Practical problems and application problems of one dimensional quadratic equations will not be? Ask

Yes. But the question is not whether you will or not, but that you don't give a question.

(1) What is the length and width of a rectangle that is 1 cm longer than the width and the area is 132 cm squared?

2) There is a 1m long iron wire, so that he besieges a rectangle with an area of square?

3) How many people attended the meeting shook hands for every two people who attended the meeting, and everyone shook hands 10 times in total?

1) Solution: Let the width of the rectangle be xcm, then the length of the rectangle is (x+1) cm. From the meaning of the title, x(x+1)=132, i.e.:

x 2 + x - 132 = 0, the solution is x1 = -12 (not in line with the topic, discarded), x2 = 11. At this time, x+1=12 Answer: The length and width of the rectangle are 12cm and 11cm respectively.

2) Solution: Let the length of the enclosed rectangle be xm, then the width is (1 2-x)cm. From the meaning of the title x(1 2-x)=, i.e.:

x 2-1 2x+, solution x1=, x2=. When x1=, 1 2-x=, is not on topic, discarded. When x2=, 1 2-x=. Answer: The length and width of the enclosed rectangle are.

3) Solution: There are x people to attend the party. From the meaning of the question, n(n-1) 2=10, that is: n 2-n-20=0, the solution is n1=5, n2= - 4 (not in line with the topic, discarded). A: There were 5 people in attendance.

The turnover of a shopping center in our city in March was 5 million, and the turnover in April decreased by 10 percent compared with March, gradually rising from May to 6.48 million in June, seeking the average growth rate of turnover in May and June.

Let's set the growth rate to x. Solution: (500-500 10%) (1+x) square = 648

450 (1 + x) square = 648 450 (x square + 2x + 1) = 648

450x square + 900x + 450 = 648 450x square + 900x-198 = 0 Divide by 450 on both sides to get :

x squared + solved by the formula method of solving a quadratic equation: a=1 b=2 c=

x=-2±√2×2-4×1× /2

x=-2± /2 x1=-2+ /2=

x2= discarded if it doesn't fit the topic.

Calculation: May: 450 + 450 20% = 450 + 90 = 540

June: 540 + 540 20% = 540 + 108 = 648.

A: The average monthly growth rate is 20%.

Question 1: There are x male students among the students who set up the game, because everyone can't see their hats, according to one male student who said, "I see that the number of yellow hats is equal to the number of red hats." It can be seen that there is one more male student than female classmate, and there is x-1 female student among the students who play, according to the equation of a female student who said, "I see the number of yellow hats is the square of the number of red hats".

x-1-1）²=x

x²-4x+4=x

x²-5x+4=0

x=1 or 4x=1, then x-1=0 is not realistic, and it is rounded.

x=4, x-1=3, there are 4 male students and 3 female students in the game Question 2: Let the two-digit single digit number be x, the ten-digit number is x-2, and the column equation gets:

x-2)×x×3=10(x-2)+x

3x²-6x=11x-20

3x²-17x+20=0

3x-5)(x-4)=0

x=5 3 or 4

x=5 3, non-integers do not conform to reality, rounded off.

x=4, and this two-digit number is 24

1: Set male x and female y. We get x-1=y and (y-1) 2=x. Solved four men and three women.

2: Let this number be 10x+y, and x y is less than 10 to get x+2=y and 3xy=10x+y. This number is 24.

Give it points, forget the uncle for 6 minutes.

1.Solution: There are male students x and female students y.

From the title: { x-1=y

x=(y-1)^2

Substitute y=x-1 into x=(y-1)2

Get { x=4 or {x=1

y=3 y=0 (rounded).

A: There are 4 male students and 3 female students.

2.Solution: Let the single digit be x and the ten digit be y

From the meaning of the title: { x-2=y

3xy=10y+x

Substitute y=x-2 into 3xy=10y+x

Get { x=4 or { x=5 3

y=2 y=-1 3 (rounded).

Answer: This two-digit number 24.

Question 1 Four males and three females The square of the equation x-1=y, x=(y-1), bring the latter into the former, solve y, and then x

Question 2 Let the ten digits be x, the single digit is y, and this number is z, then 10x+y=z, y-x=2, 3xy=z, and the elimination solution yields 3x 2-5x-2=0, and the factorization equation (3x+1)(x-2)=0, and x>0, so x=2, y=4, and this two-digit number is 24

1) Set up x male and y female. Then the square of x-1=y and (y-1) is obtained by the condition=x, and then the system of unary linear equations is solved to obtain x=4 and y=3

2) Let the ten digits be x and the single digit be y, so we get x+2=y and 3xy=10x+y, and then we get x=2 and y=4, so this number is 24

Yellow-1=Red, (Red-1) 2=Yellow, Red=3, Yellow=4 Let the ten digits be x and the single digit be y

y-x=2,3*y*x=10x+y

x=2,y=4

The first question lists the equality relationship, and then solves it: the number of male students = the number of female students (male students can see) the square of the number of female students = the number of male students, and then solves it·· Also, your question doesn't seem right. In the second question, let the single digit be x and the ten digit be (x-2), and from the title, 10 (x-2) + x = 3x (x-2) is obtained

Set: On average, one computer will infect x computers. Solution: (1+x) square = 81 open squared: 1+x=9 x=9-1 x=8 units.

1 infection in the first round: 1 + 8 = 9 in total 2 infection: 9 + 9 8 = 9 + 72 = 81 units.

Third wave of infection: 81 + 81 8 = 81 + 648 = 729 units.

A: An average of 8 infections per unit per round; More than 700 units were infected in 3 rounds, and it was 729 units.

