How do you solve the equation #(2(x+3)^2)/3-4/9=1/3#?

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Answer 1 · 2017-10-26T12:40:40

# x = -4.0801 #
Or
#x= -1.9198#

#(2(x+3)^2)/3-4/9=1/3#

#=>(2(x+3)^2)/3=1/3+4/9#

#=>(2(x+3)^2)/3=1/3 xx3/3+4/9#

#=>(2(x+3)^2)/3=3/9+4/9#

#=>2/3xx(x+3)^2=7/9#

#=>(x+3)^2=7/9xx 3/2#

#=>(x+3)^2=7/cancel9^3xx cancel3^1/2#

#=>x^2 + 6x + 9 = 7/6#

#=>x^2 + 6x + 9 - 7/6 = 0#

#=>x^2 + 6x + (9xx6)/6 - 7/6 = 0#

#=>x^2 + 6x + (54 - 7)/6 = 0#

#x^2 + 6x + 47/6 = 0#

#6x^2 + 36x + 47 = 0#

Using quadratic formula :
# x=( -b +-sqrt(b^2 -4ac))/(2a)#

Where #a=6, b=36 and c =47#

We get the truncated approximate values of #x# as
# x = -4.0801 #
Or
#x= -1.9198#