4 + A + (a-2b) quadratic then what is a + 2B equal to

4 + A + (a-2b) quadratic then what is a + 2B equal to

Is it 4 + A + (a-2b) & sup2; = 0
If one is greater than 0, the other is less than 0
So both are equal to zero
So 4 + a = 0, a-2b = 0
a=-4.2b=a-4
So a + 2B = - 8
The absolute value term and the square term are nonnegative. If sum = 0, then all terms = 0
4+a=0 a=-4
a-2b=0 b=-2
a+2b=-4+(-4)=-8
If the sum of nonnegative numbers is 0, then each item is 0
So 4 + a = 0, a-2b = 0
a=-4.2b=a-4
So a + 2B = - 8
A ^ 2 + B ^ 2 + 2a-4b + 5 = 0, find the value of 2A ^ 2 + 4b-3
Two a ^ 2 + B ^ 2 + C ^ 2 = AB + AC + BC, prove a = b = C
Three a ^ 2 + B ^ 2 + 20 = m, 4 (2a + b) = n, find the size relationship of M and n
Do it with factorization. If factorization can't be done, write the process. Thank you
One a ^ 2 + B ^ 2 + 2a-4b + 5 = 0 = (a + 1) ^ 2 + (b-2) ^ 2 = 0, the sum of two squares must be greater than or equal to 0, that is, a = - 1, B = 22a ^ 2 + 4b-3 = 2 + 8-3 = 7, two a ^ 2 + B ^ 2 + C ^ 2 = AB + AC + BC2 (a ^ 2 + B ^ 2 + C ^ 2) = 2 (AB + AC + BC) move to a ^ 2-2ab + B ^ 2 + B ^ 2-2bc + C ^ 2 + A ^ 2-2ac + C ^ 2 = (a-b) ^ 2 + (B-C) ^ 2 +
In fact, the three questions are the same.
1. The original formula can be written as A2 + 2A + 1 + b2-4b + 4 = 0
(a+1)2+(b-2)2=0
So a = - 1, B = 2
It can be obtained by substituting.
2. The two sides of the original formula are the same as × 2, and the term is transferred to get
2a2+2b2+2c2-2ab-2ac-2bc=0
... unfold
In fact, the three questions are the same.
1. The original formula can be written as A2 + 2A + 1 + b2-4b + 4 = 0
(a+1)2+(b-2)2=0
So a = - 1, B = 2
It can be obtained by substituting.
2. The two sides of the original formula are the same as × 2, and the term is transferred to get
2a2+2b2+2c2-2ab-2ac-2bc=0
So, (a-b) 2 + (A-C) 2 + (B-C) 2 = 0
So, a = b = C
3. This problem can be compared with zero difference method.
M-N=a2+b2-8a-4b+20
=(a-4)2+(b-2)2
Because (A-4) 2 + (b-2) 2 is greater than or equal to 0
So m is greater than or equal to n
Given that equation (A-2) x | a | - 1 + 4 = 0 is a linear equation with one variable, then a=______ ，x=______ .
According to the characteristics of the linear equation of one variable, we get that | a | - 1 = 1a-2 ≠ 0, the solution is a = - 2, so we substitute a = - 2 into the original equation to get (- 2-2) x | - 2 | - 1 + 4 = 0, the solution is x = 1
Given a = - 1 / 2, B = 3, find the value of [(2a = b) square - (2a = b) (2a-b)] / 2B
[(2a + b) (2a-b)] / 2B
=(4a²+4ab+b²-4a²+b²)÷2b
=(4ab+2b²)÷2b
=2a+b
=2*(-1/2)+3
=-1+3
=2
That is to bring the values of a and B into the original formula
【（2a+b)²-(2a+b)(2a-b)】÷2b
=【（-1+3）²-（-1+3）（-1-3）】÷6
=（4+8）÷6
=12÷6
=2
9（m+n)2-(m-n)2,169(a+b)2-121(a-b)2
Note: the following 2 refer to square
Original formula = [3 (M + n) + (m-n)] [3 (M + n) - (m-n)]
=(4m+2n)(2m+4n)
=4(2m+n)(m+2n)
Original formula = [13 (a + b) + 11 (a-b)] [13 (a + b) - 11 (a-b)]
=(24a+2b)(2a+24b)
=4(12a+b)(a+12b)
9（m+n)2-(m-n)2=[3(m+n)+(m-n)][3(m+n)-(m-n)]=(4m+2n)(2m+4n)=4(2m+n)(m+2n)
169 (a + b) 2-121 (a-b) 2 = [13 (a + b) + 11 (a-b)] [13 (a + b) - 11 (a-b)] = (24a + 2b) (2a + 24b) = 4 (12a + b) (a + 12b) why 9 (m + n) 2 = [3 (M + n) + (m-n)]? Thank you for your help
Quadratic equation with two variables
Square of X + square of Y - 6x + 10Y + 34 = 0
X^2+Y^2-6X+10Y+34=0
(X-3)^2+(Y+5)^2=0
X=3 Y=-5
Square of (x-3) + square of (y + 5) = 0
x-3=0
y+5=0
X=3
y=-5
X²+Y²-6X+10Y+9+25=0
(X-3)²+(Y+5)²=0
X-3=0
Y+5=0
X=3,,,Y=-5
X2+y2-6x+10y+34=0
X2-6x+9+y2+10y+25=0
(x-3)2+(y+5)2=0
(x-3)2=0 x=3
(y+5)2=0 y=-5
2A square + 2B square + 3 ＞ 2Ab + 2a-2b
2a²+2b²+3-2ab-2a+2b
=a²-2ab+b²+a²-2a+1+b²+2b+1+1
=(a-b)²+(a-1)²+(b+1)²+1
≥1>0
∴2a²+2b²+3>2ab+2a-2b
Six small calculation problems of factorization in Volume 1 of Grade 8
1:(x+a)²-（y+b)²
2:25(a+b)²-4(a-b)²
3:18a²-50
4: A quartic power-16
5:1/2p²-2pq
6:a²(a-3)-a+3
1:(x+a+y+b)(x+a-y-b)
2:(7a+3b)(3a+7b)
3:2(3a+5)(3a-5)
4:(a²+4)（a²-4)
5:
1:(x+a+y+b)(x+a-y-b)
2:(7a+3b)(3a+7b)
3:2(3a+5)(3a-5)
Solving problems with linear equation of two variables
A. The distance between B is 12 kilometers. A starts from a and walks to B. B starts from B and bicycles to A. B stops for 40 minutes after arriving at a, and then bicycles back at the same speed. As a result, a and B arrive at B at the same time. If B bicycles 8 kilometers more per hour than a, the speed of a and B can be calculated
Let the velocity of a be x and the velocity of B be y
y=x+8
12/x=(12+12)/y+(20/60+40/60)
Simplification: X & sup2; + 20x-96 = 0
It's a quadratic equation of one variable
x²+20x-96=0
(x+24)(x-4)=0
X + 24 = 0 x = - 24 (meaningless)
x-4=0
X=4
y=x+8=4+8=12
The square of (2a-b + 3) (2a + B-3) - (a-2b-3)
Not after today