Question 86802
note: I'm going to use "x" instead of "a" as my variable

We can convert any quadratic {{{ax^2+bx+c}}} to standard vertex form {{{a(x-h)^2+k}}} by this procedure:

{{{y=(1)*x^2-(12)*x+27}}} Start with the given quadratic



{{{y-27=(1)*x^2-(12)*x}}} Subtract {{{27}}} from both sides


{{{y-27=(1)*(x^2-(12)*x)}}} Factor out the leading coefficient{{{1}}}
Now to complete the square on the right side we must take half of the x coefficient (in {{{ax^2+bx+c}}} its b) and square it  (i.e. {{{(b/2)^2}}})


{{{y-27=(1)*(x^2-(12)*x+36)}}} Take half of {{{-12}}} and square it (ie {{{((-12)(1/2))^2}}}). Add the result({{{36}}}) just inside the parenthesis.



This completes the square on the right side. So it goes from{{{(1)*(x^2-(12)*x+36)}}} and factors to {{{(1)*(x-6)^2}}} which is a perfect square


{{{y-27=(1)*(x-6)^2}}} Factor the right side into a perfect square

Since we added {{{36}}} inside the parenthesis, we really added {{{(1)(36)}}} to the entire right side (just distribute the leading coefficient {{{1}}} and you'll see it). So we must add {{{(1)(36)}}} to the other side to balance the equation.


{{{y-27+(1)*(36)=(1)*(x-6)^2}}} Add {{{(1)(36)}}} to the other side



{{{y-27+36=(1)*(x-6)^2}}} Multiply

{{{y-27+36=(1)*(x-6)^2}}} Reduce


{{{y+9=(1)*(x-6)^2}}} Combine like terms on the left side


{{{y+9=(1)*(x-6)^2}}} Reduce any fractions left side

{{{y=(1)(x-6)^2-9}}}Subtract {{{9}}} from both sides

So the quadratic {{{y=(1)*x^2-(12)*x+27}}} is completed to {{{y=(1)*(x-6)^2-9}}} which is now in vertex form (which is {{{a(x-h)^2+k}}}) where {{{a=1}}} (the stretch/compression factor), {{{h=6}}}(the x-coordinate of the vertex), and {{{k=-9}}} is the y coordinate of the vertex. So this means the vertex is <font size=6> ({{{6}}},{{{-9}}})</font>. Also, since the axis of symmetry is the vertical line through the vertex, the axis of symmetry is {{{x=6}}} (it is equal to the x-coordinate of the vertex).

Here are the graphs of original quadratic {{{y=(1)*x^2-(12)*x+27}}} and our answer in vertex form {{{y=(1)(x-6)^2-9}}}

{{{
drawing( 500, 500, -16, 16, -19, 19,
  graph( 500, 500, -16, 16, -19, 19, 1*x^2+-12*x+27),
  red( locate( 6, -9, "(6, -9)" ), circle( 6, -9, 0.213333333333333 ) ),
  graph( 500, 500, -16, 16, -19, 19, 1000000(x-6)) )
)
}}} graph of {{{y=(1)*x^2-(12)*x+27}}} with the vertex <font size=6> ({{{6}}},{{{-9}}})</font> and the axis of symmetry {{{x=6}}} (it is the vertical line through the vertex)


{{{
drawing( 500, 500, -16, 16, -19, 19,
  graph( 500, 500, -16, 16, -19, 19, 1(x-6)^2+-9),
  red( locate( 6, -9, "(6, -9)" ), circle( 6, -9, 0.213333333333333 ) ),
  graph( 500, 500, -16, 16, -19, 19, 1000000(x-6)) )
)
}}} graph of {{{y=(1)(x-6)^2-9}}} with the vertex <font size=6> ({{{6}}},{{{-9}}})</font> and the axis of symmetry {{{x=6}}} (it is the vertical line through the vertex)


Notice the two graphs are equivalent; this verifies our answer.


Now to solve for x, we simply need to isolate x: 

{{{0=(1)*(x-6)^2-9}}} Set y equal to zero to solve for x

{{{0+9=(1)*(x-6)^2}}}Add {{{9}}} to both sides

{{{0+-sqrt(9)=x-6}}} Take the square root of both sides

*[Tex \LARGE \pm3=x-6] Take the square root

*[Tex \LARGE 6\pm3=x] Add {{{6}}} to both sides


So it breaks down to this

{{{x=6-3}}} or {{{x=6+3}}}


So our solution is


{{{x=3}}} or {{{x=9}}}



Notice if you look back at the graph, you will see the roots {{{x=3}}} and {{{x=9}}}. This verifies our answer