What are the solutions to the equation 1-y^2=y^4-y^2+y-1 ?

The solutions to the equation 1-y^2=y^4-y^2+y-1 are y=1,y\approx-1.35320…

Find the zeros of 1-y^2=y^4-y^2+y-1

The zeros of 1-y^2=y^4-y^2+y-1 are y=1,y\approx-1.35320…

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1 - y^{2} = y^{4} - y^{2} + y - 1

y = 1, y \approx - 1.35320 \dots

Solution steps

1 - y^{2} = y^{4} - y^{2} + y - 1

Switch sides

y^{4} - y^{2} + y - 1 = 1 - y^{2}

Move

y^{2}

to the left side

y^{4} + y - 1 = 1

Move

1

to the left side

y^{4} + y - 2 = 0

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y - 1 = 0 or y^{3} + y^{2} + y + 2 = 0

Solve

y - 1 = 0 : y = 1

Solve

y^{3} + y^{2} + y + 2 = 0 : y \approx - 1.35320 \dots

The solutions are

y = 1, y \approx - 1.35320 \dots

64 x^{3} + 24 x^{2} - 6 x - 1 = 0

4 x^{12} = 6

(x - 9) (x^{2} - 36) = 0

2 x^{3} - 6 x^{2} - 6 x + 18 = 0

9 x^{4} - 18 x^{2} + 5 = 0

What are the solutions to the equation 1-y^2=y^4-y^2+y-1 ?
The solutions to the equation 1-y^2=y^4-y^2+y-1 are y=1,y\approx-1.35320…
Find the zeros of 1-y^2=y^4-y^2+y-1
The zeros of 1-y^2=y^4-y^2+y-1 are y=1,y\approx-1.35320…