The relevant content of the one dimensional quadratic equation in junior high school mathematics is

My solution: a binary equation of the shape ax 2+bx+c=0(a≠0), (1) can directly determine the size of the δ and zero according to the formula δ=b 2-4ac (2) you can first judge the size of a and zero, and then calculate the value of ax 2+bx+c when x= - b 2a, and compare the magnitude with zero, if and only if a>0, ax 2+bx+c 0 or a<0, ax 2+bx+c 0 satisfy δ>0 or δ 0

But then again, the specific situation has to be analyzed in detail, and if there is anything wrong, please point it out, after all, I haven't touched a math problem in junior high school and high school for a long time, thank you here

If the problem says that the equation has 2 unequal roots, then it is greater than 0; If there are two equal roots, it is equal to 0; If there is no root, it is less than 0

ax2+bx+c=0 is a binary one-dimensional equation, and then δ=b2-4ac, b is the coefficient of the one-time x, and a is the coefficient of the quadratic x, you see.

The equation has one root to indicate that the discriminant is equal to zero, and two roots to indicate that the discriminant is greater than zero. If there is a solution in the question, but there are no solutions, then the discriminant formula is greater than or equal to zero, understand?

There are three main ways to solve the metaquadratic equation: the matching method, the formula method, and the factorization method, of which the first two methods are applicable to all unary quadratic equations, and the factorization method has certain limitations (here the direct opening of the square is classified as a special case of the matching method, and some reference materials also list the direct square as a solution).

Selection method: Factorization or direct squared is preferred, and parentheses are not easily removed; For those that cannot be factored or directly squared, if the coefficient of the quadratic term is 1 and the coefficient of the primary term is even, the matching method is used; If the coefficient of the quadratic term is not 1, or the coefficient of the primary term is odd, the formula method is used.

(1) (2) (5) (6) (8) is suitable for factoring, (1) (2) (5) (7) (8) is suitable for matching methods, and (3) (4) is suitable for formula methods

(6) Don't be in a hurry to remove the brackets, first see if you can factor it, (8) You can also move the items first, merge the same kind of terms and then factor them.

Analysis: (1) Rolling time of the ball = total distance average speed;

2) the speed at which the movement of the ball decreases per second on average = the speed at which the ball rolls in time;

3) The equivalence relationship to be used is: speed time = distance, time x, and the speed of limb posture is the solution: (1) The average speed of the ball rolling = (10 + 0) 2 = 5 (m s), and the time of the ball rolling: 20 5 = 4 (s).

3) When the ball rolls to 5m, about xs is used, according to the topic, get: x (, sorted out: x -8x+4=0, solution: x=4 2 3, so x=4-2 3=

Answer: Book (1) The ball rolls for 4s

2) The average speed of the ball per second decreases.

3) When the ball rolls to 5m, it is about used.

What can be factored down is decomposed, and what can't be factored is decomposed by the root finding formula.

First of all, you have a miscalculation: 880 + 220x-20x + 5x 2 = 1600, which should be: 880 + 220x-20x-5x 2 = 1600

Then, 880 + 200 x - 5 x 2 = 1600176 + 40x - x 2 = 320

x^2-40x+144=0

x-36）（x-4）=0

solution, x = 36 or 4

The unary quadratic equation of junior high school mathematics is solved 880+200x+5x 2=1600.

Solution: 5x 2+200x=720

x^2+40x=144

x^2+40x+400=144+400

x+20)^2=544

x+20 = plus or minus (4 roots number 34).

x1=-20+4 root number 34, x2=-20-4 root number 34

The second step was written incorrectly, it should be.

880+220x-20x-5x²=1600-5x²+200x-720=0

x²-40x+144=0

x-4）（x-36）=0

x=4 or x=36

1500（1+x）² 1815

1+x) collapsed chain =

1+x= or 1+x=

x= or accompanies x=

x(50-3x)=200

50x-3x ass scale = 200

3x-50x+200=0

3x-20)(x-10)=0

x=20 3 or x=10

1500（1+x）²=1815

1+x) Kiriyu=

1+x= or 1+x=

x= or simply wheel.

x (50-3x) = 200

3x²-50x+200=0

3x-20）（x-10）=0

x=20 3 or x=10

The second problem is solved by direct formula method.

x=[50 (2500-2400)] 6=(50 10) 6x=10 or 40 6=20 3

1500 (1+x) 1815 a

x(50-3x)=200 b

obtained by a, (1+x).

1+x=or, x=or.

From simple volts b, 50x-3x = 200

1)(1+x)^2=1815/1500

1+x)^2=

1+x= or 1+x=

x = or 2) 50x-3x 2 = 200

3x^2-50x+200=0

3x 2-(30x+20x)+20*10=0 Cross Fight Wisdom Multiply Positive Silver Fierce.

x-10)(3x-20)=0

x1=10 x2=3 Lift 20

1815 divided by 1500=

The opening of the fierce square is equal to x=

In the second question, the direct factoring results in x=20 and 3 x=10

You're in your third year of junior high school, don't you even know this question? It seems that sometimes we really can't blindly implement the so-called quality education

Because b = ( a-2 + 2-a)-3

a-2>=0,2-a>=0

So a=2, find b=-3

And because one of the roots of the quadratic equation ax 2+bx+c=0(a≠0) is 1, a+b+c=0

So c=1, so this unary quadratic equation 2x, 2-3x+1=0

Does the root number in b=( a-2 + 2-a)-3 contain a-2 or 2-a?

In this case, you just need to prove that the coefficient of the quadratic term is not 0.