What are the solutions to the equation 16m^4+40m^2+25=0 ?

The solutions to the equation 16m^4+40m^2+25=0 are m=(sqrt(5))/2 i,m=-(sqrt(5))/2 i

Find the zeros of 16m^4+40m^2+25=0

The zeros of 16m^4+40m^2+25=0 are m=(sqrt(5))/2 i,m=-(sqrt(5))/2 i

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16 m^{4} + 40 m^{2} + 25 = 0

m = \frac{5}{2} i, m = - \frac{5}{2} i

Solution steps

16 m^{4} + 40 m^{2} + 25 = 0

Rewrite the equation with

u = m^{2}

and

u^{2} = m^{4}

16 u^{2} + 40 u + 25 = 0

Solve

16 u^{2} + 40 u + 25 = 0 : u = - \frac{5}{4}

u = - \frac{5}{4}

Substitute back

u = m^{2},

solve for

m

Solve

m^{2} = - \frac{5}{4} : m = \frac{5}{2} i, m = - \frac{5}{2} i

The solutions are

m = \frac{5}{2} i, m = - \frac{5}{2} i

1 = (1 + x)^{8}

x^{5} + 9 x = 0

x^{12} = 12

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

x, s i x^{3} = 125

What are the solutions to the equation 16m^4+40m^2+25=0 ?
The solutions to the equation 16m^4+40m^2+25=0 are m=(sqrt(5))/2 i,m=-(sqrt(5))/2 i
Find the zeros of 16m^4+40m^2+25=0
The zeros of 16m^4+40m^2+25=0 are m=(sqrt(5))/2 i,m=-(sqrt(5))/2 i