What are the solutions to the equation -x^4+2x^2+8=0 ?

The solutions to the equation -x^4+2x^2+8=0 are x=sqrt(2)i,x=-sqrt(2)i,x=2,x=-2

Find the zeros of -x^4+2x^2+8=0

The zeros of -x^4+2x^2+8=0 are x=sqrt(2)i,x=-sqrt(2)i,x=2,x=-2

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- x^{4} + 2 x^{2} + 8 = 0

x = 2 i, x = - 2 i, x = 2, x = - 2

Solution steps

- x^{4} + 2 x^{2} + 8 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

- u^{2} + 2 u + 8 = 0

Solve

- u^{2} + 2 u + 8 = 0 : u = - 2, u = 4

u = - 2, u = 4

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = - 2 : x = 2 i, x = - 2 i

Solve

x^{2} = 4 : x = 2, x = - 2

The solutions are

x = 2 i, x = - 2 i, x = 2, x = - 2

x^{4} - 18 x^{3} + 103 x^{2} + 214 x + 192 = 0

x, x^{5} + y^{4} = 500

h, [h^{3} o] = 1.1 \cdot 1 0^{- 1} 3

x^{3} = - 216

10 x^{4} - 13 x^{3} + 15 x^{2} - 18 x - 24 = 0

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