How do you complete the square to solve #0=9x^2 + 6x - 8#?

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Answer 1 · 2015-05-23T22:28:46

#0 = 9x^2+6x-8 = 9x^2+6x+1-1-8#

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

Add #9# to both ends to get:

#(3x+1)^2 = 9#

So

#3x+1 = +-sqrt(9) = +-3#

Subtract #1# from both sides to get:

#3x = -1+-3#

Divide both sides by #3# to get:

#x=-1/3+-1#

That is #x = -4/3# or #x = 2/3#

In general #ax^2+bx+c = a(x+b/(2a))^2+(c-b^2/(4a))#