What are the solutions to the equation (7k-1)^2=-125 ?

The solutions to the equation (7k-1)^2=-125 are k= 1/7+i(5sqrt(5))/7 ,k= 1/7-i(5sqrt(5))/7

Find the zeros of (7k-1)^2=-125

The zeros of (7k-1)^2=-125 are k= 1/7+i(5sqrt(5))/7 ,k= 1/7-i(5sqrt(5))/7

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(7 k - 1)^{2} = - 125

k = \frac{1}{7} + i \frac{5 5}{7}, k = \frac{1}{7} - i \frac{5 5}{7}

Solution steps

(7 k - 1)^{2} = - 125

For

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

the solutions are

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

Solve

7 k - 1 = - 125 : k = \frac{1}{7} + i \frac{5 5}{7}

Solve

7 k - 1 = - - 125 : k = \frac{1}{7} - i \frac{5 5}{7}

The solutions to the quadratic equation are:

k = \frac{1}{7} + i \frac{5 5}{7}, k = \frac{1}{7} - i \frac{5 5}{7}

4 x^{2} - 2 x + 15 = 0

2 x^{2} - 13 x = 7

(\frac{1}{2} + \frac{3}{2} i)^{6}

10 x^{2} - 20 x - 10 = 0

(- \frac{4}{7})^{4}

What are the solutions to the equation (7k-1)^2=-125 ?
The solutions to the equation (7k-1)^2=-125 are k= 1/7+i(5sqrt(5))/7 ,k= 1/7-i(5sqrt(5))/7
Find the zeros of (7k-1)^2=-125
The zeros of (7k-1)^2=-125 are k= 1/7+i(5sqrt(5))/7 ,k= 1/7-i(5sqrt(5))/7