It is known that a and B are the two roots of the equation x, where the square minus 5x minus 3 is equal to 0. A new bivariate linear equation is obtained, and its two roots are a + 2B and a + B

It is known that a and B are the two roots of the equation x, where the square minus 5x minus 3 is equal to 0. A new bivariate linear equation is obtained, and its two roots are a + 2B and a + B

a. B is the square of the equation x minus 5x minus 3 equals two roots of 0
a+b=5 ab=-3
a^2+2ab+b^2=25
a^2-2ab+b^2=25-4ab
(a-b)^2=37
a-b=√37 a-b=-√37
A = (5 + √ 37) / 2 or a = (5 - √ 37) / 2
(1) When a = (5 + √ 37) / 2 2B = 5 - √ 37 A + 2B = (15 - √ 37) / 2 A + B = 5
The new equation is x ^ 2 - (5 + (15 - √ 37) / 2) x + 5 (15 - √ 37) / 2 = 0
2x^2-(25-√37)x+5(15-√37)=0
(2) When a = (5 - √ 37) / 2 2B = 5 + √ 37 A + 2B = (15 + √ 37) / 2 A + B = 5
The new equation is x ^ 2 - (5 + (15 + √ 37) / 2) x + 5 (15 + √ 37) / 2 = 0
2x^2-(25+√37)x+5(15+√37)=0

(1) Given that one root of the equation 18x ^ 3 + 9x ^ 2-74x + 40 = 0 is twice the other root, find the solution set of the equation in complex number
(2) Given that the four roots of equation x ^ 4 + 4x ^ 3 + 10x ^ 2 + 12x + 9 = 0 in complex number have two multiple roots a and two multiple roots B, find the value of a and B
I hope the prawns can solve the problem in as much detail as possible

There are many methods: 1) according to the condition, a, 2a is its root, which is substituted into 18 * a ^ 3 + 9 * a ^ 2-74 * a + 40 = 0144 * a ^ 3 + 36 * a ^ 2-148 * a + 40 = 0, and the third and second-order terms are deleted; {X1 = a = 2 / 3}, {x2 = 2A = 4 / 3}, {X3 = - 5 / 2};
2) Four roots have two multiple roots a and two multiple roots B, which indicates that the original equation is a square formula; it can be set as (x ^ 2 + p * x + Q) ^ 2 = x ^ 4 + 4x ^ 3 + 10x ^ 2 + 12x + 9, then 2p = 4; Q ^ 2 = 9; 2pq = 12, so p = 2, q = 3; so there are:
X ^ 2 + 2 * x + 3 = 0, so a = - 1 + sqrt (2) * I, B = - 1-sqrt (2) * I; or a = - 1-sqrt (2) * I, B = - 1 + sqrt (2) * I

A problem of quadratic equation with one variable
It is known that two of the equations x (square) - 3mx + 2 (m-1) = 0 about X are X1 and X2, and 1 / X1 + 1 / x2 = - 3 / 4 is used to find M

1/X1+1/X2=X1+X2/X1X2
From Veda's theorem: X1 + x2 = 3m / 2, x1x2 = 2 (m-1)
So 1 / X1 + 1 / x2 = X1 + x2 / x1x2 = (3m / 2) / 2 (m-1) = - 3 / 4
m=8/25

If K is an integer, then the solution of equation (k-2012) x = 2014-2013x is also an integer. How many values of K are there?

(k-2012)x=2014-2013x
(k-2012)x+2013x=2014
(k+1)x=2014=1×2014=2×1007=19×106=38×53
∴k+1=±1 k+1=±2 k+1=±19 k+1=±38 k+1=±53 k+1=±106 k+1=±1007 k+1=±2014
There are 16 values

The solution of (2-k) x = 2013-x is a positive integer
Hurry, hurry, hurry!!!!

﹙2－k﹚x＋x＝2013
﹙3－k﹚x＝2013
x＝2013/﹙3－k﹚
∵ 2013＝3×11×61
X is a positive integer
∴ ①3－k＝1, k＝2
②3－k＝3, k＝0
③3－k＝11, k＝﹣8
④3－k＝61, k＝﹣58
⑤3－k＝33, k＝﹣30
⑥3－k＝183, k＝﹣180
⑦3－k＝671, k＝﹣668
⑧3－k＝2013, k＝﹣2010.

Let a and B be two unequal real roots of the equation x2-x-2012 = 0, then the value of A2 + 2A + B is______ ．

∵ A is the real root of equation x2 + x-2012 = 0, ∵ A2 + a-2012 = 0, that is, A2 = - A + 2012, ∵ A2 + 2A + B = - A + 2012 + 2A + B = 2012 + A + B, ∵ a, B are two unequal real roots of equation X2 + x-2012 = 0, ∵ a + B = - 1, ∵ A2 + 2A + B = 2012-1 = 2011

It is known that the sum of three consecutive integers is - 15. If the number in the middle is x, then the equation is

（x-1）+x+(x+1)=-15

Given 2aX = (a + 1) x + 6, when a is an integer, the solution of the equation is a positive integer

To solve the equation 2aX = (a + 1) x + 6 about X, we can get ax-x = 6, that is, (A-1) x = 6, so its solution is x = 6a − 1. To make the solution of the equation a positive integer, that is, 6a − 1 must be a positive integer, then (A-1) should be a positive divisor of 6, then A-1 = 1, 2, 3, 6, then a = 2, 3, 4, 7. So a = 2, 3, 4, 7

For the system of quadratic equations 4x + 7Y = - 19
4x-5y = 17 the equation obtained by adding and subtracting x is?

4x+7y-(4x-5y)=-19-17
12y=-36
y=-3
Substituting the original formula:
4x+7*(-3)=-19
4x-21=-19
4x=2
X = 1 / 2 (half)

Given the quadratic equation 4x-7y = 3, if y is expressed by the algebraic expression of X, then y=______ ．

From 4x-7y = 3, we can get y = 4x-37. So the answer is y = 4x-37