The solution of the inequality ax & # 178; + BX + C ＜ 0 about X is x ＜ - 2 or X ＞ - # 189; find the solution of the inequality ax & # 178; - BX + C ＞ 0 about X

The solution of the inequality ax & # 178; + BX + C ＜ 0 about X is x ＜ - 2 or X ＞ - # 189; find the solution of the inequality ax & # 178; - BX + C ＞ 0 about X

The solution is X-1 / 2
The corresponding inequality (x + 2) (x + 1 / 2) > 0
So ax ^ 2 + BX + C can be decomposed into a (x + 2) (x + 1 / 2), and A0
A
x> - 2 or X
Given that the solution set of inequality ax & # 178; + BX + 2 ＞ 0 is {x │ & # 189; < x ＜ 1 / 3}, find the value of a and B
Simple solution: because the solution set is {x │ & # 189; < x ＜ 1 / 3}, then 1 / 2 and 1 / 3 are the two roots of the original equation,
1 / 2 + 1 / 3 = - B / A. 1 / 2 times 1 / 3 = 2 / A, then a = 12, B = - 10
Excuse me, is this topic wrong?
Dizzy, this topic is wrong, solution set is wrong! I can't find such a value of X!
The solution set of inequality ax & # 178; + BX-1 > 0 is {x| 3
The solution set of inequality ax & # 178; + BX-1 > 0 is {x| 3
-1/a=3*4=12，a=-1/12
-b/a=7
b=7/12
So a + B = - 1 / 12 + 7 / 12 = 1 / 2
Have a good time!
-b/a=7
-1/a=12
a= -1/12
b=7/12
a+b=1/2
-The solution set of ax & # 178; - BX + 10 is {x | 3
A2 + 4A + k is a complete square, K should be ()
A. 2B. 4C. ±4D. -4
∵ 4A = 2 × 2 · a, ∵ k = 22 = 4
"Term shifting", "merging" and "coefficient to 1" all change a more complex one variable linear equation, such as 2x-19 = 7x + 31, into the simplest one variable linear equation, such as x = - 10. Can you change the equation AX + B = CX + D (x is unknown, a, B, C, D is known, and a ≠ C) into the simplest one variable linear equation?
The results show that: ax CX = D-B, x = D-B, ∵ a ≠ C, x = D − BA − C
Set a = {y | y = x square}, B = {y | y = 1 / 2 x power, X ＞ 1}, what is a ∩ B
y = x² ≥ 0
∴A = { y | y ≥ 0 }
X>1
y = 1/2^x < 1/2
1 / 2 ^ x > 0
∴ 0 < y < 1/2
B= { y | 0 < y < 1/2 }
A∩B = { y | 0 < y < 1/2 }
A polynomial a minus the quadratic power of 4x - 3Y + 5's errand 2Y + 6, then a
A - （4 x ² - 3 y + 5）= 2 y + 6A = 2 y + 6 + （4 x ² - 3 y + 5）A = 2 y + 6 + 4 x ² - 3 y + 5A = 4 x ² +（2 - 3）y + 6 + 5A = 4 x ² - y + 11
A=4x²+y+1
If the difference of a polynomial a minus 4x & # 178; - 3Y + 5 is 2Y + 6, then a
The relationship between the subtracted, the subtracted and the handicap is used to solve the problem.
A =2y+6+(4x²-3y+5)
=4x²-3y+2y+5+6
=4x²-y+11
How is the circumference formula of a circle obtained?
It's urgent
C = π * D π = 3.1412926 D is diameter
This is calculated by countless previous scientists
Circle cutting is to approach the circumference of a circle by the circumference of the inscribed regular polyhedron of the circle
If the square of 4A + 4A + k is a complete square, then k =?
（2a）²=4a²
The original formula = (2a + k) ²
=4a²+4ak+k²
Because 4ak = 4A, K & # 178; = k, so k = 1
（2a+1）^2=4a^2+4a+1 k=1
Let (2a + b) ^ 2 = 4A ^ 2 + 2 * 2A * B + B ^ 2 = 4A ^ 2 + 4A * B + B ^ 2
4b=4,b=1
k=b^2=1
Let this complete square be (2a + b) ^ 2, then (2a + b) ^ 2 = 4A ^ 2 + 2 * 2A * B + B ^ 2 = 4A ^ 2 + 4A * B + B ^ 2
4b=4,b=1
k=b^2=1
4a²+4a+k
=(2a+1)²
=4a²+4a+1
So k = 1
a²+2ab+b²=（a+b）²
The square of 4A + 4A + k = (2a + b) & #178;
4a=2*2a*b
b=1=k
If the square of X + 3x + 1 = 0, solve the equation and find the value of one part of X + X
X & # 178; + 3x + 1 = 0 divided by X at the same time
x+3+1/x=0
x+1/x=-3
-3
Solution 1: X & # 178; + 3x + 1 = 0
x²+1=-3x
(x²+1)×1/x=-3x×1/x
x+1/x=-3
Solution 2: (x + 1 / x) &# 178; = (- 3) &# 178;
x²+1/x²+2=9
x²+1/x²=9-2=7