Step-by-step explanation:
k² + 5k + 13 = 0
Using the quadratic formula which is
[tex]x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ [/tex]
From the question
a = 1 , b = 5 , c = 13
So we have
[tex]k = \frac{ - 5 \pm \sqrt{ {5}^{2} - 4(1)(13) } }{2(1)} \\ = \frac{ - 5 \pm \sqrt{25 - 52} }{2} \\ = \frac{ - 5 \pm \sqrt{ - 27} }{2} \: \: \: \: \: \: \\ = \frac{ - 5 \pm3 \sqrt{3} \: i}{2} \: \: \: \: \: \: [/tex]
Separate the solutions
[tex]k_1 = \frac{ - 5 + 3 \sqrt{3} \: i }{2} \: \: \: \: or \\ k_2 = \frac{ - 5 - 3 \sqrt{3} \: i}{2} [/tex]
The equation has complex roots
Separate the real and imaginary parts
We have the final answer as
[tex]k_1 = - \frac{5}{2} + \frac{3 \sqrt{3} }{2} \: i \: \: \: \: or \\ k_2 = - \frac{5}{2} - \frac{3 \sqrt{3} }{2} \: i[/tex]
Hope this helps you
