x6 + 3x4 - x2 - 3 = 0
(x6 + 3x4) - (x2 + 3) = 0
x4(x2 + 3) - (x2 + 3) = 0
(x2 + 3)(x4 - 1) = 0
(x2 + 3)[(x2)2 - 12] = 0
(x2 + 3)(x2 + 1)(x2 - 1) = 0
(x2 + 3)(x2 + 1)(x + 1)(x - 1) = 0
x2 + 3 = 0 or x2 + 1 = 0 or x + 1 = 0 or x - 1 = 0
x2 + 3 = 0
x2 = -3
x = ±√-3 = ±i√3 ≈ ±1.7i
x2 + 1 = 0
x2 = -1
x = ±√-1 = ±i√1 ≈ ±i
x + 1 = 0
x = -1
x - 1 = 0
x = 1
The solutions are x = ±1, ±i, ±1.7i.
A=0 b=0 c=0
2x^2 + 8x + 3 = 0
x2-5-4x2+3x = 0 -3x2+3x-5 = 0 or as 3x2-3x+5 = 0
y equals negative 8.
Zero
Yes because if 1+0=1 than 0 plus b equals b
I would think -5 plus 5 equals 0.
A=0 b=0 c=0
No.
x = y = 0 ?
many solutions
It equals 7
It equals 6
1 plus 1 plus 1 plus 1 equals 1 times 4. 1 times 4 equals 4. 4 minus 4 equals 0. 0
x=0
The sum of 2 plus 0 plus 0 plus 0 plus 0 plus 0 plus 0 plus 0 plus 0 plus 0 plus 0 plus 0 plus 0 plus 0 plus 0 plus 0 equals 2.
0. Any equation that has times 0 equals 0.