Solving quadratic inequality with one variable (1) 16x ^ 2-8x + 1 is greater than 0 (2) - 2x ^ 2 + X-1 / 8 ≥ 0 (3) 3x ^ 2-5x + 1 ≥ 0 (1) 16x ^ 2-8x + 1 greater than 0 （2）-2x^2+x-1/8≥0 （3）3x^2-5x+1≥0

Solving quadratic inequality with one variable (1) 16x ^ 2-8x + 1 is greater than 0 (2) - 2x ^ 2 + X-1 / 8 ≥ 0 (3) 3x ^ 2-5x + 1 ≥ 0 (1) 16x ^ 2-8x + 1 greater than 0 （2）-2x^2+x-1/8≥0 （3）3x^2-5x+1≥0

1. X is not equal to 1 / 4
2.x=1/4
3. X > = (5 ＋ 13) / 6 or X
utuututut
1、X≠1/4
2、0＜X＜1/2
3、（5-√13）/6＜X＜（5+√13）/6
Let Tana and tanb be the two real roots of the quadratic equation AX square + BX + C = 0, and B is not equal to 0, then find the value of cot (a + b)
Because Tana and tanb are two real roots of the quadratic equation AX square + BX + C = 0
So Tana + tanb = - B / A, tana.tanB=c/a
tan(a+B)=sin(a+B)/cos(a+B)=(tana+tanB)/(1-tanatanB)
cot(a+B)=1/tan(a+B)=(1-tanatanB)/(tana+tanB)=（1-c/a)/(-b/a)=(c-a)/b
According to Weida's theorem, Tana + tanb = - B / A, tanatanb = C / A
∴cot(A+B)=1/tan(A+B)=(1-tanAtanB)/(tanA+tanB)=-(a-c)/b=(c-a)/b
3x-2x + 1 ≥ 1, 4-1-3x / 3 ≤ 5x-3 / 2 to solve inequality system
Firstly, it is divided into two inequality systems: (1): 3x-2x + 1 ≥ 14-1-3x / 3
(2):1 4-1-3x/3≤5x-3/2
The results are as follows: (1): 2x ≥ 12
（2）：11x≥29
Further simplification: X ≥ 6 (the same big takes the big)
Given that Tana and tanb are the two roots of the equation 3x & # 178; - 4x-5 = 0, find cot (a + b)
Answer: the two parts of the equation 3x & # 178; - 4x-5 = 0 are Tana and tanb. According to Weida's theorem, Tana + tanb = 4 / 3tan * tanb = - 5 / 3, so: COTA * cotb = - 3 / 5cot (a + b) = (COTA * cotb-1) / (COTA + cotb) = (- 3 / 5-1) / (COTA + cotb) = (- 8 / 5) / (1 / Tana + 1 / tanb) = (- 8 / 5) (Tana * tanb)
In this paper, we solve the system of inequalities and express its solution set on the number axis
Solve the system of inequalities, and express its solution set on the number axis: < 2 / 2 x-3 is greater than or equal to x, 1-3 (x-1) < 8-x,
According to the meaning of the problem, x-3 is greater than or equal to x, 1-3 (x-1) < 8-x, the solution is x ≤ 1 / 3, x > 2
Its solution set is expressed on the number axis as follows:
Known 0
The given equation should be x ^ 2 - KX + K + 1 = 0! In this case, the method is as follows: by Weida's theorem, there are: sin α + cos α = k, sin α cos α = K + 1. From the square of sin α + cos α = k, we can get: (sin α) ^ 2 + (COS α) ^ 2 + 2Sin α cos α = k ^ 2, ｜ 1 + 2Sin α cos α = k ^
sinα+cosα＝k
sinαcosα＝k+1
∵sinαcosα＝[(sinα+cosα﹚²-1]/2
∵sinαcosα-(sinα+cosα)=1
∴(sinα+cosα)²-2(sinα+cosα)-3=0
Ψ sin α + cos α = k = 3 or - 1..... 3 is rounded off (because sin α + cos α = √ 2Sin β ≤ √ 2)
∴k= -1
∴y=x^2+kx-k/4=x²-x+1/4=(x-1/2)²≥0
There are sin α + cos α = k, sin α * cos α = K + 1
(sinα)^2+(cosα)^2=(sinα+cosα)^2-2sinα*cosα=k^2-2(k+1)=1
So k = 3 or K = - 1
When k = 3, y = x ^ 2 + 3x-3 / 4 = (x + 3 / 2) ^ 2-3 > = - 3, that is, the value range is Y > = - 3
When k = - 1, y = x ^ 2-x + 1 / 4 = (x-1 / 2) ^ 2 > = 0, that is, the value range is Y > = 0
