#color(blue)("First; in generic algebraic form")#
Standard form : #ax^2+bx+c=0#
Change to: #a(x^2+b/a) +c=0#
Note: in your question a=1
You determine #1/2 xx b/a# and include it in the form:
#a(x+b/(2a))^2 +c->0#
But this will accurately give you the #bx# term but will produce an error because #(b/(2a))# results in an #color(brown)("additional constant")# over and above #c#. To very simply get over this you #color(blue)("'remove' it. In other words subtract.")#
#color(green)(a(x+b/(2a))^2+c ->) ax^2 +bx color(brown)(+(b/(2a))^2 color(blue)(-(b/(2a))^2) +c#
So you completed square should look like:
#a(x+b/(2a))^2 color(blue)(-(b/(2a))^2)+c-=0#
Then your #color(green)(x_("vertex") = (-1)xx b/(2a))#
#color(blue)("~~~~~~~~~ The process of completing the square ~~~~~~~~~~~~~~~~~~~~~~~")#
#color(brown)("How does this play out with your question?")#
Write as: #color(brown)(x^2+3x+1=0)#
#underline(color(green)(Step 1))#
With the inherent error we have: #(x+3/2)^2+1 !=0#
#color(green)(x_("vertex")= 3/2= 1 1/2#
We can not equate this to zero as we have changed the left but not the right
#underline(color(green)(Step 2))#
#color(brown)(color(blue)("Corrected")" we have:"[(x+3/2)^2+1]color(blue)( - (3/2)^2) =0)#
#color(brown)(=>(x+3/2)^2 +1 - 9/4) color(green)(->(x+3/2)^2 -5/4=0)#
By means of the subtraction we have now returned the left back to its original value. So we can at this point quite correctly equate it to zero
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Quick way to find "x_("vertex")#
#color(red)("Shortcut: Write as"color(white)(.)a(x^2+b/a)+c=0" then just")#
#color(red)("halve the "b/a". This is exactly what completing the square")#
#color(red)("has done")#
