What are the solutions to the equation (x^2-x)^2-3(x^2-x)+2=0 ?

The solutions to the equation (x^2-x)^2-3(x^2-x)+2=0 are x=-1,x=2,x=(1+sqrt(5))/2 ,x=(1-sqrt(5))/2

Find the zeros of (x^2-x)^2-3(x^2-x)+2=0

The zeros of (x^2-x)^2-3(x^2-x)+2=0 are x=-1,x=2,x=(1+sqrt(5))/2 ,x=(1-sqrt(5))/2

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

x = - 1, x = 2, x = \frac{1 + 5}{2}, x = \frac{1 - 5}{2}

Decimal Notation

x = - 1, x = 2, x = 1.61803 \dots, x = - 0.61803 \dots

Solution steps

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

Expand

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

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 1 = 0 : x = - 1

Solve

x - 2 = 0 : x = 2

Solve

x^{2} - x - 1 = 0 : x = \frac{1 + 5}{2}, x = \frac{1 - 5}{2}

The solutions are

x = - 1, x = 2, x = \frac{1 + 5}{2}, x = \frac{1 - 5}{2}

2 x^{3} - 12 x^{2} + 18 x = 0

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

(4^{2} - X)^{4} = 16

\frac{1}{50} = b^{3}

x^{5} - x^{4} = 0

What are the solutions to the equation (x^2-x)^2-3(x^2-x)+2=0 ?
The solutions to the equation (x^2-x)^2-3(x^2-x)+2=0 are x=-1,x=2,x=(1+sqrt(5))/2 ,x=(1-sqrt(5))/2
Find the zeros of (x^2-x)^2-3(x^2-x)+2=0
The zeros of (x^2-x)^2-3(x^2-x)+2=0 are x=-1,x=2,x=(1+sqrt(5))/2 ,x=(1-sqrt(5))/2