How do you solve using the completing the square method #x^2 + 12x + 20 = 0#?

1 Answer
Mar 27, 2016

Answer:

For an in depth explanation of method see
http://socratic.org/s/at7MQHg2

#color(blue)(x~~-1.101" "x~~-10.899)#

Explanation:

Using Shortcuts

Given:#" "y=x^2+12x+12#

Let #k# be the error adjustment constant

#y=x^2+12x+12+k#

Step 1:#" "y=(x^color(magenta)(2)+12x)+12+k#

Step 2:#" "y=(x+color(green)(12)x)^(color(magenta)(2))+12+k#

Step 3:#" use "(1/2)xx(color(green)(12))=color(magenta)(6)#

#=> y=(xcolor(magenta)(+6))^2+12+k#

Step 4:#" "color(magenta)((+6))^2+k=0#

#=>k=color(green)(-36)#

#=>y=(x+6)^2+12-36#

#=>y=(x+6)^2-24#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~

At #y=0"; " x=+- sqrt(24)-6#

#color(blue)(x~~-1.101" "x~~-10.899)#

#color(magenta)("Notice that plots for both equations match")#
Tony B