Unary quadratic equations solve more complex practical problems No. 10

<> just count the number in the band.

1) Time is the distance divided by the average rate, i.e. 25 [(20+0) 2] =2) The change in speed divided by the time: (20-0) 3) Let the time taken to glide for 16m be t: The speed is:

20-8t, so the average speed of the product is [20+(20-8t)] 2=20-4t

Then according to the distance = average speed * time:

16=(20-4t)*t i.e.: t 2-5t+4=0, t-1)(t-4)=0 t=1 or 4

Because it is 4 >, so if it is rounded, it will take 1s to glide for 16m

This is a physics question!!

1. The braking process is a uniform deceleration movement, and the average speed is (20+0) 2 10m s, so the time is 25 10 seconds.

2. The initial speed is 20m s, the time, decreases by 20 per second 3, the acceleration is (0-20), the speed after gliding 16m is v, then there is (20*20-v 2)=2*8*16, the solution is v 12m s, and the time is (20-12—) 8 1 second.

1) x^2-√5*x+1/4=0

x1=(√5+2)/2, x2=(√5-2)/2

2) x^2-x=0

3) 4-4*6*a>=0

a<=1/6

x1+x2=-1/3，x1*x2=a/6

x1-x2|=2

x1-x2)^2=4

x1+x2)^2-4x1*x2=4

1/9-4a/6=4

a=-35/6

4) b^2-4c>=0

b^2>=4c

b,c)=(2,1)(3,1)(3,2)(4,1)(4,2)(4,3)(4,4)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)

5) (1) 6x 2-3x-2=0 are a, b

a+b=1/2，ab=-1/3

x1+x2=-1/a-1/b=-(a+b)/(ab)=3/2

x1*x2=(-1/a)(-1/b)=1/(ab)=-3

Equation: x 2-3 2*x-3=0 i.e. 2x 2-3x-6=0

2) 4x 2-3x-2=0 is a, b

a+b=3/4，ab=-1/2

x1+x2=a^2+b^2=(a+b)^2-2ab=9/16+1=25/16

x1*x2=a^2*b^2=(ab)^2=1/4

Equation: x 2-25 16*x+1 4=0 i.e. 16x 2-25x+4=0

6) x1+x2=(2m-1)/m，x1*x2=(m-2)/m

x1-3)(x2-3)=5m

x1*x2-3(x1+x2)+9=5m

m-2)/m-3(2m-1)/m+9=5m

m-2)-3(2m-1)+9m=5m^2

5m^2-4m-1=0

m-1)(5m+1)=0

m=1 or m=-1 5 (rounded).

7) 4(m+1)^2-4(4m+1)>=0 (1)

2(m+1)<0 (2)

4m+1<0 (3)

Solve m<-1

I'm tired to death, let's get fewer questions in the future, and it's your blessing to meet a kind person like me. Hope it helps. Good luck with your studies! ）

1. According to Vedic theorem [x 2 + (x1 + x2) x + x1x2 = 0], write a one-dimensional quadratic equation: x 2 + 5x+1 4 = 0 Use the formula method to solve the equation to obtain two numbers.

2.According to Vedder's theorem [(x1+x2=-b a) x1x2=c a] 0<-b a<2 -1< c a<1 Write a quadratic equation based on [x 2+b ax+c a=0]: e.g. x 2-x=0

b=2 c=a because x1-x2=2 x1+x2=-1 3 x1x2=a 6

So (x1-x2) 2=x1 2-2x1x2+x2 2=x1 2+2x1x2+x2 2-4x1x2=(x1+x2 ) 2-4x1x2=4

1 9+4*a 6=4 solution gives a=35 6

Piece. 5, x 2 + 3 2x + root number 2 2 = 0

6、a=m b=-(2m-1) c=m-2 x1+x2=(2m-1)/m x1x2=(m-2)/m

x1-3)*(x2-3)=x1x2-3(x1+x2)+9 =5m substitution and solution m

7. x1+x2=2(m+1) x1x2=4m+1 2(m+1) <0 to get m<-1 4m+1<0 to get m<-1 4 finally get m<-1

Ask the teacher directly, the Internet is unreliable.

Solution: Let the velocity of A be xkm h and B be ykm h

4x+150=4y

Solution x=y=

Suppose A's velocity is x, and B's velocity is y

1) If two cars start at the same time and go in the same direction, car B can catch up with car A in 4 hours. Equation 4y=4x+150, so y-x=35

2) Go in the opposite direction, and the two cars meet in an hour. So x+y=100, solve the system of equations x=65 2=, y=35+

So the speed of car A is, and the speed of car B is.

Solution: Let the average speed of cars A and B be x kilometers and y kilometers respectively, according to the question, 4 (y-x) = 150

Solving the system of equations yields, x=, y=A:

Suppose the speed of vehicles A and B is x y kilometers per hour.

4y=4x+150

Let the speed of A and B be x y

then 150 x + 4 = 150 y

The solution is 75 25

Untie; Let the speed of A and B be x kilometers and y kilometers.

