The square of X + 3x-130 = 0 is calculated by formula

The square of X + 3x-130 = 0 is calculated by formula

x ² + 3 x - 130 = 0
a = 1 ,b = 3 ,c = - 130
∵ △ = b ² - 4 a c
= 3 ² - 4 × 1 × （- 130）
= 9 + 520
= 529 > 0
The equation has two unequal real roots
∴ x = 【 - b ± √（n ² - 4 a c）】 / 2 a
= （- 3 ± √529）/ 2 × 1
= （- 3 ± √ 23 ² ）/ 2
= （- 3 ± 23）/ 2
x1 = （- 3 + 23）/ 2 = 20 / 2 = 10
x2 = （- 3 - 23）/ 2 = - 36 / 2 = - 13
{2x+y+Z=10 {2x+4y+3z=9 {2x+2y+2=16{3x+2y+5z=11 {x+y+2z=8{5x+6y+7z=13
The title should be
{2x+y+Z=10 （1）
{2x+4y+3z=9 （2）
{2x+2y+z=16 （3）
From (2) - (1)
3y+2z=-1
From (3) - (1)
Y=6
z=-19/2
x=23
{3x+2y+5z=11 （1）
{x+y+2z=8 （2）
{5x+6y+7z=13 （3）
From (1) - (2) * 2
x+z=-5 （4）
From (3) - (2) * 6
x+5z=35 (5)
From (5) - (4)
4z=40
z=10
x=-15
Y=3
Factorization-x ^ 5Y ^ 3 + x ^ 3Y ^ 5
-x^5*y^3+x^3*y^5
=-x^3*y^3(x^2-y^2)
=-x^3*y^3(x+y)(x-y)
-x^5*y^3+x^3*y^5
=-x^3*y^3(x^2-y^2)
=-(x+y)(x-y)x^3*y^3
-x^5y^3+x^3y^5=x^3y^3(y^2-x^2)=x^3y^3(y+x)(y-x)
Solving equation 6x to the second power - (2x + 3) (3x-2) = 0
6X to the second power - (2x + 3) (3x-2) = 0
6x^2-(6x^2-4x+9x-6)=0
6x^2-(6x^2+5x-6)=0
6x^2-6x^2-5x+6=0
5x=6
x=6/5
）=(4x + 2Y) (2x-4y) = 4 (2x + y) (x-2y) why? I think the common factor should be 2 A
The common factor of both brackets is 2
So it's 2 * 2 = 4
2 (a + b) & # 178; - 5 (a + b) + 3 is factorized by cross phase multiplication
2﹙a＋b﹚²－5﹙a＋b﹚＋3
1 -1
2 -3
=[(a+b)-1][2(a+b)-3]
=(a+b-1)(2a+2b-3)
The original formula = [2 (a + b) - 3] [(a + b) - 1]
=(2a + 2b-3) (a + B-1) can you be more detailed?
How to calculate x ^ 3-3x-2 = 0? What is the formula for solving the equation with cubic power?
Hello, I'm glad to answer your question
x^3-3x-2=0
x^3-1-3x+3=0
(x-1)(x^2+x+1)-3(x-1)=0
(x-1)(x^2+x-2)=0
(x-1)^2*(x+2)=0
x1=1,x2=-2
x³-x-2x+2=0x(x²-1)-2x+2=0x(x+1)(x-1)-2(x-1)=0(x²+x)(x-1)-2(x-1)=0(x-1)(x²+x-2)=0(x-1)(x+2)(x-1)=0(x-1)²(x+2)=0x=1,x=-2
X^3-3X-2=X^3-2X^2+2X^2-4X+X-2
=X^2(X-2)+2X(X-2)+(X-2)
=(X-2)(X^2+2X+1)
=(X-2)(X+1)^2=0
X1=2; X2=-1
How to calculate 4x * 4y-2x * 2Y
=2X*2Y(2-1)=4XY
Factorization (X & # 178; + Y & # 178;) &# 178; - 4x & # 178; Y & # 178;
Original formula = x & # 8308; + 2x & # 178; Y & # 178; + Y & # 8308; - 4x & # 178; Y & # 178;
=x⁴-2x²y²+y⁴
=（x²-y²）²
The equation is solved by formula: (1) x2-4x + 1 = 0 (2) 5x2 = 4x-1 (3) 2x2-2x-1 = 0 (4) 4x (X-52) = 8
(1) Here a = 1, B = - 4, C = 1, ∵ △ 16-4 = 12, ∵ x = 4 ± 232 = 2 ± 3; (2) the equation is sorted out as follows: 5x2-4x + 1 = 0, here a = 5, B = - 4, C = 1, ∵ △ 16-20 = - 4 < 0, ∵ the equation has no solution; (3) here a = 2, B = - 2, C = - 1, ∵ △ 4 + 8 = 12, ∵ △ x = 1 ± 32, the solution is: X1 = 1 + 32