How do you solve the equation #3x^2+6x=2x^2+3x-4# by completing the square?

1 Answer
Jul 10, 2017

Answer:

# x = -3/2 -sqrt(7)/2 \ i, -3/2 +sqrt(7)/2 \ i #

Explanation:

We have:

# 3x^2+6x = 2x^2+3x-4 #

First let us collect terms on the LHS

# 3x^2 - 2x^2 + 6x - 3x -4 = 0 #
# :. x^2 + 3x +4 = 0 #

Now we complete the square, by forming a term with half the "#b#" value; that is

# :. (x + 3/2)^2 - (3/2)^2 +4 = 0 #
# :. (x + 3/2)^2 - 9/4 +4 = 0 #

Which we can now re-arrange and solve;

# :. (x + 3/2)^2 = -7/4 #

# :. x + 3/2 = +-sqrt(7)/2 \ i#

# :. x = -3/2 +-sqrt(7)/2 \ i#

Leading to the two complex conjugate solutions:

# x = -3/2 -sqrt(7)/2 \ i, -3/2 +sqrt(7)/2 \ i #

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