How do you solve #3x^2+6x-2=0# by completing the square?

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Answer 1 · 2017-07-01T23:24:16

# x = -1 +-sqrt(5/3)#

We have:

# 3x^2+6x-2=0 #

The standard steps to complete the square on quadratic expression are as follows:

Step 1 - Factor out the quadratic coefficient, thus:

# 3x^2+6x-2=3{x^2+2x-2/3} #

Step 2 - Factor #1/2# of the #x# coefficient to form a perfect square, and subtract its square, and simplifiy, thus:

# 3x^2+6x-2 = 3{(x+2/2)^2-(2/2)^2-2/3} #
# " " = 3{(x+1)^2-1-2/3} #
# " " = 3{(x+1)^2-5/3} #

So now returning to the quadratic equation , we have;

# 3x^2+6x-2=0 #

# :. 3{(x+1)^2-5/3} = 0#
# :. (x+1)^2-5/3 = 0#
# :. (x+1)^2 = 5/3#
# :. x+1 = +-sqrt(5/3)#
# :. x = -1 +-sqrt(5/3)#