How do you solve 4x^2 +12x=294x2+12x=29 by completing the square?

1 Answer
Jan 25, 2017

Please see the explanation.

Explanation:

Divide both sides by 4:

x^2 + 3x = 29/4x2+3x=294

Add a^2a2 to both sides:

x^2 + 3x + a^2= 29/4 + a^2" [1]"x2+3x+a2=294+a2 [1]

Using the pattern (x+a)^2 = x^2 + 2ax + a^2(x+a)2=x2+2ax+a2, set the middle term of the pattern equal to the middle term on the left of equation [1]:

2ax = 3x2ax=3x

Divide both sides of the equation 2x:

a = 3/2a=32

Substitute 3/232 for "a" on both sides of equation [1]:

x^2 + 3x + (3/2)^2= 29/4 + (3/2)^2" [2]"x2+3x+(32)2=294+(32)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 = 3/2a=32 into equation [2]:

(x + 3/2)^2= 29/4 + (3/2)^2" [3]"(x+32)2=294+(32)2 [3]

Simplify the right side of equation [3]:

(x + 3/2)^2= 38/4" [4]"(x+32)2=384 [4]

Use the square root on both sides:

x + 3/2= +-sqrt38/2" [5]"x+32=±382 [5]

Subtract 3/232 form both sides:

x = -3/2 +-sqrt38/2" [6]"x=32±382 [6]

Split equation [6] into two equations:

x = -(3 +sqrt38)/2 and x = -(3 -sqrt38)/2x=3+382andx=3382