SOLUTION: solve by completing the square. x^2+4x-12=0

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: solve by completing the square. x^2+4x-12=0      Log On


   

Question 453794: solve by completing the square.
x^2+4x-12=0
Found 3 solutions by mananth, Alan3354, ikleyn:
Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+4x-12=0
x^2+4x=12
x^2+4x+4=12+4
(x+2)^2=16
take the square root
x+2=4
x=2

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+4x-12=0
x^2+4x = 12
x^2+4x+4 = 12+4
(x+2)^2=16
x+2= +/-4
x = -6, 2

Answer by ikleyn(53570) About Me  (Show Source):
You can put this solution on YOUR website!
.
Solve by completing the square.
x^2+4x-12=0
~~~~~~~~~~~~~~~~~~~~~~~~


        In his post, @mananth found one root, but missed/lost another root.
        So, his solution is incomplete.
        I came to bring a complete solution. 


x^2 + 4x - 12 = 0

x^2 + 4x = 12

x^2 + 4x + 4 = 12 + 4

(x+2)^2 = 16

x + 2 = +/- sqrt%2816%29

x + 2 = +/- 4


Case (a).  x + 2 = 4  --->  x = 4 - 2 = 2.


Vase (b).  x + 2 = -4 --->  x = -4 - 2 = -6.


Thus, equation  x^2 + 4x - 12 = 0  has two roots:  x = 2  and  x = -6.


Substitute these values into the equation to make sure that they do satisfy this equation.

Solved correctly and completely.