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`