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

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Answer 1 · 2018-05-31T14:29:29

#x = 0# and #-16/3#

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.

Answer 2 · 2018-05-31T14:29:33

#x=0, x=-16/3#

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#