What are the solutions to the equation (x^2+x+1)^2=x^4+8x^2+1 ?

The solutions to the equation (x^2+x+1)^2=x^4+8x^2+1 are x=0,x= 1/2 ,x=2

Find the zeros of (x^2+x+1)^2=x^4+8x^2+1

The zeros of (x^2+x+1)^2=x^4+8x^2+1 are x=0,x= 1/2 ,x=2

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(x^{2} + x + 1)^{2} = x^{4} + 8 x^{2} + 1

x = 0, x = \frac{1}{2}, x = 2

Decimal Notation

x = 0, x = 0.5, x = 2

Solution steps

(x^{2} + x + 1)^{2} = x^{4} + 8 x^{2} + 1

Expand

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

x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 1 = x^{4} + 8 x^{2} + 1

Move

1

to the right side

x^{4} + 2 x^{3} + 3 x^{2} + 2 x = x^{4} + 8 x^{2}

Move

8 x^{2}

to the left side

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

Move

x^{4}

to the left side

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

Factor

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

x (2 x - 1) (x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 2 x - 1 = 0 or x - 2 = 0

Solve

2 x - 1 = 0 : x = \frac{1}{2}

Solve

x - 2 = 0 : x = 2

The solutions are

x = 0, x = \frac{1}{2}, x = 2

x^{3} + x^{3} = 2

(x - 1)^{3} = 1

6 x^{3} + 13 x^{2} - 4 = 0

2 x^{3} - 7 x^{2} - 6 x + 8 = 0

x, x^{4} - 6 x^{2} + 12 = 7

What are the solutions to the equation (x^2+x+1)^2=x^4+8x^2+1 ?
The solutions to the equation (x^2+x+1)^2=x^4+8x^2+1 are x=0,x= 1/2 ,x=2
Find the zeros of (x^2+x+1)^2=x^4+8x^2+1
The zeros of (x^2+x+1)^2=x^4+8x^2+1 are x=0,x= 1/2 ,x=2