What are the solutions to the equation 3x^4+2x^2-8=0 ?

The solutions to the equation 3x^4+2x^2-8=0 are x= 2/(sqrt(3)),x=-2/(sqrt(3)),x=sqrt(2)i,x=-sqrt(2)i

Find the zeros of 3x^4+2x^2-8=0

The zeros of 3x^4+2x^2-8=0 are x= 2/(sqrt(3)),x=-2/(sqrt(3)),x=sqrt(2)i,x=-sqrt(2)i

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

x = \frac{2}{3}, x = - \frac{2}{3}, x = 2 i, x = - 2 i

Solution steps

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

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

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

Solve

3 u^{2} + 2 u - 8 = 0 : u = \frac{4}{3}, u = - 2

u = \frac{4}{3}, u = - 2

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = \frac{4}{3} : x = \frac{2}{3}, x = - \frac{2}{3}

Solve

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

The solutions are

x = \frac{2}{3}, x = - \frac{2}{3}, x = 2 i, x = - 2 i

x, x^{3} + y^{3} = a^{3}

x^{4} + x^{3} + x^{2} + x = - 1

z^{3} + 32 + 32 i 3 = 0

81 = 1. 5^{2} \times x^{9}

x^{4} + 5 x^{3} + 5 x^{2} - 5 x - 6 = 0

What are the solutions to the equation 3x^4+2x^2-8=0 ?
The solutions to the equation 3x^4+2x^2-8=0 are x= 2/(sqrt(3)),x=-2/(sqrt(3)),x=sqrt(2)i,x=-sqrt(2)i
Find the zeros of 3x^4+2x^2-8=0
The zeros of 3x^4+2x^2-8=0 are x= 2/(sqrt(3)),x=-2/(sqrt(3)),x=sqrt(2)i,x=-sqrt(2)i