Step 1: (X4 = 2)Step 2: /4 (X = 2/4 or 1/2)If you meant x raised to the 4th power = 2, then the answer is x = 2^(1/4) = ±1.189207115, and the two imaginary roots are ±1.189207115i
f(x) = 2*(x-3)*(x+2)/(x-1) for x ≠1
f(x)=x+1 g(f(x))=x f(x)-1=x g(x)=x-1
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
X^4 ? 4x^3
4X + 2x = 1. Where x = 0.166666666666666666666666666666666666666 recurring.
N = x4 x8 = x4+4 = (x4)2 = N2
9 x 4 = 36
If: x4 = 5 Then: x = 1.4953487812
x8-1 becomes (x4 + 1) (x4 - 1) - perfect square and then (x4 + 1) (x2 - 1)(x2 + 1) - perfect square again. and then (x4 + 1)(x2 + 1)(x + 1)(x - 1) - perfect square again
1 - x4 = (1 - x2)(1 + x2) = (1 - x)(1 + x)(1 + x2) (difference of squares)
No.
Not necessarily.
(x - 1)(x + 1)(x2 + 1)
So x4 - 1 is the difference of squares, so x4 - 1 = (x2 - 1)(x2 + 1) = (x + 1)(x - 1)(x2 + 1).
x4 - 1 = (x2 + 1)(x + 1)(x - 1).
x4 + 3x3 - x2 - 9x - 6 = 0 x4 + x3 + 2x3 + 2x2 - 3x2 - 3x - 6x - 6 = 0 x3(x + 1) + 2x2(x + 1) - 3x(x + 1) - 6(x + 1) = 0 (x + 1)(x3 + 2x2 - 3x - 6) = 0 (x + 1)[x2(x + 2) - 3(x + 2)] = 0 (x + 1)(x + 2)(x2 - 3) = 0 So x + 1 = 0 so that x = -1 or x + 2 = 0 so that x = -2 or x2 - 3 = 0 so that x = +/- sqrt(3)