3x2 -6x - 4 =0 how to complete the square?

1 Answer
LucaGasparro
Feb 16, 2018

See below...

Explanation:

We have the quadratic #3x^2-6x-4=0#

First of all, we take out a factor of 3. Don't take it out from the constant however as it can lead to some unnecessary fraction work.

#3x^2-6x-4 => 3[x^2-2x]-4#

Now we write out our initial bracket. To do this we have

#(x+b/2)^2# #=># in this case #b# is #-2#. Note that we don't include an #x# after the #b#...
Once we have our initial bracket, we subtract the square of #b/2#

#therefore 3[x^2-2x]-4 => 3[(x-1)^2 -1]-4#

Now we must remove the square brackets by multiplying whats in it by the factor on the outside, in this case #3#.

#therefore# we get # 3(x-1)^2 -3-4 = 3(x-1)^2 -7#

Final answer #3(x-1)^2 - 7 =0#

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