What are the solutions to the equation 2x^4-x^2-6=0 ?

The solutions to the equation 2x^4-x^2-6=0 are x=sqrt(2),x=-sqrt(2),x=sqrt(3/2)i,x=-sqrt(3/2)i

Find the zeros of 2x^4-x^2-6=0

The zeros of 2x^4-x^2-6=0 are x=sqrt(2),x=-sqrt(2),x=sqrt(3/2)i,x=-sqrt(3/2)i

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2 x^{4} - x^{2} - 6 = 0

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

Solution steps

2 x^{4} - x^{2} - 6 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

2 u^{2} - u - 6 = 0

Solve

2 u^{2} - u - 6 = 0 : u = 2, u = - \frac{3}{2}

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

Substitute back

u = x^{2},

solve for

x

Solve

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

Solve

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

The solutions are

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

(x^{2} + 2 x - 14)^{5} = 1

m^{3} - 2 m + 1 = 0

\frac{z ^{3}}{2} = 8 i

x, a = (- \frac{1}{2} \cdot x^{2} \cdot y^{3})^{1} \cdot (- x \cdot y^{2})

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

What are the solutions to the equation 2x^4-x^2-6=0 ?
The solutions to the equation 2x^4-x^2-6=0 are x=sqrt(2),x=-sqrt(2),x=sqrt(3/2)i,x=-sqrt(3/2)i
Find the zeros of 2x^4-x^2-6=0
The zeros of 2x^4-x^2-6=0 are x=sqrt(2),x=-sqrt(2),x=sqrt(3/2)i,x=-sqrt(3/2)i