What are the solutions to the equation x^4+x=x^3+1 ?

The solutions to the equation x^4+x=x^3+1 are x=1,x=-1,x= 1/2+i(sqrt(3))/2 ,x= 1/2-i(sqrt(3))/2

Find the zeros of x^4+x=x^3+1

The zeros of x^4+x=x^3+1 are x=1,x=-1,x= 1/2+i(sqrt(3))/2 ,x= 1/2-i(sqrt(3))/2

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x^{4} + x = x^{3} + 1

x = 1, x = - 1, x = \frac{1}{2} + i \frac{3}{2}, x = \frac{1}{2} - i \frac{3}{2}

Solution steps

x^{4} + x = x^{3} + 1

Move

1

to the left side

x^{4} + x - 1 = x^{3}

Move

x^{3}

to the left side

x^{4} + x - 1 - x^{3} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

x^{4} - x^{3} + x - 1 = 0

Factor

x^{4} - x^{3} + x - 1 : (x - 1) (x + 1) (x^{2} - x + 1)

(x - 1) (x + 1) (x^{2} - x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x + 1 = 0 or x^{2} - x + 1 = 0

Solve

x - 1 = 0 : x = 1

Solve

x + 1 = 0 : x = - 1

Solve

x^{2} - x + 1 = 0 : x = \frac{1}{2} + i \frac{3}{2}, x = \frac{1}{2} - i \frac{3}{2}

The solutions are

x = 1, x = - 1, x = \frac{1}{2} + i \frac{3}{2}, x = \frac{1}{2} - i \frac{3}{2}

x^{3} + 4 x^{2} - 9 x - 36 = 0

x^{5} = 0

z^{3} - 4 = 0

2 x^{3} - 5 x^{2} - x + 6 = 0

x^{5} - 3 x^{2} = 0

What are the solutions to the equation x^4+x=x^3+1 ?
The solutions to the equation x^4+x=x^3+1 are x=1,x=-1,x= 1/2+i(sqrt(3))/2 ,x= 1/2-i(sqrt(3))/2
Find the zeros of x^4+x=x^3+1
The zeros of x^4+x=x^3+1 are x=1,x=-1,x= 1/2+i(sqrt(3))/2 ,x= 1/2-i(sqrt(3))/2