What are the solutions to the equation 4z^4+2z^2+1=0 ?

The solutions to the equation 4z^4+2z^2+1=0 are

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4z^4+2z^2+1=0

4 z^{4} + 2 z^{2} + 1 = 0

z = \frac{1}{2 2} + \frac{3}{2 2} i, z = - \frac{1}{2 2} - \frac{3}{2 2} i, z = \frac{1}{2 2} - \frac{3}{4} 2 i, z = - \frac{1}{2 2} + \frac{3}{4} 2 i

Solution steps

4 z^{4} + 2 z^{2} + 1 = 0

Rewrite the equation with

u = z^{2}

and

u^{2} = z^{4}

4 u^{2} + 2 u + 1 = 0

Solve

4 u^{2} + 2 u + 1 = 0 : u = - \frac{1}{4} + i \frac{3}{4}, u = - \frac{1}{4} - i \frac{3}{4}

u = - \frac{1}{4} + i \frac{3}{4}, u = - \frac{1}{4} - i \frac{3}{4}

Substitute back

u = z^{2},

solve for

z

Solve

z^{2} = - \frac{1}{4} + i \frac{3}{4} : z = \frac{1}{2 2} + \frac{3}{2 2} i, z = - \frac{1}{2 2} - \frac{3}{2 2} i

Solve

z^{2} = - \frac{1}{4} - i \frac{3}{4} : z = \frac{1}{2 2} - \frac{3}{4} 2 i, z = - \frac{1}{2 2} + \frac{3}{4} 2 i

The solutions are

z = \frac{1}{2 2} + \frac{3}{2 2} i, z = - \frac{1}{2 2} - \frac{3}{2 2} i, z = \frac{1}{2 2} - \frac{3}{4} 2 i, z = - \frac{1}{2 2} + \frac{3}{4} 2 i

x^{3} - 3 x - 18 = 0

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

y, x = y^{5} - 2

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

24 x^{4} - 2 x^{2} - 1 = 0

What are the solutions to the equation 4z^4+2z^2+1=0 ?
The solutions to the equation 4z^4+2z^2+1=0 are
Find the zeros of 4z^4+2z^2+1=0
The zeros of 4z^4+2z^2+1=0 are