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

2 Answers
May 31, 2018

x = 0 and -16/3

Explanation:

As per the question, we have

(3x + 8)^2 - 64 = 0

:.(3x)^2 + (8)^2 + 2(3x)(8) - 64 = 0

:. 9x^2 + 64 + 48x - 64 = 0

:.9x^2 + 48x + cancel64 cancel(- 64) = 0

:. 9x^2 + 48x = 0

:.3(3x^2 + 16x) = 0

:. 3x^2 + 16x = 0/3

:. x(3x + 16) = 0

:. x = 0, -16/3

Hence, the answer.

May 31, 2018

x=0, x=-16/3

Explanation:

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

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

and if x=-8
3x=-8-8
3x=-16
thus
x=-16/3