The solution of equation 5x-2a = - 6-x is an integer

The solution of equation 5x-2a = - 6-x is an integer

5x-2a=-6-x
5x+x=-6+2a
6x=2a-6
x=a/3-1
Because a / 3-1 is an integer
So a / 3 is an integer, that is to say, a can be divided by 3

If x 1 and x 2 are the two roots of the quadratic equation x 1-6x-7 = 0, then the value of the square of x 1 and the square of x 2
7 = 0 are X1 and X2 respectively, then the square of X1 + the square of X2 is

X 1, x 2 are the two roots of the quadratic equation x with one variable
x1+X2=6,x1x2=-7
x1^2+x2^2
=x1^2+2x1x2+x2^2-2x1x2
=(x1+x2)^2-2x1x2
=36+14
=50

If X1 and X2 are quadratic equations of one variable and the square of x-6x-2 = two real roots of 0, then what is X1 + x2?

According to Weida's theorem:
x1+x2=-b/a=6.

Known: the square of X - 5x + 1 = 0, find the square of (x4 + 1) parts of X

The square of X - 5x + 1 = 0 1 1 X-5 - - = 0 X - - = 5 x x (x ^ 4 + 1) times the square of X, first calculate its reciprocal: x ^ 4 + 1 1 1 - = x ^ 2 + - = (x - -) ^ 2-2 substitute the above simplified formula: (5) ^ 2-2 = 25-2 = 23. X ^ 2 x ^ 2 x, then the square of (x ^ 4 + 1) times x = 23

If the square of polynomial X4 - (A-1) x & # 179; + 5x & # 178; - (B + 3) x + 6 contains no terms of X & # 179; and X, then a = [], B = []

If the square of polynomial X4 - (A-1) x & # 179; + 5x & # 178; - (B + 3) x + 6 contains no terms of X & # 179; and X, then a = [1], B = [- 3]
According to the meaning of the title:
a-1=0,a=1
b+3=0,b=-3
So, a = [1], B = [- 3]

How to solve 5x + (x + 10) X4 = 400?
Answer in a few minutes. The answer must be 40

How to solve 5x + (x + 10) X4 = 400?
5x+4x+40=400;
9x=360;
x=40;
I'm very glad to answer your questions. Skyhunter 002 will answer your questions
If you don't understand this question, you can ask,

7. Given x4-5x3 + 8x2-5x + 1 = 0, then the value of X + is__________ .
cxccxxx

The steps of x = 1 or x = (3 + radical 5) / 2 or x = (3-radical 5) / 2 are a little difficult to explain. It's a small skill to use a remainder theorem... You can understand it as a piece of work. First, according to the 0 of X, this term = 1 and the remainder theorem, we can deduce that the value of X must be 1 or - 1, or any two numbers that multiply by 1. First, we can substitute 1, etc

It is known that x2 + 5x-990 = 0, X3 + 6x2-985x + 1020=
X2 is the square of X and X3 is the third power of X

2010. The calculation method is as follows: because x2 + 5x-990 = 0, so x2 + 5x = 990
x3+6x2-985x+1020
＝x（x2+6x-985）+1020
＝x（x2+5x+x-985）+1020
＝x（990+x-985）+1020
＝x（5+x）+1020
＝5x+x2+1020
＝990+1020
＝2010

Given x2 + 5x-4 = 0, find the value of X3 + 6x2 + X + 2
Given x2 + X-1 = 0, find the value of X3 + 2x2 + 3

x2+5x-4=0
therefore
x2+5x=4
x3+6x2+x+2
=x(x2+5x)+x2+x+2
=4x+x2+x+2
=x2+5x+2
=4+2
=6
x3+2x2+3
=x(x2+x)+x2+3
=x+x^2+3
=1+3
=4

If x2 + 5x-990 = 0, what is the value of X3 + 6x2-985x + 1012?
(the numbers after X are all square.) thank you!

x^2+5x-990=0
x^2+5x=990
x^3+6x^2-985x+1012
=x（x^2+5x+x-985）+1012
=x(990+x-985)+1012
=x(x+5)+1012
=x^2+5x+1012
=990+1012
=2002