How do you solve ( 3x + 8) ^ { 2} - 64= 0?

May 31, 2018

$x = 0$ and $- \frac{16}{3}$

Explanation:

As per the question, we have

${\left(3 x + 8\right)}^{2} - 64 = 0$

$\therefore {\left(3 x\right)}^{2} + {\left(8\right)}^{2} + 2 \left(3 x\right) \left(8\right) - 64 = 0$

$\therefore 9 {x}^{2} + 64 + 48 x - 64 = 0$

$\therefore 9 {x}^{2} + 48 x + \cancel{64} \cancel{- 64} = 0$

$\therefore 9 {x}^{2} + 48 x = 0$

$\therefore 3 \left(3 {x}^{2} + 16 x\right) = 0$

$\therefore 3 {x}^{2} + 16 x = \frac{0}{3}$

$\therefore x \left(3 x + 16\right) = 0$

$\therefore x = 0 , - \frac{16}{3}$

May 31, 2018

$x = 0 , x = - \frac{16}{3}$

Explanation:

Take 64 to the right, square root both sides then you'll have $3 x + 8 = \pm 8$
...therefore solving for that, $x = 0$

you gonna have 2 equations
if $x = 8$
$x = 0$

and if $x = - 8$
$3 x = - 8 - 8$
$3 x = - 16$
thus
$x = - \frac{16}{3}$