A math problem for the second year of junior high school about equations!

y^2=2x

y=kx+1

There is substitution. k 2x 2+(2k-2)x+1=0=(2k-2) 2-4k 2>0 to k<1 2x1x2=1 k 2 x1+x2=-(2k-2) k 2x1+x1x2+x2=1 k 2-(2k-2) k 2=11-(2k-2)=k 2

k^2+2k-3=0

k1 = 1 and k2 = -3

And k<1 2

So eligible k exists, k=-3

Solution: Simultaneous equations y 2=2x, y=kx+1 are simplified to k x +2k-2)+1=0

From the meaning of the problem, the equation has two unequal solutions to real numbers, so 0 solves k 1 2 (we will use this range to expel the value of a k later).

It is also known that there are two unequal real number roots, x1 and x2

then it is derived from Vedic theorem, x1+x2 = - 2k-2) k x1x2 =1 k Here is the core of this problem, the application of Veda's theorem is examined, once memorized, it is very useful, Veda's theorem is often used, must be remembered).

Substituting the formula into x1+x1x2+x2=1 and simplifying it yields k +2k - 3 = 0

Solution k = 1 or - 3

And because from the meaning of the title it is K 1 2

So k= - 3

Solution: Set x yuan per kilogram of oranges and y yuan per kilogram of apples.

6x 5y 3x+7y (the total amount of money caught remains the same).

y = total amount of money: 6x+5*

Because oranges are X yuan per kilogram, Juanjuan uses all the money to buy only oranges and can buy kilograms: kilograms.

1.Let the pedestrian speed be x km h

5 x = 5 (x+7) +40-5) 60x1= 5 -- walking speed , x2 = -12 (rounded) x+7 = 7 -- cycling speed.

2.Solution: It takes X days for A to do this project alone, and it takes Y days for B to do this project alone.

Analysis, this project can be regarded as unit "1", A's daily workload is 1 x, and B's daily workload is 1 y).

1/x+1/y)×12=1①

5 1 x+(1 x+1 y) 9=1 (system of equations, parentheses should be enlarged) to get 12y+12x=xy

Finishing gets 14y+9x=xy

2y-3x=0 from - to x=2y 3 can be found by substituting y=30 x=20

A: It takes 20 days for A to complete the project alone, and 30 days for B to complete the project alone.

3.At the beginning, it is a pure pesticide, assuming that the volume is x, then after the first time it is filled with water: the medicine is (x-8) the water is 8, and the second time it is filled, the medicine is and the water is, and the ratio of the two is 18:7, just solve the equation yourself, and the result is the barrel volume: 40l

"There are a few brothers and there are sisters," he said. ”

Because he himself is a boy, the number of boys = the number of girls + 1 She said, "My brothers are twice as many as my sisters." Because she herself is a girl, it means that the number of boys = (number of girls - 1) * 2

So the number of boys is 4 and the number of girls is 3

Solution: Let's set off a walker x hour and meet a car.

Analysis: When the car met the pedestrian, it took x-1 hour.

The distance by car is: 60 (x-1) km.

The distance traveled by the walker is 5 x km.

Equivalence relation: the sum of the distance between the pedestrian and the car is equal to 2 times the distance of the whole journey.

60(x-1)+5x=2*60

60x-60+5x=120

65x=180

x=36/13

A: The walker set off 36 hours after 13 hours and met with the car.

Solution: After setting off, the walkers will meet the car that picks them up after X hours.

60(x-1)+5x=2*60

60x-60+5x=120

65x=180

x=36/13

A: The walker set off 36 hours after 13 hours and met with the car.

Solution: Suppose the walkers set off and meet the car that picks them up after X hours.

5x+60（x-1）=60×2

65x-60=120

65x=180

x=13/36

x hours.

From the problem can be solved, it takes an hour for the car to reach the destination.

It means that the car and the pedestrian are moving in the same direction at the same time.

Set to take time x, get.

60+5）*（x-1）=60

Solve x as well.

As can be seen from the title, the time it takes for the car to send some people to their destination is 60 divided by 60 = 1 hour, and there are walkers who leave an hour earlier than the car, so when the car arrives at the destination, the walker has already left.

1+1)*5=10 km. The next question is the encounter between the car and the pedestrian. At this time, if they meet at t hours, then there is, (5+60) x t = 60-10=50 km to get t = 10/13.

Add to that the original two hours to the answer asked for 36/13

Let the car arrive at the destination and then meet the pedestrian through xh, 5x+60x=60-10, x=10 13, because the pedestrian starts one hour earlier and walks with the car for another hour, so it shares 2+10 13=36 13h

｛2x-3y＝8，① y＝1-3x.②

Solution: Put the substitution in, get.

2x-3(1-3x)=8

2x-3+9x=8

11x=11

x=1 solution: Substitute x=1 into to get.

y=1-3×1

y=1-3y=-2

x=1，y=-2.

Solve by substitution: 2x 3y=8 y=1-3x

Substituting the formula into the formula, we get x=1;y=－2

Solution: (20-x)(32-x)=540Polynomial multiplication: 20*32-32x-20x+x 2=540 Merge similar terms:

640-52x+x 2=540 shift: x squared - 52x+100=0 factor: (x-50)(x-2)=0

x=2 or x=50

It's hard for you to be a master! Such an order of a question.

20-x）(32-x)=540

640-20x-32x+x squared = 540

x squared - 52x + 100 = 0

Use the square of the discriminant b of the root to -4ac

The latter is up to you, it's very simple, and the result is a quadratic radical.

Solution: 640-20x-32x+x = 540

x²-52x+640-540=0

x²-52x+100=0

x-2）（x-50）=0

x=2 or x=50

If you don't understand, you can hi me.

Left: 640-32x-20x+x 2=540 merge homogeneous terms: x 2-52x+100=0 cross factor:

x-2)(x-50)=0 to solve: x1=2, x2=50

The square side length x

2x*(x-2)-xsquared = 96

This gives x=12

So the side length of the square is 12cm

The rectangle is 24 cm long and 10 cm wide