What are the solutions to the equation 36t^4+29t^2-7=0 ?

The solutions to the equation 36t^4+29t^2-7=0 are t=(sqrt(7))/6 ,t=-(sqrt(7))/6 ,t=i,t=-i

Find the zeros of 36t^4+29t^2-7=0

The zeros of 36t^4+29t^2-7=0 are t=(sqrt(7))/6 ,t=-(sqrt(7))/6 ,t=i,t=-i

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36 t^{4} + 29 t^{2} - 7 = 0

t = \frac{7}{6}, t = - \frac{7}{6}, t = i, t = - i

Solution steps

36 t^{4} + 29 t^{2} - 7 = 0

Rewrite the equation with

u = t^{2}

and

u^{2} = t^{4}

36 u^{2} + 29 u - 7 = 0

Solve

36 u^{2} + 29 u - 7 = 0 : u = \frac{7}{36}, u = - 1

u = \frac{7}{36}, u = - 1

Substitute back

u = t^{2},

solve for

t

Solve

t^{2} = \frac{7}{36} : t = \frac{7}{6}, t = - \frac{7}{6}

Solve

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

The solutions are

t = \frac{7}{6}, t = - \frac{7}{6}, t = i, t = - i

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

x^{5} = - 1

x, x^{3} = y^{3}

x^{8} - 16 = 15 x^{4}

x^{3} = 25

What are the solutions to the equation 36t^4+29t^2-7=0 ?
The solutions to the equation 36t^4+29t^2-7=0 are t=(sqrt(7))/6 ,t=-(sqrt(7))/6 ,t=i,t=-i
Find the zeros of 36t^4+29t^2-7=0
The zeros of 36t^4+29t^2-7=0 are t=(sqrt(7))/6 ,t=-(sqrt(7))/6 ,t=i,t=-i