Question

4x^2+64=0 solve by any method

Answer 1

For this case we have the following quadratic equation:

$4x ^ 2 + 64 = 0$

The solutions are given by:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}$

Where:

$a = 4\\b = 0\\c = 64$

Substituting we have:

$x = \frac {-0 \pm \sqrt {(0) ^ 2-4 (4) (64)}} {2 (4)}\\x = \frac {\pm \sqrt {0-1024}} {8}\\x = \frac {\pm \sqrt {-1024}} {8}$

We have to by definition:

$i ^ 2 = -1\\x = \frac {\pm \sqrt {1024i ^ 2}} {8}\\x = \frac {\pm i \sqrt {1024}} {8}\\x = \frac {\pm32i} {8}$

We have two roots:

$x_ {1} = + \frac {32i} {8} = 4i\\x_ {2} = - \frac {32i} {8} = - 4i$

Answer:

We have two complex roots:

$x_ {1} - = + 4i\\x_ {2} = - 4i$