What are the solutions to the equation (x-1)^4-(x-1)^2-12=0 ?

The solutions to the equation (x-1)^4-(x-1)^2-12=0 are x=-1,x=3,x=1+sqrt(3)i,x=1-sqrt(3)i

Find the zeros of (x-1)^4-(x-1)^2-12=0

The zeros of (x-1)^4-(x-1)^2-12=0 are x=-1,x=3,x=1+sqrt(3)i,x=1-sqrt(3)i

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(x - 1)^{4} - (x - 1)^{2} - 12 = 0

x = - 1, x = 3, x = 1 + 3 i, x = 1 - 3 i

Solution steps

(x - 1)^{4} - (x - 1)^{2} - 12 = 0

Expand

(x - 1)^{4} - (x - 1)^{2} - 12 : x^{4} - 4 x^{3} + 5 x^{2} - 2 x - 12

x^{4} - 4 x^{3} + 5 x^{2} - 2 x - 12 = 0

Factor

x^{4} - 4 x^{3} + 5 x^{2} - 2 x - 12 : (x + 1) (x - 3) (x^{2} - 2 x + 4)

(x + 1) (x - 3) (x^{2} - 2 x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 3 = 0 or x^{2} - 2 x + 4 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 3 = 0 : x = 3

Solve

x^{2} - 2 x + 4 = 0 : x = 1 + 3 i, x = 1 - 3 i

The solutions are

x = - 1, x = 3, x = 1 + 3 i, x = 1 - 3 i

x, (x^{2} + y^{2})^{3} = y^{4}

x (x - 3) (x - 2) = 72

x^{3} + x^{2} + x + 1

x^{3} + 3 x^{2} - 4

z^{4} = - 1 - i 3

What are the solutions to the equation (x-1)^4-(x-1)^2-12=0 ?
The solutions to the equation (x-1)^4-(x-1)^2-12=0 are x=-1,x=3,x=1+sqrt(3)i,x=1-sqrt(3)i
Find the zeros of (x-1)^4-(x-1)^2-12=0
The zeros of (x-1)^4-(x-1)^2-12=0 are x=-1,x=3,x=1+sqrt(3)i,x=1-sqrt(3)i