The equation x2 + Y2 + ax + 2ay + 2A2 + A-1 = 0 represents a circle, then the value range of a is () A. a＜-2B. -23＜a＜0C. -2＜a＜0D. -2＜a＜23

The equation x2 + Y2 + ax + 2ay + 2A2 + A-1 = 0 represents a circle, then the value range of a is () A. a＜-2B. -23＜a＜0C. -2＜a＜0D. -2＜a＜23

The equation x2 + Y2 + ax + 2ay + 2A2 + A-1 = 0 indicates that the circle 〈 A2 + 4a2-4 (2A2 + A-1) ＞ 0 〉 3a2 + 4a-4 ＜ 0, 〈 (a + 2) (3a-2) ＜ 0, 〈 − 2 ＜ a ＜ 23, so D is selected
What is the factorization of the square of 10A (X-Y) - 5B (Y-X)?
[10a(x-y)+5b](x-y)
(x-y)[10a(x-y)+5b]
X + 3 is less than - 1; 3a is more than 27; - 3 molecule x is more than 5; 5x is less than 4x-6
x+39
-x/3>5
X
(x + 2Y) & # 178; - (x-2y) & # 178; factorization results
Use the square difference formula
（x+2y）²-（x-2y）²
=（x+2y+x-2y）（x+2y-x+2y）
=8xy
Original formula = (x + 2Y + x-2y) (x + 2y-x + 2Y)
=(2x)(4y)
=8xy
8xy
（x + 2 y）² - （x - 2 y）²
= 【（x + 2 y）+ （x - 2 y）】【（x + 2 y）- （x - 2 y）】
= （x + 2 y + x - 2 y）（x + 2 y - x + 2 y）
= 2 x · 4 y
= （2 × 4）x y
= 8 x y
The equation x ^ 2 + y ^ 2 + ax + 2ay + 2A ^ 2 + a -- 1 = 0 represents a circle, then the value range of a is (please write the procedure)
x^2+y^2+ax+2ay+2a^2+a--1=0
(x+a/2)^2+(y+a)^2=1-a-(3/4)a^2>=0
3a^2+4a-4
Higher Algebra: finding the common root of polynomial f (x) = x ^ 3 + 2x ^ 2 + 2x + 1 and G (x) = x ^ 4 + x ^ 3 + 2x ^ 2 + X + 1
f(x)=x^3+2x^2+2x+1＝g(x)=x^4+x^3+2x^2+x+1
x^4-x=0
x(x^3-1)=0
x(x-1)(x^2+x+1)=0
There are two roots of real numbers
x=0,x=1
-1 / 2 + I (radical 3) / 2
Because there is a common factor X ^ 2 + X + 1
First, we try to find that f (x) has a root x = - 1, and f (x) is decomposed into (x + 1) * (x ^ 2 + X + 1)
It is found that ^ G + 2 can be divided into (x + 1) *
So their common roots are two roots of the quadratic equation x ^ 2 + X + 1 = 0.
I don't need to tell you the root formula of quadratic equation.. )
Let H (x) = g (x) - f (x) = x ^ 4 + x ^ 3 + 2x ^ 2 + X + 1 - (x ^ 3 + 2x ^ 2 + 2x + 1) = x ^ 4-x = x (x ^ 3-1). When x = 0 or x = 1, H (x) = 0, that is, f (x) = g (x)
Let f (x) = g (x), and solve the equation x ^ 4-x = 0 to get x = 0 or 1
System of inequalities 2x ≤ 4 + X + 2
2X
X3, x > 1.2, 3, 4, a total of three
4
For the quadratic equation of one variable x, the two of square + PX + q = 0 are 2 and - 3 respectively, then p = [], q = []
It can be seen from the meaning that (X-2) (x + 3) = 0
x2+x-6=0
So p = 1, q = - 6
The equation x2 + Y2 + ax + 2ay + 2A2 + A-1 = 0 represents a circle, then the value range of a is ()
A. a＜-2B. -23＜a＜0C. -2＜a＜0D. -2＜a＜23
The equation x2 + Y2 + ax + 2ay + 2A2 + A-1 = 0 indicates that the circle 〈 A2 + 4a2-4 (2A2 + A-1) ＞ 0 〉 3a2 + 4a-4 ＜ 0, 〈 (a + 2) (3a-2) ＜ 0, 〈 − 2 ＜ a ＜ 23, so D is selected
If the sum of a polynomial and X2 - 2x + 1 is 3x - 2, find the polynomial
(3x-2) - (x2-2x + 1), = 3x-2-x2 + 2x-1, = - x2 + 5x-3