SOLUTION: Bill used completing the square to find the zeroes of the function {{{y=9x^2-12x-33}}}. How is this done ?

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Question 120255: Bill used completing the square to find the zeroes of the function y=9x%5E2-12x-33. How is this done ?
Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=9+x%5E2-12+x-33 Start with the given equation



y%2B33=9+x%5E2-12+x Add 33 to both sides



y%2B33=9%28x%5E2%2B%28-4%2F3%29x%29 Factor out the leading coefficient 9



Take half of the x coefficient -4%2F3 to get -2%2F3 (ie %281%2F2%29%28-4%2F3%29=-2%2F3).


Now square -2%2F3 to get 4%2F9 (ie %28-2%2F3%29%5E2=%28-2%2F3%29%28-2%2F3%29=4%2F9)





y%2B33=9%28x%5E2%2B%28-4%2F3%29x%2B4%2F9-4%2F9%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 4%2F9 does not change the equation




y%2B33=9%28%28x-2%2F3%29%5E2-4%2F9%29 Now factor x%5E2%2B%28-4%2F3%29x%2B4%2F9 to get %28x-2%2F3%29%5E2



y%2B33=9%28x-2%2F3%29%5E2-9%284%2F9%29 Distribute



y%2B33=9%28x-2%2F3%29%5E2-4 Multiply



y=9%28x-2%2F3%29%5E2-4-33 Now add %2B33 to both sides to isolate y



y=9%28x-2%2F3%29%5E2-37 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=9, h=2%2F3, and k=-37. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=9x%5E2-12x-33 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C9x%5E2-12x-33%29 Graph of y=9x%5E2-12x-33. Notice how the vertex is (2%2F3,-37).



Notice if we graph the final equation y=9%28x-2%2F3%29%5E2-37 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C9%28x-2%2F3%29%5E2-37%29 Graph of y=9%28x-2%2F3%29%5E2-37. Notice how the vertex is also (2%2F3,-37).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
SOLUTION BY EDWIN:
Bill used completing the square to find the zeroes of the function
y=9x%5E2-12x-33. How is this done ?

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%2F9x² - 12%2F9x = 33%2F9

Simplifying:

    x² - 4%2F3x = 11%2F3

3. To the side multiply the coefficient of x by 1%2F2

     %28-4%2F3%29%281%2F2%29 = -4%2F6 = -2%2F3

4.  Square the result of step 3:

     %28-2%2F3%29%5E2 = 4%2F9

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

    x² - 4%2F3x + 4%2F9 = 11%2F3+4%2F9

6.  Factor the left side:

    {x - 2%2F3)(x - 2%2F3) = 11%2F3+4%2F9

7.  Combine the terms on the right sides

    Write 11%2F3 as 33%2F9

    {x - 2%2F3)(x - 2%2F3) = 33%2F9+4%2F9

    {x - 2%2F3)(x - 2%2F3) = 37%2F9

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

    {x - 2%2F3)² = 37%2F9
  
10. Use the principle of square roots:

       x - 2%2F3 = ±sqrt%2837%29%2F3

11. Solve for x.

                   x = 2%2F3 ± sqrt%2837%29%2F3

That can be written:

                   x = %282+%2B-sqrt%2837%29%29%2F3

Edwin