Just put this one in again

The most annoying thing about the Olympiad Horse!!

Solution 1: If one right-angled side is a centimeter, then the other right-angled side is (14-a) centimeters; According to the meaning of the question, the equation can be listed:

a(14-a)×1/2=24

a(14-a)=48

14a-a²=48

a²-14a+48=0

a-6)(a-8)=0

a-6=0 or a-8=0

a=6 or a=8

When a=6, 14-a=8

When a=8, 14-a=6

A: The length of the two right-angled sides is 6 cm and 8 cm respectively.

Solution 2: Let each branch grow x small branches, 1 main stem grow x branches, there are main trunks and branches (x+1), each branch grows x small branches, x branches can grow x small branches in total, the total number is (x +x+1); According to the meaning of the question, the equation can be listed:

x²+x+1=91

x²+x-90=0

x-9)(x+10)=0

x-9=0 or x+10=0

x=9 or x=-10 (not on topic, should be discarded) Answer: Each branch grows 9 small branches.

1、x*（14-x）=24

The solution is x=6 or 8

2、x²+x+1=91

The solution is x=9 or -10 (negative rounding).

The average growth rate problem --- one of the types of application questions (let the growth rate be x):

1) A in the first year, A(1+x) in the second year, and A (1+x) in the third year

2) The following equality relation series equations are often used: Year 3 = Year 3 or Year 1 + Year 2 + Year 3 = Sum.

This type of question is done in this way

3（1+x）²=

Solution x= how much.

It's good that I don't know how to add me again: 1095705037

1.(a+b）x^2-2ax+(a-b)=01 -1

a+b -a+b

x-1)[(a+b)x-a+b]=0

x=1 x=(a-b)/(a+b)

2. Knowing that x1 and x2 are the two real roots of the unary quadratic equation x 2+(m+1)x+m+6=0 about x, and x1 2+x2 2=5, what is the value of m?

x1+x2=-m-1

x1x2=m+6

x1^2+x2^2

x1+x2)^2-2x1x2

m^2+2m+1-2m-12

m^2-11=5

m=±4x1x2=1/2

1/x1+1/x2=(x1+x2)/x1x2=51/x1*1/x2=2

x^2-5x+2=0

4. If x1 and x2 are the two roots of the equation x 2-x=5, then x1 2+x2 2=?

x1+x2=1

x1x2=-5

x1^2+x2^2

x1+x2)^2-2x1x2=11

1. Solve the equation about x: (a+b)x 2-2ax+(a-b)=0[(a+b)x-(a-b)](x-1)=0x=1 x=(a-b) (a+b) You can discuss it yourself, three cases 2, we know that x1 and x2 are the two real roots of the unary quadratic equation about x x 2+(m+1)x+m+6=0, and x1 2+x2 2=5, what is the value of m?

x1+x2=-m-1

x1*x2=m+6

x1²+x2²=(x1+x2)²-2x1x2=(m+1)²-2(m+6)=5

m=4 or -4

m=4 discriminant = -25<0 rounded.

m=-43, do not solve the equation, find a new equation, so that its two roots are the reciprocal of the equation 2x 2-5x+1=0.

x1+x2=5/2

x1x2=1/2

1/x1+1/x2=(x1+x2)/x1x2=51/x1*1/x2=2

The new equation is x -5x + 2 = 0

4. If x1 and x2 are the two roots of the equation x 2-x=5, then x1 2+x2 2=?

x1+x2=1

x1x2=-5

x1²+x2²=(x1+x2)²-2x1x2=1+10=11

x 2-3x+2=0 x=[3 (9-8)] 2 =(3 1) 2 So x1=2, x2=1x 2+x-6=0 x=[-1 (1+24)] 2 =(1 5) 2 So x1=2,x2=-3y 2-2y-3=0 y=[2 (4+12)] 2 =(2 4) 2 So y1=3,y2=-1

x 2+3x=4 x 2+3x-4=0 x=[-3 (9+16)] 2 =(3 5) 2 So x1=1,x2=-2 x 2+9x+20=.

1.Let A run x laps every minute and B run y laps per minute, then.

2(x+y)=1

6(x-y)=1

The solution is: x=1 3, y=1 6

2.Set up the A-shaped steel plate to imitate the defeat of the X block for the first tremor, and the B-shaped steel plate Y block, then.

2x+y=15

x+2y=18

Solution. x=4,y=7

3.The spring should be taken xcm, and the elongation of each 1kg object is ycm.

x+2y=x+5y=

Solution x=y=

The spring should be long. 4.If you take out x 1 dime coin and 5 jiao coin Qin Hu y, then there are (15-x-y) pieces of 1 yuan coin.

So. x+5y+10(15-x-y)=70 gives x=(80-5y) 9

Because x,y is an integer, and x,y<=15, so if and only if y=7, x=5 there are 5 coins for 1 dime, 7 coins for 5 jiao, and 3 coins for 1 yuan.