It's a simple quadratic equation of one variable. Today A piece of white paper can be made into two sides or three bottoms, and a carton can be made from one side or two bottoms. Then how many pieces of white paper can be used as sides and how many pieces of white paper can be used as bottoms, and the sides and bottoms can match each other?

The ratio of white paper for side and bottom is 3:4
Quadratic equation of two variables!!!
X paper for side and Y paper for bottom
2X sides for 2x boxes
3Y bottoms for 3Y / 2 boxes
If necessary, it is necessary to set up a complete set
2X = 3Y / 2 and 3Y / 2, 2 is an integer
Solving quadratic equation of one variable 2x (x-3) - (x-3) = 0
(2x-1)(x-3)=0
x1=1/2,x2=3
Univariate quadratic equation (1) 1-2 / 3 (X-2) ^ 2 = 0 (2) (3x - √ 2) ^ 2-3 = 0 (3) (1-2x) ^ 2-x ^ 2 = 0 (4) (x-1) ^ 2-2 (x ^ 2-1) = 0
（1）1-2\3(x-2)^2=0
（2）(3x-√2)^2-3=0
(3)(1-2x)^2-x^2=0
（4）(x-1)^2-2(x^2-1)=0
1-2 / 3 (X-2) &# = 0 shift
2 / 3 (X-2) ² = 1 multiply both sides of the equation by 3 / 2 at the same time
(X-2) ² = 3 / 2 square root
x-2=±√6/2
X-2 = √ 6 / 2 or X-2 = - √ 6 / 2
x1=2 + √6/2 ,x2=2 - √6/2
(3x - √ 2) &# 178; - 3 = 0
(3x - √ 2) &# 178; = 3 square root
3x-√2=±√3
3x - √ 2 = √ 3 or 3x - √ 2 = - √ 3
x1=(√2+√3)/3
x2=(√2-√3)/3
(1-2x) &# 178; - X & # 178; = 0 shift
(1-2x) &# 178; = x & # 178; root
1-2x=±x
1-2x = x or 1-2x = - x
x1=1/3 ,x2=1
(x-1) &# 178; - 2 (X & # 178; - 1) = 0
X & # 178; - 2x + 1-2x & # 178; + 2 = 0 merge
-X & # 178; - 2x + 3 = 0 multiply both sides of the equation by - 1
X & # 178; + 2x-3 = 0 cross phase multiplication decomposition
x -1
X
x +3
(x-1)(x+3)=0
X-1 = 0 or x + 3 = 0
x1=1 ,x2=-3
Solving one variable quadratic equation 25 = 2x ^ 2 + x-3
Just write the answer
25=2x^2+x-3
2x^2+x-28=0
(2x-7)(x+4)=0
x1=7/2,x2=-4
2x²+x-28=0
(x+4)(2x-7)=0
x=-4
x=7/2
When m is a value, the solution of the inequality system 3x + 2Y = m satisfies x > y > 0? 2x-3y = M-1
It is 3x + 2Y = m, 2x-3y = M-1
To solve this system of equations:
3x+2y=m,
2x-3y=m-1
The solutions of the equations are as follows
x=(5m-2)/13
y=(3-m)/13
Then: (5m-2) / 13 > (3-m) / 13 > 0
By solving this inequality, we obtain:
5/6
Solution: from 3x + 2Y = m, and 2x-3y = M-1
The solution is as follows
x=(5m-2)/13
y=(3-m)/13
Then: (5m-2) / 13 > (3-m) / 13 > 0
By solving this system of inequalities, we obtain the following results:
5/6
If the set a = {X-2 ≤ x ≤ a}, B = {y y = 2x + 3, x = ∈ a}, C = {y | y = x ^ x ∈ A and B contains C, what is the value range of real number a?
B={y｜-1≤y≤2a+3}
∵ B contains C
∴x^≤2a+3
∴2a+3≥4 ① a^≤2a+3②
∴①a≥1/2
②a^-2a-3≤0
(a-3)(a+1)≤0
-1≤a≤3
To sum up
1/2≤a≤3
Solve the equations: 3x − y + 2Z = 32x + y − 4Z = 117x + y − 5Z = 1
3x−y+2z＝3 ①2x+y−4z＝11 ②7x+y−5z＝1 ③ The results show that: ① + ②, 5x-2z = 14, ④, ③ - ②, 5x-z = - 10, ⑤, ④ - ⑤, - z = 24, z = - 24; substituting ④ into solution, x = - 345; substituting z = - 24, x = - 345 into ①, y = - 3575; so the solution of the original equations is x = - 345y = - 3575z = - 24
(2008 · Yangzhou) if there are two and only one of the two quadratic equations AX2 + 2x-5 = 0 about X between 0 and 1 (excluding 0 and 1), then the value range of a is ()
A. a＜3B. a＞3C. a＜-3D. a＞-3
According to the meaning of the question: when x = 0, the function y = AX2 + 2x-5 = - 5; when x = 1, the function y = a + 2-5 = A-3. There is only one of the two quadratic equations AX2 + 2x-5 = 0 of X between 0 and 1 (excluding 0 and 1), so when x = 1, the function image must be above the X axis, so y = A-3 > 0, that is, a > 3
Nonnegative integer solutions of (1-3y) ^ 2 + (2y-1) ^ 2 > 13 (Y-1) (y + 1)
Expand the complete square and the square difference of the equation, and the left and right sides of the quadratic term are eliminated
Simplify to y
1 and 0, expand the solution of the equation, the first and second terms all cancel out
Given the quadratic function f (x) = 2x ^ 2 + MX + 2m, if f (m ^ 2-m + 1) > F (3m ^ 2 + 2m + 2), then the value range of real number m is
F (x) = 2x ^ 2 + MX + 2m is an increasing function when x > - M / 4, m ^ - M + 1 - (- M / 4) = (M-3 / 8) ^ + 55 / 64 > 0,3m ^ + 2m + 2 - (- M / 4) = 3 (M + 3 / 8) ^ + 101 / 64 > 0, х m ^ - M + 1 > - M / 4,3m ^ + 2m + 2 > - M / 4, х from F (m ^ 2-m + 1) > F (3m ^ 2 + 2m + 2), we get m ^ 2-m + 1 > 3M ^ 2 + 2m + 2, х 2m ^ + 3M + 1

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