What are the solutions to the equation 2x^2-4x^4=5 ?

The solutions to the equation 2x^2-4x^4=5 are

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2x^2-4x^4=5

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

x = \frac{1 + 2 5}{2 2} - \frac{19}{2 2 1 + 2 5} i, x = - \frac{1 + 2 5}{2 2} + \frac{19}{2 2 1 + 2 5} i, x = \frac{1 + 2 5}{2 2} + \frac{19}{2 2 1 + 2 5} i, x = - \frac{1 + 2 5}{2 2} - \frac{19}{2 2 1 + 2 5} i

Solution steps

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

Move

5

to the left side

2 x^{2} - 4 x^{4} - 5 = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

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

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

- 4 u^{2} + 2 u - 5 = 0

Solve

- 4 u^{2} + 2 u - 5 = 0 : u = \frac{1}{4} - i \frac{19}{4}, u = \frac{1}{4} + i \frac{19}{4}

u = \frac{1}{4} - i \frac{19}{4}, u = \frac{1}{4} + i \frac{19}{4}

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = \frac{1}{4} - i \frac{19}{4} : x = \frac{1 + 2 5}{2 2} - \frac{19}{2 2 1 + 2 5} i, x = - \frac{1 + 2 5}{2 2} + \frac{19}{2 2 1 + 2 5} i

Solve

x^{2} = \frac{1}{4} + i \frac{19}{4} : x = \frac{1 + 2 5}{2 2} + \frac{19}{2 2 1 + 2 5} i, x = - \frac{1 + 2 5}{2 2} - \frac{19}{2 2 1 + 2 5} i

The solutions are

x = \frac{1 + 2 5}{2 2} - \frac{19}{2 2 1 + 2 5} i, x = - \frac{1 + 2 5}{2 2} + \frac{19}{2 2 1 + 2 5} i, x = \frac{1 + 2 5}{2 2} + \frac{19}{2 2 1 + 2 5} i, x = - \frac{1 + 2 5}{2 2} - \frac{19}{2 2 1 + 2 5} i

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

- 1 + v^{3} = - 3

x (x + 3) (x - 2) = 0

x^{3} + 6 x^{2} - 36 x + 2 = 0

10 (2 x - 1)^{4} = 0

What are the solutions to the equation 2x^2-4x^4=5 ?
The solutions to the equation 2x^2-4x^4=5 are
Find the zeros of 2x^2-4x^4=5
The zeros of 2x^2-4x^4=5 are