What are the solutions to the equation (2k-1)^2=-27 ?

The solutions to the equation (2k-1)^2=-27 are k= 1/2+i(3sqrt(3))/2 ,k= 1/2-i(3sqrt(3))/2

Find the zeros of (2k-1)^2=-27

The zeros of (2k-1)^2=-27 are k= 1/2+i(3sqrt(3))/2 ,k= 1/2-i(3sqrt(3))/2

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(2 k - 1)^{2} = - 27

k = \frac{1}{2} + i \frac{3 3}{2}, k = \frac{1}{2} - i \frac{3 3}{2}

Solution steps

(2 k - 1)^{2} = - 27

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

2 k - 1 = - 27 : k = \frac{1}{2} + i \frac{3 3}{2}

Solve

2 k - 1 = - - 27 : k = \frac{1}{2} - i \frac{3 3}{2}

The solutions to the quadratic equation are:

k = \frac{1}{2} + i \frac{3 3}{2}, k = \frac{1}{2} - i \frac{3 3}{2}

y, x = 4 - y^{2}

0 = (x - 1) (x + 1)

w^{2} + 4 w - 96 = 0

16 = (x - 4) (x + 2)

t^{2} - 8 t - 20 = 0

What are the solutions to the equation (2k-1)^2=-27 ?
The solutions to the equation (2k-1)^2=-27 are k= 1/2+i(3sqrt(3))/2 ,k= 1/2-i(3sqrt(3))/2
Find the zeros of (2k-1)^2=-27
The zeros of (2k-1)^2=-27 are k= 1/2+i(3sqrt(3))/2 ,k= 1/2-i(3sqrt(3))/2