In order to solve #x^2 + 12x = 2# by completing the square, what value must be added to both sides?

1 Answer
Dec 13, 2015

#color(red)("Question interpretation 1")#

Given: #color(brown)(x^2+12x=2)#

#color(red)(underline("Add")) color(white)(.)color(blue)((-2))# to both sides:

#color(brown)(x^2+12x + color(blue)((-2))=2+color(blue)((-2)))#

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#color(red)("Question interpretation 2")#

Write as #color(blue)(x^2+12x-2=0) ..........................(1)#

Put a bracket round the first 2 terms giving:

#color(blue)((x^2+12x)-2=0)............................(2)#

Known:
#1/2xx(+12)=color(red)((+6))#

#sqrt(x^2)=color(green)(x)#

Using these value write equation (2) as

#(color(green)(x)color(red)(+6))^2-2=0..............................(3)#

We have now introduced an error. If we square the bracket you end up with:

#x^2+12xcolor(purple)(+36) -2 =0#

The #color(purple)(+36)# was not in the original equation so this is the error. Wee allow for this by mathematically removing it from equation (3) giving:

#{(x+6)^2-2} color(purple)(-36).....................(3_a)#

So the number that must be added is (-36)

#color(blue)("Another way of looking at "underline(color(purple)("add"))" is" color(red)(" 'put with'")color(white)(.) "so you" color(red)(" put (-36) with") " the equation")#

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#color(Green)("Technically: You add -2 to the right and -38 to the left!!!!")#