What are the solutions to the equation 16y^4-55y^2-36=0 ?

The solutions to the equation 16y^4-55y^2-36=0 are y=2,y=-2,y= 3/4 i,y=-3/4 i

Find the zeros of 16y^4-55y^2-36=0

The zeros of 16y^4-55y^2-36=0 are y=2,y=-2,y= 3/4 i,y=-3/4 i

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16 y^{4} - 55 y^{2} - 36 = 0

y = 2, y = - 2, y = \frac{3}{4} i, y = - \frac{3}{4} i

Solution steps

16 y^{4} - 55 y^{2} - 36 = 0

Rewrite the equation with

u = y^{2}

and

u^{2} = y^{4}

16 u^{2} - 55 u - 36 = 0

Solve

16 u^{2} - 55 u - 36 = 0 : u = 4, u = - \frac{9}{16}

u = 4, u = - \frac{9}{16}

Substitute back

u = y^{2},

solve for

y

Solve

y^{2} = 4 : y = 2, y = - 2

Solve

y^{2} = - \frac{9}{16} : y = \frac{3}{4} i, y = - \frac{3}{4} i

The solutions are

y = 2, y = - 2, y = \frac{3}{4} i, y = - \frac{3}{4} i

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

x^{4} = 1

x^{3} + x + 1 = 0

x^{3} = - 1

x^{4} - 2 x^{2} = - 1

What are the solutions to the equation 16y^4-55y^2-36=0 ?
The solutions to the equation 16y^4-55y^2-36=0 are y=2,y=-2,y= 3/4 i,y=-3/4 i
Find the zeros of 16y^4-55y^2-36=0
The zeros of 16y^4-55y^2-36=0 are y=2,y=-2,y= 3/4 i,y=-3/4 i