Grade 9 one dimensional quadratic equations, velocity, to complete the process, to be in a standard

9 x squared - 5 = 3

9x^2=8

x^2=8/9

x= root number (8 9) = (2 root number 2) 3

x+6) squared - 9=0

x+6)^2=9

x+6=±3

x1=-6-3=-9

x2=-6+3=-3

The square of 3 (x-1) is -6 = 0

x-1)^2=2

x-1= root number 2

x1 = 1 - root number 2

x2 = 1 + root number 2

x squared - 4x + 4 = 5

x-2)^2=5

x-2= root number 5

x1 = 2 - root number 5

x2 = 2 + root number 5

9x squared + 6x + 1 = 4

3x+1)^2=4

3x+1)=±2

3x=-1±2

x1=(-1-2)/3=-1

x2=(-1+2)/3=1/3

1]9x"-5=3 9x"=8 x=+-(2 roots, numbers, 2) 3 2](x+6)."-9=0 x=+-3 x1=-3 x2=-9 3]x"-4x+4=5 x"-4x-1=0 x1=2+root3 x2=2-root3 4]9x"+6x+1=4 3x"+2x-1=0 (3x-1)(x+1)=0 x1=1 3 x2=-1 FYI! ）

Seems complicated? You care about being teasing, right? It's an 'irrational number'. It's so unreasonable, it's so unreasonable, I don't even bother to pay attention to you!

9x 2-5=3 <==> x 2=8 9 ==> x = plus or minus (2 under 2) 3

x+6) squared -9=0 <==> (x+6) 2=9 ==>x+6=3, i.e., x=-3 or x+6=-3 i.e., x=-9

3 (x-1) squared -6 = 0 <==> (x-1) 2 = 2 ==> x-1 = 2 under the root number, i.e. x=1 + 2 under the root number or x-1=

2 under the root number is x=1 - 2 under the root number

x squared -4x+4=5 <==> x 2-4x-1=0 x1,2=[4+(20)] 2=2+(5) according to the formula for the root of a quadratic equation

9x squared +6x+1=4 <==> 9x 2+6x-3=0 <==> 3x 2+2x-1=0 <==> (3x-1)(x+1)=0 gives x=1 3 or x=-1

9x^2-5=3

9x^2=8

x^2=8/9

x=2√2/3

x+6)^2=9

x+6=3 x=3

or x+6=-3 x=-9

3(x-1)^2-6=0

x-1)^2=2

So x-1= 2 x=1+ 2

or x-1=- 2 x=1- 2

x^2-4x+4=5

x-2)^2=5

So x-2= 5 x=2+ 5

or x-2=- 5 x=2- 5

9x^2+6x+1=4

3x+1)^2=4

So 3x+1=2 x=1 3

or 3x+1=-2 x=-1

9x^2=8

x^2=8/9

x=2 3*sqrt(2) or x=-2 3*sqrt(2)2(x+6)^2-9=0

x+6)^2=9

x+6=3 or x+6=-3

x=3 or x=-9

3(x-1)^2=6

x-1)^2=2

x-1=sqrt(2) or x-1=-sqrt(2)x=1+sqrt(2) or x=1-sqrt(2)x 2-4x-1=0

The discriminant formula is 4 2+4=20>0

x=(4+sqrt(20)) 2 or x=(4-sqrt(20)) 2

x=2+2sqrt(5) or x=2-2sqrt(5)9x 2+6x+3=0

3x^2+2x+1=0

The discriminant formula is 2 2-3*4=-8<0

So the original equation has no real roots.

1 : The formula for the area of a circle is r, so r =21 can solve the equation, and you can take an approximate value.

2: The width and length after expansion is (200-x), so (100+x)*(200-x)=20000 solves x=100 so the area can be 20000m. The second (100+x)*(200-x)=23000 Because b -4ac is less than 0, the unsolved area of the secondary equation cannot reach 23000m

3: Is it a mistake, it should be cut off four small quadrilaterals on each of the four corners of the square to get a rectangle.

4 : Let the length be xcm and the width be (11 x)cm so x*(11-x)=30 can solve the value of x, and the length is 6, the width is 5, and the second equation is unsolved because b -4ac is less than 0. Fold into a rectangle of 32.

I learned this last year, and I'm pretty good at math, I hope it will help you.

1。。。2πr=21 r=12/2π

2.。。2. Give you a formula, it's a bit messy, the equation is not very good at solving (100+x)*(300-100-x)=20000

3...What does this mean? How many cm is the length and width of the rectangular iron plate by 20 cm?

4...x*(11-x)=30 x=6 or 5 I set the length So it's 6 long and 5 wide 5 and the 6 is rounded off.

2) You can't set the length x x*(11-x)=32 to find b 2-4ac is greater than 0 The result is less than 0 so the equation is unsolved, so it can't.

The area of the circle: 7 6 1 2=42 1 2=21m Solution: Let the radius of the circle be xm.

then x =21

x²=21/π

x = 21 under the root number

A: ...The second one is a bit messy、、、 please be a master.、、

Solution: If the width is x, the length is (x+20).

x（x+20）=800

x²+20x=800

x²+20x+100=800+100

x+10）²=900

x+10=±30

x+10=30 or x+10=-30

x1 = 20, x2 = -20 (rounded).

x+20=20+20=40

A: The length is 40cm and the width is 20cm.

One length + one width = 22 2 = 11cm

Solution: If the length is x, then the width is (11-x).

x(11-x)=30

11x-x²=30

x²-11x=-30

x²-11x+121/4=-30+121/4(x-11/2)²=1/4

x-11/2=±1/2

x-11 2 = 1 2 or x-11 2 = -1 2x1 = 6, x2 = 5

Okay, that's it、、、

Take a look、、If it's right, add points、、I'm in a hurry、、、

Extract the coefficient before the 2nd term, and then make the formula, and it's OK.

1 Suppose one bacterium multiplies x bacteria per round, x = 256 x = 16 so an average of 16 bacteria multiply in each round of reproduction.

2 Let the average growth rate be x 1000 (1+x) = 1210 x = the average monthly growth rate of the plant in April and May is 10.

3 Set the width to x meters 30x+20x-x =20 30-551 x=1 or x=49 (rounded) So x=1 The width of the road is 1 meter.

Solution: (1)(x-3)(x+4)=0

x-3=0,x+4=0

x1=3，x2=-4

2) Move items to: x -2x = 3

Formula: x -2x + 1 = 3 + 1

x-1)²=4

Straight squared: x-1 = 2

x1=3,x2=-1

1.Solution: Set.

The average monthly growth rate in April and May is x, according to the question, 1000 (1 + x) 2 = 1210, the solution is x1 = , x2 = negative value is rounded off), so the plant.

The average monthly growth rate in April and May was 10%.

2.Solution: Set the width of the road to be xm

30－x）（20－x）＝551

x²－50x＋49＝0

x－1）（x－49）＝0

x1 1, x2 49 (untitled, discarded).

The width of the road should be 1m

I can't see it clearly. You can only faintly see the graph, as if calculating the area relationship between the cross and the rectangle.

Set the road width to x.

The length and width of a rectangular ground are a, b. Area ab

The area of the road is: ax+bx-x 2

Do the math yourself.