What are the solutions to the equation 216=1r^{4-1} ?

The solutions to the equation 216=1r^{4-1} are r=6,r=-3+3sqrt(3)i,r=-3-3sqrt(3)i

Find the zeros of 216=1r^{4-1}

The zeros of 216=1r^{4-1} are r=6,r=-3+3sqrt(3)i,r=-3-3sqrt(3)i

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216 = 1 r^{4 - 1}

r = 6, r = - 3 + 33 i, r = - 3 - 33 i

Solution steps

216 = 1 \cdot r^{4 - 1}

Switch sides

1 \cdot r^{4 - 1} = 216

Multiply:

1 \cdot r^{4 - 1} = r^{4 - 1}

r^{4 - 1} = 216

For

x^{3} = f (a)

the solutions are

Simplify

r = 6, r = - 3 + 33 i, r = - 3 - 33 i

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

27 y^{3} - 8 = 0

(49 x^{2} - 4) (x^{2} - 7) = 0

x^{2} + x^{3} - 5 x = 0

x^{5} + x^{3} + x = 0

What are the solutions to the equation 216=1r^{4-1} ?
The solutions to the equation 216=1r^{4-1} are r=6,r=-3+3sqrt(3)i,r=-3-3sqrt(3)i
Find the zeros of 216=1r^{4-1}
The zeros of 216=1r^{4-1} are r=6,r=-3+3sqrt(3)i,r=-3-3sqrt(3)i