3x^2=4x-1 、 y^2+2=2√2y 6x^2+13x=-6 4x^2+5x+30=0 How to do it? It needs specific steps. Thank you

3x^2=4x-1 、 y^2+2=2√2y 6x^2+13x=-6 4x^2+5x+30=0 How to do it? It needs specific steps. Thank you

1) 3x²=4x-1
3x²-4x+1=0
(3x-1)(x-1)=0
x1=1/3 x2=1
2) y²+2=2√2y
y²-2√2y+2=0
(y-√2)²=0
y1=y2=√2
3) 6x²+13x=-6
6x²+13x+6=0
(6x+1)(x+6)=0
x1=-1/6 x2=-6
4) 4x²+5x+30=0
Δ=25-480

Given x ^ 2 + X + 1 = 0, find the value of x ^ 4 + 3x ^ 3 + 5x ^ 2 + 4x + 1997. Given x ^ 2-13x + 1 = 0, find the number of x ^ 4 + 1 / x ^ 4

X ^ 3 = 1 is obtained by multiplying both sides of x ^ 2 + X + 1 = 0 by X-1, so: x ^ 4 + 3x ^ 3 + 5x ^ 2 + 4x + 1997 = x + 3 + 5x ^ 2 + 4x + 1997 = 5x ^ 2 + 5x + 2000 = 5 (x ^ 2 + X + 1) + 1995 = 1995x ^ 4 + 1 / x ^ 4 = (X & # 178; + 1 / X & # 178;)) &# 178; - 2 = [(x + 1 / x) &# 178; - 2] &# 178; - 2 is obtained by dividing both sides of x ^ 2-13x + 1 = 0 by X ≠ 0

3x-2=x+4
3x-2 = x + 4, which is a simple equation of first degree with one variable,

3x-2=x+4
You can move the item with X to the left: 3x-x-2 = 4 -- > 2x-2 = 4
Move the constant term to the right: 2x = 4 + 2 ----- > 2x = 6
The solution is: x = 3

How to solve the equation 3x + 1 / 3x = x + 2 (x + 10),

3x+1/3x=x+2(x+10)
10/3x=x+2x+2*10
10/3x=3x+20
(10/3-3)x=20
1/3x=20
1/3x*3=20*3
x=60