resources.

1.Solution: Let the width of the horizontal color bar be xcm, then the width of the vertical color bar is 32x, and the area of a horizontal color bar can be seen from the figure:

x 20, the area of a vertical color bar is: 32x 30, there are four overlapping parts, the overlapping area is: x 32x 4, because the area of all color bars is one-third of the total area, so the column equation is:

2 x 20+2 32x 30-x 32x 4= 13 20 30, solution: x1 = 53, x2 = 20 (twice greater than 30, rounded), should be designed horizontal color bar width of 53cm, vertical color bar width of.

2.Solution: (1) If the length AB of the enclosed rectangular ABCD is X meters, then the width of the AD is 12 (80-X) meters

According to the title, x 12 (80-x) = 750

That is, x2-80x+1500=0, solve this equation to get x1=30, x2=50

The length of the wall should not exceed 45m, x2=50 is not in line with the topic and should be discarded

When x=30, 12(80-x)=12(80-30)=25, so when the length of the enclosed rectangle is 30m and the width is 25m, the area of the rectangle can be 750m2

2) No, because x12 (80-x) = 810 gives x2-80x + 1620 = 0

and b2-4ac=(-80)2-4 1 1620=-80 0, and the above equation has no real roots

Therefore, it is not possible to make the area of the enclosed rectangular site 810 m2

3.(1) From the title, 400 (2x + 400x) + 300 400x + 200 80 = 47200

That is, 800x+ 400 700x+200 80=47200

Simplification yields x2-39x+350=0

The solution is x=14, x=25

It is tested that it is the solution of the original equation, but x=25 16 (discarded if it does not fit the topic).

So when the total cost of the pool is 47,200 yuan, the pool is 14 meters long

2.(1) The wall can use x m, x*(80-x) 2=750, that is, x 2-80x=1500, x=50 x=30Find the area of 750 when the wall is 30m

2) x*(80-x) 2=810, no solution.

Question 2: (1) How to make the area 750 m 2 The length of the strong side is 15 meters.

2) Can the area of the rectangular field be 810m2 No.