Question 120255
<font size = 6 color = "red"><b>SOLUTION BY EDWIN:</b></font>

Bill used completing the square to find the zeroes of the function 
{{{y=9x^2-12x-33}}}. How is this done ?
<pre><font size = 4><b>
It's long! Sorry!

y = 9x² - 12x - 33

To find the zeros we substitute 0 for y and solve for x:

0 = 9x² - 12x - 33

Let's put the 0 on the right:

9x² - 12x - 33 = 0

1. Isolate the terms in x on the left:

That is, add 33 to both sides:

    9x² - 12x = 33

2. Divide every term through by the coefficient
   of x², since it is not 1.

    {{{9/9}}}x² - {{{12/9}}}x = {{{33/9}}}

Simplifying:

    x² - {{{4/3}}}x = {{{11/3}}}

3. To the side multiply the coefficient of x by {{{1/2}}}

     {{{(-4/3)}}}{{{(1/2)}}} = {{{-4/6}}} = {{{-2/3}}}

4.  Square the result of step 3:

     {{{(-2/3)^2}}} = {{{4/9}}}

5.  Add that to both sides of the equation we 
    had at the end of step 2:

    x² - {{{4/3}}}x + {{{4/9}}} = {{{11/3}}}+{{{4/9}}}

6.  Factor the left side:

    {x - {{{2/3}}})(x - {{{2/3}}}) = {{{11/3}}}+{{{4/9}}}

7.  Combine the terms on the right sides

    Write {{{11/3}}} as {{{33/9}}}

    {x - {{{2/3}}})(x - {{{2/3}}}) = {{{33/9}}}+{{{4/9}}}

    {x - {{{2/3}}})(x - {{{2/3}}}) = {{{37/9}}}

8. Write the left side as a square.  We always can.  That's
   why the method is called "completing the square".

    {x - {{{2/3}}})² = {{{37/9}}}
  
10. Use the principle of square roots:

       x - {{{2/3}}} = ±{{{sqrt(37)/3}}}

11. Solve for x.

                   x = {{{2/3}}} ± {{{sqrt(37)/3}}}

That can be written:

                   x = {{{(2 +-sqrt(37))/3}}}

Edwin</pre>