What are the solutions to the equation x+4=((x^2-4x)-2)^2 ?

The solutions to the equation x+4=((x^2-4x)-2)^2 are x=0,x=5,x=(3+sqrt(21))/2 ,x=(3-sqrt(21))/2

Find the zeros of x+4=((x^2-4x)-2)^2

The zeros of x+4=((x^2-4x)-2)^2 are x=0,x=5,x=(3+sqrt(21))/2 ,x=(3-sqrt(21))/2

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x + 4 = ((x^{2} - 4 x) - 2)^{2}

x = 0, x = 5, x = \frac{3 + 21}{2}, x = \frac{3 - 21}{2}

Decimal Notation

x = 0, x = 5, x = 3.79128 \dots, x = - 0.79128 \dots

Solution steps

x + 4 = ((x^{2} - 4 x) - 2)^{2}

Expand

((x^{2} - 4 x) - 2)^{2} : x^{4} - 8 x^{3} + 12 x^{2} + 16 x + 4

x + 4 = x^{4} - 8 x^{3} + 12 x^{2} + 16 x + 4

Switch sides

x^{4} - 8 x^{3} + 12 x^{2} + 16 x + 4 = x + 4

Move

4

to the right side

x^{4} - 8 x^{3} + 12 x^{2} + 16 x = x

Move

x

to the left side

x^{4} - 8 x^{3} + 12 x^{2} + 15 x = 0

Factor

x^{4} - 8 x^{3} + 12 x^{2} + 15 x : x (x - 5) (x^{2} - 3 x - 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 5 = 0 : x = 5

Solve

x^{2} - 3 x - 3 = 0 : x = \frac{3 + 21}{2}, x = \frac{3 - 21}{2}

The solutions are

x = 0, x = 5, x = \frac{3 + 21}{2}, x = \frac{3 - 21}{2}

x^{3} - x^{2} - 8 x + 10 = 0

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

x^{3} - 13 x^{2} - 12 = 0

(3 x^{2} - 4 x - 5)^{2} = - 3 x^{2} + 4 x + 5

(1 + x)^{4} = 1.4

What are the solutions to the equation x+4=((x^2-4x)-2)^2 ?
The solutions to the equation x+4=((x^2-4x)-2)^2 are x=0,x=5,x=(3+sqrt(21))/2 ,x=(3-sqrt(21))/2
Find the zeros of x+4=((x^2-4x)-2)^2
The zeros of x+4=((x^2-4x)-2)^2 are x=0,x=5,x=(3+sqrt(21))/2 ,x=(3-sqrt(21))/2