For this case we have the following quadratic equation:
[tex]4x ^ 2 + 64 = 0[/tex]
The solutions are given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Where:
[tex]a = 4\\b = 0\\c = 64[/tex]
Substituting we have:
[tex]x = \frac {-0 \pm \sqrt {(0) ^ 2-4 (4) (64)}} {2 (4)}\\x = \frac {\pm \sqrt {0-1024}} {8}\\x = \frac {\pm \sqrt {-1024}} {8}[/tex]
We have to by definition:
[tex]i ^ 2 = -1\\x = \frac {\pm \sqrt {1024i ^ 2}} {8}\\x = \frac {\pm i \sqrt {1024}} {8}\\x = \frac {\pm32i} {8}[/tex]
We have two roots:
[tex]x_ {1} = + \frac {32i} {8} = 4i\\x_ {2} = - \frac {32i} {8} = - 4i[/tex]
Answer:
We have two complex roots:
[tex]x_ {1} - = + 4i\\x_ {2} = - 4i[/tex]