Solve the system of inequalities ① (x-3 of 2) + 3 ≥ x + 1 ② 1-3 (x-1) < 8-x and express the solution set on the number axis
x-3/2+3≥x+1①
1-3（x-1）＜8-x ②
On the solution of inequality 1
X is not greater than 1
On the solution of inequality 2
x＜-5
X ≤ 1 is obtained from ①, and x > 2 is obtained from ②
So 1 ≤ x > 2
It is known that the sum of two univariate quadratic equations a ^ 2x ^ 2 + B ^ 2x + C ^ 2 = 0 is the sum of two squares of the univariate quadratic equation AX ^ 2 + BX + C = 0
It is known that the sum of the two univariate quadratic equations a ^ 2x ^ 2 + B ^ 2x + C ^ 2 = 0 is the sum of the squares of the two univariate quadratic equations ax ^ 2 + BX + C = 0, then the relationship between a and B.C is
The sum of the two equations a ^ 2x ^ 2 + B ^ 2x + C ^ 2 = 0 is - B ^ 2 / A ^ 2, which is based on the relationship between root and coefficient
Let two of the equations ax ^ 2 + BX + C = 0 be x1, X2, then X1 + x2 = - B / A, x1x2 = C / A
x1^2+x2^2=(x1^2+2x1x2+x2^2)-2x1x2=(x1+x2)^2-2x1x2=(-b/a)^2-2*(c/a)=b^2/a^2-2c/a
It should be - B ^ 2 / A ^ 2 = B ^ 2 / A ^ 2-2c / A, 2b ^ 2 / A ^ 2 = 2C / A
Multiply both sides by a / 2 to get B ^ 2 / a = C, B ^ 2 = AC
This is the relationship
A = b > or = 4C
AC = B ^ 2 let a ^ 2x ^ 2 + B ^ 2x + C ^ 2 = 0 be a and B, ax ^ 2 + BX + C = 0 be C and D, a + B = C ^ 2 + D ^ 2
A + B = - B ^ 2 / A ^ 2, C ^ 2 + D ^ 2 = (c + D) ^ 2-4c * d = (- B / a) ^ 2-4c / A, and - B ^ 2 / A ^ 2 = (- B / a) ^ 2-4c / A, so
ac=b^2
A equals B, B = 4
Solve the following inequality and express the solution set on the number axis. 2 (x-1) + 2 is less than 5-3 (x + 1)
Solve the following inequality and express the solution set on the number axis
(1) 2 (x-1) + 2 ＜ 5-3 (x + 1) (2) 2 of (1 + x) ≤ 3 of (2x-1) + 1
Answer: 1) 2 (x-1) + 2 & lt; 5-3 (x + 1) 2x-2 + 2 & lt; 5-3x-32x & lt; - 3x + 22x + 3x & lt; 25X & lt; 2x & lt; 2 / 52) (1 + x) / 2 & lt; = (2x-1) / 3 + 13 (x + 1) & lt; = 2 (2x-1) + 63x + 3 & lt; = 4x-2 + 63x + 3 & lt; = 4x + 44x-3x & gt; = 3-4x & gt; = - 1
Given the equation 2x ^ 2 + ax-2a + 1 = 0, when the value of a is, the sum of the squares of the two is equal to 29 / 4
I don't know if it's right. By the way, what's the sum of the squares of two
X1 + x2 = - A / 2, x1x2 = (- 2A + 1) / 2, so the sum of squares = X1 ^ 2 + x2 ^ 2 = (x1 + x2) ^ 2-2x1x2 = a ^ 2 / 4 - (- 2A + 1) = 29 / 4A ^ 2 / 4 + 2a-1 = 29 / 4A ^ 2 + 8a-33 = 0 (a + 11) (A-3) = 0A = - 11, a = 3, the discriminant is greater than or equal to 0A ^ 2-8 (-)
Sum of squares of two = X1 ^ 2 + x2 ^ 2
x1+x2=a/2
x1x2=(1-2a)/2
x1^2+x2^2=(x1+x2)^2-2x1x2
=a^2/4-(1-2a)=29/4
a^2-4+8a=29
a^2+8a-33=0
（a+11）（a-3）=0
a=-11,a=3
a^2-8(1-2a)>=0
a^2+16a-8>=0
so,a=-11,a=3
Yes, the answer is 3
Let two be x1, x2,
The sum of squares of the two is X1 ^ 2 + x2 ^ 2
x1+x2=-a/2
x1x2=(1-2a)/2
x1^2+x2^2=(x1+x2)^2-2x1x2=a^2/4-1+2a=29/4
a^2+8a-33=0
The solution is a = - 11 or a = 3
The discriminant is a ^ 2-8 + 16A
When a = - 11, the discriminant is less than 0,
So a = 3