The relationship between the root and coefficient of quadratic equation with one variable We know the equation x ^ 2 - (K + 1) x + 1 / 4K ^ 2 + 1 = 0 about X, if two real roots x, y of the equation satisfy |X | = y, find the value of K

The relationship between the root and coefficient of quadratic equation with one variable We know the equation x ^ 2 - (K + 1) x + 1 / 4K ^ 2 + 1 = 0 about X, if two real roots x, y of the equation satisfy |X | = y, find the value of K

There are two real roots, so the discriminant (K + 1) & sup2; - 4 (1 / 4K & sup2; + 1) > = 0k & sup2; + 2K + 1-k & sup2; - 4 > = 0k > = 3 / 2|x| = x or - x, so x = y or - x = y, that is, x + y = 0x = y, the discriminant is equal to 0, k = 3 / 2x + y = 0, then the relation between root and coefficient X + y = - [- (K + 1) / 1] = K + 1 = 0, k = - 1 and k > = 3 / 2 with root
Algebraic factorization uses a simple grouping method
（1）2ax-10ay+5by-bx （2）9x²-6x+2y-y²
Note that the simple grouping method is used. Help me write out the grouping step
(1) The original formula = (2aX BX) - (10ay-5by)
=（2a-b）x-5y(2a-b)
=（2a-b）(x-5y)
(2) Original = (9x & # 178; - Y & # 178;) - (6x-2y)
=(3x-y)(3x+y)-2(3x-y)
=(3x+y)(3x-y-2)
1 =2a(x-5y)+b(5y-x)=(2a-b)(x-5y)
2 3x(3x-2)+y(2-y)
（1）2ax-10ay+5by-bx
=(2ax-bx)-(10ay-5by)
=(2a-b)x-(10a-5b)y
=(2a-b)x-5(2a-b)y
=(2a-b)(x-5y)
（2）9x²-6x+2y-y²
=(9x & # 178; - 6x) + (2y-y & # 1... Expansion
（1）2ax-10ay+5by-bx
=(2ax-bx)-(10ay-5by)
=(2a-b)x-(10a-5b)y
=(2a-b)x-5(2a-b)y
=(2a-b)(x-5y)
（2）9x²-6x+2y-y²
=(9x²-6x)+(2y-y² )
=(9x²-6x)-(y²-2y )
=(3x-1)^2-(y-1 )^2-1+1
=(3x-1+)y-1 (3x-1-y+1)
=(3x + Y-2) (3x-y) fold up
1/3x=1/2x+2
1/3x=1/2x+2
1/3*x-1/2*x=2
-1/6*x=2
x=-12
Math problem?
-2=1/2x-1/3x
-2=1/6x
x=-12
The relationship between the root and coefficient of quadratic equation with one variable
1. Given that X1 and X2 are the two real roots of the equation x * 2 + 3x + 1 = 0, find the value of X1 * 3 + 8x2 + 20
2. The lengths of the two right sides of a right triangle a and B are exactly two of the equation x * 2 - (M + 2) x + 4m = 0, where a and B are integers. Try to find the value of M and the lengths of the three sides of the right triangle
According to the relationship between ^ 2 and the root of the equation, ^ 3x1 + 2 + 3x1-20 + 3x1-8x1 + 2x1 + 3x1-20 + 3x1-8x1 + 3x1-20 + 3x1 + 2x1 + 3x1 + 2x2 = 3x1 + 3x1-20 + 3x1 + 3x1-2x2 + 3x1 + 3x1-20 + 3x1 + 3x1 + 2x2 + 3x1-2x2 + 3x1 + 3x1 + 3x1-20 + 3x1 + 3x1 + 3x1 + 2x2 + 3x1-2x2 + 3x1-20 + 3x1 + 3x1 + 3x1 + 3x1-2x2 + 3x1 + 3x1-20 + 3x1 + 3x1-3x1 + 2x2 + 3x1 + 3x1 + 2x2 + 3x1 + 3x1-20 + 3x1 + 3x1 + 2x2 + 3x1 + 2x2 + 3x1-2x2 + 3x1 + 3x1 + 2x2 + 3x1-20 + 3x
