How do you solve #4x^2 +12x=29# by completing the square?

1 Answer
Jan 25, 2017

Answer:

Please see the explanation.

Explanation:

Divide both sides by 4:

#x^2 + 3x = 29/4#

Add #a^2# to both sides:

#x^2 + 3x + a^2= 29/4 + a^2" [1]"#

Using the pattern #(x+a)^2 = x^2 + 2ax + a^2#, set the middle term of the pattern equal to the middle term on the left of equation [1]:

#2ax = 3x#

Divide both sides of the equation 2x:

#a = 3/2#

Substitute #3/2# for "a" on both sides of equation [1]:

#x^2 + 3x + (3/2)^2= 29/4 + (3/2)^2" [2]"#

We know that the left side of equation [2] is a perfect square, therefore, we can substitute the left side of the pattern with #a = 3/2# into equation [2]:

#(x + 3/2)^2= 29/4 + (3/2)^2" [3]"#

Simplify the right side of equation [3]:

#(x + 3/2)^2= 38/4" [4]"#

Use the square root on both sides:

#x + 3/2= +-sqrt38/2" [5]"#

Subtract #3/2# form both sides:

#x = -3/2 +-sqrt38/2" [6]"#

Split equation [6] into two equations:

#x = -(3 +sqrt38)/2 and x = -(3 -sqrt38)/2#