How do you solve #(x + 4) ^ { 2} + 12( x + 4) + 35= 0#?

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Answer 1 · 2017-07-20T11:59:23

Solution : #x= -11 or x =-9#

# (x+4)^2 +12 (x+4) + 35 =0 # or

# (x+4)^2 +7 (x+4) + 5(x+4) + 35 =0 # or

#(x+4)(x+4+7) +5 (x+4+7) =0# or

#(x+4)(x+11) +5 (x+11) =0# or

#(x+11)(x+4+5)=0# or

#(x+11)(x+9)=0# , either #x= -11 or x =-9#

Solution : #x= -11 or x =-9# [Ans]