In the following formulas, the one that cannot be solved by the square difference formula is ()
A. 9x2n-36y2nB. a3n-a5nC. （x+y）2-4xyD. （x2-y2）2-4x2y2
A. 9x2n-36y2n conforms to the characteristics of the square difference formula and can be decomposed by the square difference formula, so it is wrong; B, a3n-a5n conform to the characteristics of the square difference formula and can be decomposed by the square difference formula, so it is wrong; 4xy in C, (x + y) 2-4xy cannot be written in the form of the square term and can not be decomposed by the square difference formula, so
1/3x+(1/2x+1/3x)×2=1
Solve the equation in detail
1/3x+(1/2x+1/3x)×2=1
1/3x+x+2x/3=1
2x=1
x=1/2
Is there a formula for finding the root of an equation of degree n of one variable? If so, please write it. If not, can it be proved
Up to 4 times, there is a formula
Then there is no formula solution
Several factoring problems in mathematics
1、(x+p)^2-(x+q)^2
2、(a+b)^2-1
3、x-1+b^2(1-x)
4、12x^2-3y^2
1、(x+p)^2-(x+q)^2
=(x+p+x+q)(x+p-x-q)
=(2x+p+q)(p-q)
2、(a+b)^2-1
=(a+b+1)(a+b-1)
3、x-1+b^2(1-x)
=b^2(1-x)-(1-x)
=(b^2-1)(1-x)
=(b+1)(b-1)(1-x)
4、12x^2-3y^2
=3(4x^2 -y^2)
3(2x+y)(2x-y)
3x-1 in 2 = 2x + 1 in 1-6
Double six on both sides
3(3x-1)=6-(2x+1)
9x-3=6-2x-1
9x+2x=6-1+3
11x=8
x=8/11
（3X-1）/2=1-（2X+1）/6
3（3X-1）=6-（2X+1）
9X-3=6-2X-1
11X=8
X=8/11
Matlab solution of multivariate higher order equations
A, B, C, D, e is a system of equations of five variables. I can't debug the results all the time. When I run it, it's busy. Is there something wrong with the program?
syms A B C D E;
eq1='1/4*(110+220*A)*(3/25+B)*(69/50+A)*(11/100+C)-1377380*B*A*(11/400+5/4*C)*(1/2+A)-305/4*A*(11/400+5/4*C)*(3/25+B)*(69/50+A)=0';
eq2='(271698404112853371/46116860184273879040+33962300514106665/1152921504606846976*D+18411/31250*A)*(3/25+B)*(69/50+A)*(9/10+B)*((49/2+49*E)*(1/2+E)+81/625)-10901624/125*B*A*(6712548807494023/18446744073709551616+4214072214526101626067/360287970189639680000000*D+6498/390625*A)*(9/10+B)*((49/2+49*E)*(1/2+E)+81/625)-3715146/78125*B*(671254880749402 3/18446744073709551616+4214072214526101626067/360287970189639680000000*D+6498/390625*A)*(3/25+B)*(69/50+A)=0';
eq3='4516*B*A*(3/20+C)+(21/1250+7/50*B)*(69/50+A)*(3/20+C)-9/50*C*(3/25+B)*(69/50+A)-1/500*C*(3/25+B)*(69/50+A)*(3/20+C)=0';
eq4='6156/625*B*(1/20+D)-17/200*D*(1/20+D)*(9/10+B)*((49+98*E)*(1+2*E)+81/625)+9/50*D*(9/10+B)*((49+98*E)*(1+2*E)+81/625)=0';
eq5='17/200*D-7/100*E=0';
[A,B,C,D,E]=solve(eq1,eq2,eq3,eq4,eq5,'A,B,C,D,E')
Matlab7.0 is used to solve the equations of degree 6 with variables a, B, C, D and E
The problem is that I have no results after busy! If normal, you send the results!