What are the solutions to the equation 16x^4+24x^2+9=0 ?

The solutions to the equation 16x^4+24x^2+9=0 are x=(sqrt(3))/2 i,x=-(sqrt(3))/2 i

Find the zeros of 16x^4+24x^2+9=0

The zeros of 16x^4+24x^2+9=0 are x=(sqrt(3))/2 i,x=-(sqrt(3))/2 i

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16 x^{4} + 24 x^{2} + 9 = 0

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

Solution steps

16 x^{4} + 24 x^{2} + 9 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

16 u^{2} + 24 u + 9 = 0

Solve

16 u^{2} + 24 u + 9 = 0 : u = - \frac{3}{4}

u = - \frac{3}{4}

Substitute back

u = x^{2},

solve for

x

Solve

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

The solutions are

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

w^{3} = - 2 - 23 i

r, v = \frac{4}{3} π r^{3}

2 x - 13 = - x^{4} + 6

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

3 x^{4} - 48 x^{2} = 0

What are the solutions to the equation 16x^4+24x^2+9=0 ?
The solutions to the equation 16x^4+24x^2+9=0 are x=(sqrt(3))/2 i,x=-(sqrt(3))/2 i
Find the zeros of 16x^4+24x^2+9=0
The zeros of 16x^4+24x^2+9=0 are x=(sqrt(3))/2 i,x=-(sqrt(3))/2 i