What are the solutions to the equation 144x^4+24x^2+1=0 ?

The solutions to the equation 144x^4+24x^2+1=0 are x= 1/(2sqrt(3))i,x=-1/(2sqrt(3))i

Find the zeros of 144x^4+24x^2+1=0

The zeros of 144x^4+24x^2+1=0 are x= 1/(2sqrt(3))i,x=-1/(2sqrt(3))i

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144 x^{4} + 24 x^{2} + 1 = 0

x = \frac{1}{2 3} i, x = - \frac{1}{2 3} i

Solution steps

144 x^{4} + 24 x^{2} + 1 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

144 u^{2} + 24 u + 1 = 0

Solve

144 u^{2} + 24 u + 1 = 0 : u = - \frac{1}{12}

u = - \frac{1}{12}

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = - \frac{1}{12} : x = \frac{1}{2 3} i, x = - \frac{1}{2 3} i

The solutions are

x = \frac{1}{2 3} i, x = - \frac{1}{2 3} i

77 = 16 \cdot (1 + r)^{4}

- 9 = t^{8} - 12

x^{3} - x^{2} - 11 x + 3 = 0

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

(x - 2)^{3} = 4

What are the solutions to the equation 144x^4+24x^2+1=0 ?
The solutions to the equation 144x^4+24x^2+1=0 are x= 1/(2sqrt(3))i,x=-1/(2sqrt(3))i
Find the zeros of 144x^4+24x^2+1=0
The zeros of 144x^4+24x^2+1=0 are x= 1/(2sqrt(3))i,x=-1/(2sqrt(3))i