What are the solutions to the equation 25t^4+22t^2-3=0 ?

The solutions to the equation 25t^4+22t^2-3=0 are t=(sqrt(3))/5 ,t=-(sqrt(3))/5 ,t=i,t=-i

Find the zeros of 25t^4+22t^2-3=0

The zeros of 25t^4+22t^2-3=0 are t=(sqrt(3))/5 ,t=-(sqrt(3))/5 ,t=i,t=-i

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25 t^{4} + 22 t^{2} - 3 = 0

t = \frac{3}{5}, t = - \frac{3}{5}, t = i, t = - i

Solution steps

25 t^{4} + 22 t^{2} - 3 = 0

Rewrite the equation with

u = t^{2}

and

u^{2} = t^{4}

25 u^{2} + 22 u - 3 = 0

Solve

25 u^{2} + 22 u - 3 = 0 : u = \frac{3}{25}, u = - 1

u = \frac{3}{25}, u = - 1

Substitute back

u = t^{2},

solve for

t

Solve

t^{2} = \frac{3}{25} : t = \frac{3}{5}, t = - \frac{3}{5}

Solve

t^{2} = - 1 : t = i, t = - i

The solutions are

t = \frac{3}{5}, t = - \frac{3}{5}, t = i, t = - i

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

z^{2} (1 - z^{2}) = 16

x + 8 x^{4} = 2 x^{2} - \frac{1}{6}

x, - f = - x^{3}

- 24 x - 24 = x^{3} + 8 x^{2}

What are the solutions to the equation 25t^4+22t^2-3=0 ?
The solutions to the equation 25t^4+22t^2-3=0 are t=(sqrt(3))/5 ,t=-(sqrt(3))/5 ,t=i,t=-i
Find the zeros of 25t^4+22t^2-3=0
The zeros of 25t^4+22t^2-3=0 are t=(sqrt(3))/5 ,t=-(sqrt(3))/5 ,t=i,t=-i