A summary of the quadratic equation of junior high school mathematics

There are x teams.

Then each team will play against the other (x-1) team.

Since it is a single cycle.

So x*(x-1) 2=10{ divided by 2 because there is a repetition, e.g. A and B play a game, and B plays a game with A, because it is a single round-robin.

So just a and b or b and a}

So x*(x-1)=20

So x 2 - x - 20 = 0

x+4)(x-5)=0

x=-4 (rounded), x=5

So there are 5 teams.

If you don't understand, you can ask!! Thank you!

There are 5 teams in this group, and I do it with arithmetic, not equations.

Let the percentage increase be x

If it was 200 in the first quarter, it would be 200 (1+x) in the second quarter and 200 (1+x) 2 in the third quarter

Equation: 200 (1 + x) + 200 (1 + x) 2 = 5281 + x + 1 + 2x + x 2=

x 2 + solution gives x =.

That is, the percentage is: 20%.

Let the percentage increase be a

According to the title: 200x(1+a)+200x(1+a)x(1+a)=528

Solution: a = i.e. 20%.

Let the percentage increase be x

200*（1+x)+200*(1+x)*(1+x)=5281+x+x*x+2x+1=

x*x+3x=

Because x 0 so x=

That is, the percentage increase is 20%.

Let the raised one be x

200*（1+x）+200*（1+x）²=528200+200x+200*（1+2x+x²）=528400+600x+200x²=528

25x²+75x-16=0

5x-1）（5x+16）=0

x=1 5 or.

Because x 0 so x=1 5=20%.

1 (x+1) square = 4

x+1 = positive hail and negative 2

x = 1 (i.e. 100 percent) (negative values are not in line with the topic, the source of the meditation house goes) has a formula: a(1+x) squared = b or a(1-x) squared.

A represents the original gross value, and B represents the total value of the planned grandchildren.

b by the question, there is a+b+c=0

Substituting 1 into 0 gives you 0, so one must be 1

Substituting the root x=1 into the original equation yields a+b+c=0, which is consistent with the sum of the quadratic coefficients, the primary coefficients, and the constant terms equal to zero. In this kind of problem, you can substitute the options into the equation to find the law.

b a=0 is a one-time function with only one root: b+c=0 and bx=-c x=1