What are the solutions to the equation 2x^4-3x^2+10=0 ?

The solutions to the equation 2x^4-3x^2+10=0 are

Find the zeros of 2x^4-3x^2+10=0

The zeros of 2x^4-3x^2+10=0 are

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2 x^{4} - 3 x^{2} + 10 = 0

x = \frac{3 + 4 5}{2 2} + \frac{71}{2 2 3 + 4 5} i, x = - \frac{3 + 4 5}{2 2} - \frac{71}{2 2 3 + 4 5} i, x = \frac{3 + 4 5}{2 2} - \frac{71}{2 2 3 + 4 5} i, x = - \frac{3 + 4 5}{2 2} + \frac{71}{2 2 3 + 4 5} i

Solution steps

2 x^{4} - 3 x^{2} + 10 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

2 u^{2} - 3 u + 10 = 0

Solve

2 u^{2} - 3 u + 10 = 0 : u = \frac{3}{4} + i \frac{71}{4}, u = \frac{3}{4} - i \frac{71}{4}

u = \frac{3}{4} + i \frac{71}{4}, u = \frac{3}{4} - i \frac{71}{4}

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = \frac{3}{4} + i \frac{71}{4} : x = \frac{3 + 4 5}{2 2} + \frac{71}{2 2 3 + 4 5} i, x = - \frac{3 + 4 5}{2 2} - \frac{71}{2 2 3 + 4 5} i

Solve

x^{2} = \frac{3}{4} - i \frac{71}{4} : x = \frac{3 + 4 5}{2 2} - \frac{71}{2 2 3 + 4 5} i, x = - \frac{3 + 4 5}{2 2} + \frac{71}{2 2 3 + 4 5} i

The solutions are

x = \frac{3 + 4 5}{2 2} + \frac{71}{2 2 3 + 4 5} i, x = - \frac{3 + 4 5}{2 2} - \frac{71}{2 2 3 + 4 5} i, x = \frac{3 + 4 5}{2 2} - \frac{71}{2 2 3 + 4 5} i, x = - \frac{3 + 4 5}{2 2} + \frac{71}{2 2 3 + 4 5} i

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

x^{4} + 3 x^{3} - 5 x - 15 = 0

z, z^{2} (p^{2} z^{2} + q^{2}) = 1

(x^{2} - 6) (x^{2} - 1) = 0

2 x^{4} - 81 = 0

What are the solutions to the equation 2x^4-3x^2+10=0 ?
The solutions to the equation 2x^4-3x^2+10=0 are
Find the zeros of 2x^4-3x^2+10=0
The zeros of 2x^4-3x^2+10=0 are