It can also be done as follows: solution (1+x)(1+x)=81 x squared + 2x+1=81x squared + 2x-80=0 Cross multiplication: x1=8 x2=-10 (discarded if not in place).

Assuming that each computer in each round of infection will infect X computers on average, according to the title:

1 x (1 x) x = 81, (1 x ) = 81, x 1 = 9 or x 1 = 9

x1 = 8 or x2 = 10 (rounded), 1 x) = (1 8) = 729>700

A: In each round of infection, an average of 8 computers will be infected per computer, and after 3 rounds of infection, more than 700 computers will be infected

The average infection rate of each round is x, which is derived from the title.

1+x) =81, that is, 1+x= 9, then x=8, (x=-10 rounded), then after 3 rounds of infection, the number of infected computers is.

1+x) =9 =729 (sets), more than 700 units.

x+1) = 81

On average, 1 infected 2 per round.

It will be more than 700

Solution: Let the side length of the square be x

then the rectangle is long x+5 and wide x-3

So (x+5)(x-3)=(4 5)x

x²+2x-15=(4/5)x²

x²+10x-75=0

x+15)(x-5)=0

x>0x=5, so the side length is 5cm

Solution: Length shortened by 5 cm = width increased by 3 cm.

So length = width + 8 cm.

If the length is x, then the width is x-8

It turns out that the side lengths of the rectangle are x x-8

Later the side length of the square was x-5

x(x-8)=4/5*(x-5)²

5x²-40x=4x²-40x+100

x²-100=0

x=10, so square side length = x-5=5 cm.

A: The side length of the square is 5 cm.

Dissolution: The original rectangle is x cm long and y cm wide.

..x－5＝y＋3

..x y (x-5) (y 3) 4/5 by x y 8....

Substituting in yields (y 8) y (y 8-5) (y 3) 4/5....

Solution y 2 x 10

So the side length of the square is 5

Let the original rectangle be long x and wide y

According to the question: x-5=y+3

5xy=4（x-5）(y+3)

Merge into one and two is that.

5（x+8）x=4(x+3)(x+3)

Solve it.

Each pair earns a -21 yuan.

So (a-21) (350-10a) = 400 (a-21) (a-35) = -40

a²-56a+775=0

a-31)(a-25)=0

a=31,a=25

The selling price must not exceed 120% of the purchase price

a≤21×120%=

So a=25

Therefore, the price of each pair of shoes should be set at 25 yuan.

Answer: If the price is x, you can sell 350-10x of this sneaker

350-10x）×（x-21）=400

10x^2+560x-7750=0

x^2-56x+775=0

The solution is x=25 or x=31

Because the selling price shall not exceed 120% of the purchase price, that is, 21 120%=, so x=25 yuan.

1. Let one side length be x, and the other is 64 4-x, which is 16-x.

The equation is x 2 + (16-x) 2 = 160. x=4 or x=12.

So one is 4 and one is 12.

2. Set pour out x

Pour out all of them (x 63) each time by the title, i.e

x 63) (x 63) * 63 = 28, the solution is x = 42 or x = -42.

So pour out 42 each time.

3. (1) The adjacent edge of the wall is x, and the opposite side of the wall is 35-2x; Gotta :

x(35-2x)=160 solution (x-10)(2x-15)=0;

So x=10 or 15 2.

So the side is ; Or.

2) Limit it to the length of the fence on the edge.

3) 20 meters (because x is less than 9).

4. Set the yield per mu to be x

200 (1 + x) * [50% (1 + x 2)] = 132 explained.

1 x+y=64 x 2+y 2=160*16 x=48 y=16 with side lengths of 12 and 4, respectively

2 63x 2=28 x=2 3 63*(1-2 3)=213 Set the length to x and the width to y 150=xy x+2y=35 y=10 x=15 or y= x=20

If the length is less than a and 9 meters away from the wall, the road can only be y= x=20 and the length is greater than 20 meters, and the yield growth rate of 4 mu is x

100(1+x)(1+x/2)=132 x=

1。Solution: Let one section of the wire be x cm long, then the other section is (64-x) cm long, according to the question, (x 4) 2 + [(64-x) 4] 2 = 160 to solve this equation, x = 16 or x = 48

Answer: Omitted 2Solution: If x liters of liquid medicine are poured each time, (63-x) liters of pure liquid medicine will be left after the first pouring, and the pure liquid liquid poured out the second time is.

63-x) 63] multiplied by x

According to the title, got.

63-x)-[63-x) 63]x=28

x = 21 or x = 105 is not suitable for the topic, so x = 21 answer: omitted 3

Consider the mass fraction and set the pour x liter 63-x-((63-x) 63)*x=28 x=

Solution: (1) If DF is linked, then DF BC

ab bc, ab = bc = 200 nautical miles

ac= ab=200 nautical miles, c=45°

cd= ac=100 nautical miles.

df=cf， df=cd

df = cf = cd = 100 = 100 (nautical miles) So, islands D and F are 100 nautical miles apart

2) If the supply ship sails x nautical miles at the time of the encounter, then de=x nautical miles, ab+be=2x nautical miles, ef=ab+bc-(ab+be)-cf = (300-2x) nautical miles.

In RT def, the equation is obtained according to the Pythagorean theorem.

x2=1002+（300-2x）2

Let be x, ** will be 2 times the speed of the supply ship, and ** will sail 2 times the distance of the supply ship.

That is: the sailing distance of the supply ship = DA+AB+BE

** Distance = 2AB+2BE+CA+(BC-BE): 2(DA+AB+BE)=2AB+2BE+CA+(BC-BE)2(100+200+X)) =2x200+2x+200+（200 2-x)

Solution: x=200 2