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

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

Find the zeros of 4x^4+4x^2+1=0

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

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

x = \frac{1}{2} i, x = - \frac{1}{2} i

Solution steps

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

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

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

Solve

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

u = - \frac{1}{2}

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = - \frac{1}{2} : x = \frac{1}{2} i, x = - \frac{1}{2} i

The solutions are

x = \frac{1}{2} i, x = - \frac{1}{2} i

2 (x^{2} - 9)^{2} (2 x - 1) = 0

x, y^{3} = C x^{4} + 2

4 x^{4} - 40 x^{2} + 36 = 0

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

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

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