A quadratic problem of one variable, just write the equation A three digit number, its one digit number, ten digit number and hundred digit number are three consecutive increasing integers, and the sum of square of one digit number and hundred digit number is five times of ten digit number. Find the three digit number, let the ten digit number of this number be x, please list the equation according to the meaning of the question

A quadratic problem of one variable, just write the equation A three digit number, its one digit number, ten digit number and hundred digit number are three consecutive increasing integers, and the sum of square of one digit number and hundred digit number is five times of ten digit number. Find the three digit number, let the ten digit number of this number be x, please list the equation according to the meaning of the question

(x-1)(x-1)+(x+1)(x+1)=5x
You can't play square

Solution of quadratic equation with one variable
The water level of a reservoir has exceeded the warning line, and the upstream river still flows into the reservoir at a cubic meter per hour. In order to prevent flood, it is necessary to open the sluice. Assuming that each sluice discharges water at a cubic meter per second (a times a minus 3), it is estimated that if one sluice is opened, the water level can be lowered to the warning line in 15 hours. If two sluice are opened, the water level can be lowered to the warning line in 5 hours, and the value of a can be calculated!

According to the meaning of the question, we can get the equation
15（a＾2-3）-15a＝2×5（a＾2-3）-5a
15a＾2-45-15a＝10a＾2-30-5a
15a＾2-15a-10a＾2+5a-45+30＝0
5a＾2-10a-15＝0
a＾2-2a-3＝0
（a+1）（a-3）＝0
A1 = - 1 (rounding off), A2 = 3
A: a = 3

Two mathematical problems of quadratic equation with one variable
1.3a^=4 2.X^-4X+3=0

3a^2=4
a^2=3/4
a=√3/2 or -√3/2;
X^2-4X+3=0
(x-3)(x-1)=0
x=3 or x=1.

Finding the positive integer solution of the equation 4x + 7Y = 30

x=4,y=2

If the solution of equation 8x + 7y-6 = 0 is also the solution of equation 4x-3y + 10 = 0, find x, y

According to the meaning of the problem, we can get the equation system 8x + 7Y − 6 = 04x − 3Y + 10 = 0, and the solution is x = − 1y = 2

Mathematics problem. See the supplementary explanation of the problem for details. The calculation of linear equation with two variables. Note: there should be a specific process!
There should be a specific process:
The monthly water fee paid by a resident includes two items: the monthly water fee and the sewage treatment fee for the contracted volume of water. The unit price of the sewage treatment fee (yuan / m3) is 1 / 4 of the municipal water fee. Xiaohua family used 21 m3 of water in May and paid 42 yuan in total. How many yuan are the water fee and sewage treatment fee per m3

The water fee is x yuan per cubic meter and the sewage treatment fee is y yuan per cubic meter
21x+21*(1/4)y=42
(1/4)x=y
x=1;y=1/4=0.25
Answer: water fee is 1 yuan per cubic meter; sewage treatment fee is 0.25 yuan per cubic meter

The following equations are solved by proper methods: (1) (3x-1) 2 = 1; (2) 2 (x + 1) 2 = x2-1; (3) (2x-1) 2 + 2 (2x-1) = 3; (4) (y + 3) (1-3y) = 1 + 2Y2

(1) The original equation can be transformed into: 2 (x + 1) 2 - (x + 1) (x-1) = 0, (x + 1) (2x + 2-x + 1) = 0, that is, (x + 1) (x + 3) = 0.x + 1 = 0 or x + 3 = 0. (3) the original equation can be transformed into: (2x-1) 2 + 2 (2x-1) - 3 = 0, (2x-1-1) (2x-1 + 3) = 0, that is, (2x-2) (2x + 2) = 02x- 2 = 0 or 2x + 2 = 0.. X1 = 1x2 = - 1. (4) after finishing, 5y2 + 8y-2 = 0. ∵ a = 5, B = 8, C = - 2, b2-4ac = 82-4 × 5 × (- 2) = 104 > 0, ∵ y = − 8 ± 1042 × 5 = − 8 ± 22610 ∵ Y1 = − 4 + 265, y2 = − 4 − 265

Univariate quadratic solution equation / system of bivariate quadratic equations (just the formula of that equation, add a little more points)

1、 Multiple choice questions: (2 points for each sub question, 20 points in total)
1. The following equations are not necessarily univariate quadratic equations ()
A．（a-3）x2=8（a≠0） B．ax2+bx+c=0
C．（x+3）（x-2）=x+5 D．
2. Given the quadratic equation AX2 + C = 0 (a ≠ 0), if the equation has a solution, then there must be C equal to ()
A. - B. - 1 C.D. not sure
3. If the equation AX2 + 2 (a-b) x + (B-A) = 0 has two equal real roots, then a: B is equal to ()
A. - 1 or 2 B.1 or C. - or 1 D. - 2 or 1
4. If the quadratic equation ky2-4y-3 = 3Y + 4 with respect to y has real roots, then the value range of K is ()
A. K > - b.k ≥ - and K ≠ 0 C.K ≥ - D.K > and K ≠ 0
5. Given that the two roots of the equation are a,, respectively, then the root of the equation is ()
A． B． C． D．
6. If the sum of the two real roots of the equation x2 + 2 (K + 2) x + K2 = 0 of X is greater than - 4, then the value range of K is ()
A．k>-1 B．k

Solving equations, one variable quadratic, two
4x^2-4x-2＝0
4x^2+4x-2＝0

（1）4x^2-4x-2＝0
So 2x ^ 2-2x-1 = 0
X=(1±√3）/2
(2)4x^2+4x-2＝0
2x^2+2x-1＝0
X=(-1±√3）/2
This is my conclusion after meditation. If I can help you, I hope you will give me a chance to adopt it. If I can't, please ask, I will try my best to help you solve it

Some univariate quadratic equations (with processes)
2x^2+x=0
2x（x-3）= 5（x-3）
1/2x^2-3x=1
（x+2）^2-2x=3x^2
Another question: x ^ 2 + X-1 = 0

2x^2+x=0x(2x+1)=0x=0,x=-1/22x（x-3）= 5（x-3）2x^2-6x=5x-152x^2-11x+15=0(2x-5)(x-3)=0x=5/2,x=31/2x^2-3x=1x^2-6x=2x^2-6x+9=2+9(x-3)^2=(±√11)^2x-3=±√11x=3+√11,x=3-√11（x+2）^2-2x=3x^2 x^2+4x+4-2x=...