Two unary quadratic equations, two unary quadratic equation problems

Divide by 3 to get x 2+2x-4 3=0

Recipe x 2+2x+1-1-4 3=0 x+1) 2-7 3=0 Then move the root to get two.

Divide by 4 to get x 2-3 2x-4 3=0

Recipe x 2-3 2x+9 16-9 16-4 3=0 x-3 4) 2-9 16-4 3=0 Then move the root to get two.

The first lane: 3 (x 2 + 2x) - 4 = 0

3(x+1)^2-3-4=0

3(x+1)^2=7

x+1)^2=7/3

x = 21 -1 under the root number of 3 or 21 -1 under the root number of 3 points 2nd way: the same as the one above.

Let's go for yourself!!

The unary quadratic equation ax 2+b+c=0

Check if there is a solution:

b 2-4ac>=0 In this problem, 6 2-4*3*( 4) 0 has two unequal solutions.

x=(-b+root(4ac)) 2a -1 (root21) 3 or x=(-b-root(4ac)) 2a -1-(root21) 3

3x 2+6x-4=0 Recipe: 3(x 2-2x)-4=0

3(x^2-2x+1)-4-3=0

3(x-1) 2-7=0 solution: 1 plus or minus 3 7 under the root number

If two questions are put together, it would look like this: 3x 2+6x-4=03x 2+6x=4

3x+6=4/x---1)

4x^2-6x-3=0

4x^2-6x=3

4x-6=3/x---2)

1)+(2)=7x=7/x

So x=1 x

So x = plus or minus 1

This problem needs to be understood with Veddane, and you can find it in the math book, which is very simple.

(2x-3) squared - 2x+3=0

2x-3)2-(2x-3)=0 (2x-3)squared-2x+3=4x2-14x+12

2x-3)(2x-3-1)=0 4x2-14x+12=0x=3/2 or x=2 2x2-7x+6=0x=3/2 or x=2

2x-3) squared = 9 (2x+3) squared.

2x-3) squared = 3 square (2x+3) squared.

2x-3)+3(2x+3))(2x-3)-3(2x+3))=0(8x+6)(-4x-12)=0

x=-3/4 x=-3

2x-3) squared = 9 (2x+3) squared.

4x2-12x+9=36x2+108x+8132x2+120x+72=0

4x2+15x+9=0

x=-3/4 x=-3

ABC is an isosceles right triangle.

s△abc=1/2ab^2=1/2*12^2=72de‖bc，df‖ac

So ade and dbf are also isosceles right triangles.

After the exercise t seconds:

ad=vt=1*t=t

db=ab-ad=12-t

s△ade=1/2ad^2=1/2t^2

s△dbf=1/2db^2=1/2(12-t)^2s(dfce)=s△abcs-△ades-△dbf=72-1/2t^2-1/2(12-t)^2=20

t^2-12t+20=0

t-2)(t-10)=0

t = 2 seconds or 10 seconds.

16-3x-2x)^2+6^2=10^2(16-5x)^2=10^2-6^2=8^216-5x=±8

5x=16±8

x1=8/5=

x2=24/5=

Note: If it is "(16-3x-2x 2) 2+6 2=10 2".

16-3x-2x^2)^2=10^2-6^2=8^216-3x-2x^2=±8

a) by 16-3x-2x 2=+8

2x^2+3x-8=0

x=[-3 root number(3 2+4*2*8)] (2*2)=(-3 root number 73) 4

b) by 16-3x-2x 2=-8

2x^2+3x-24=0

x=[-3 root(3 2+4*2*24)] (2*2)=(-3 root: 201) 4

What is the first question? The second topic is.

1、x（2-3x）+（3x-2）=0

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

then 2-3x=0 or x-1=0

Suppose 2-3x=o, x=2 3

If x-1=0, x=1

So x=2 3 or 1

x-5）²=x²-25

The stupidest way is to take it apart to get the equation x -15x + 50 = 0 cross multiplication to get (x-10) (x-5) = 0 to get x = 10 or 5 simple way, use the square difference conversion to become.

3（x-5）²=x+5）（x-5）

Classification discussion, if x-5=0, the equation holds, x=5 if x-5 is not equal to 0, remove x-5 on both sides of the equation, and get.

3 (x-5) = x+5, can be sour x=10

x+y=1

x-y=3 simultaneous equation, x=2, y=-1

Substitute x=2 and y=-1.

ax+2by=4

bx+(a-1)y=3

i.e. { 2a-2b=4

2b-(a-1)=3

The solution is a=

ax+2by=4

x+y=1x-y=3

bx+(a-1)y=3

The solution is the same, x+y=1

x-y=3: x=2

y=-1 is substituted into 2a-2b=4

2b-a=2

obtained: a=6b=4

Solution: Derived from the question.

Solve the system of equations x+y=1 x-y=3 to get x=2,y=-1 and substitute x=2,y=-1 into the system of equations ax+2by=4;bx+(a-1)y=3.

System of equations 2a-2b = 4;2b-(a-1)=3 gives a=6, b=4

First pass x+y=1 ; x-y=3 solves x=2, y=-1

Then substitute the equation 2a-2b=4 with a,b 2b+(a-1)(-1)=3

The solution is a=6 b=4

A system of equations can be formed:

x+y=1x-y=3

So x=2y=-1

Bring back 2a-2b=4

2b-a+1=3

So a=6b=4

Combined, x=3+y is substituted for the first group, and the second equation gives 3+y+y=1, so y=-1 x=2

So 2a-2b=4 2b-a=2 so a=6 b=4