SOLUTION: John used completing the square to find the zeros of the function y=9x^2-12x-33. Can someone please show please show me the steps for solving this ???

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: John used completing the square to find the zeros of the function y=9x^2-12x-33. Can someone please show please show me the steps for solving this ???      Log On


   

Question 123486: John used completing the square to find the zeros of the function y=9x^2-12x-33. Can someone please show please show me the steps for solving this ???
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
John used completing the square to find the zeros of the function
y = 9x^2 - 12x - 33
:
Find the zeros:
9y^2 - 12x - 33 = 0
:
When completing the square, y^2 has to have a coefficient of 1,
divide each term by 9:
y^2 - 12%2F9x - 33%2F9 = 0
:
We are going to replace the 3rd term with a value that will make it a square
Add 33%2F9 to both sides:
y^2 - 12%2F9x + ____ = 33%2F9
:
We can determine the 3rd term as follows:
Divide the coefficient of x by 2; that would be 6%2F9
Square this value; that would be 36%2F81 reduced to 4%2F9
:
y^2 - 12%2F9x + 4%2F9 = 33%2F9 + 4%2F9; we have to add the same amt to both sides to keep an equation
:
y^2 - 12%2F9x + 4%2F9 = 37%2F9;
Now it is a square, and can be factored to:
(y - 2%2F3) (y - 2%2F3) = 37%2F9
or
(y - 2%2F3)^2 = 37%2F9
:
Find the square root of both sides and we have;
y - 2%2F3 = +/-sqrt%2837%2F9%29%29
:
y = +2%2F3 +/-sqrt%2837%2F9%29%29 add 2/3 to both sides
We can factor 1/3 out of the radical, gets rid of the denominator inside the radical
y = 2%2F3 +/-%281%2F3%29sqrt%2837%29%29
or
y = %282+%2B+sqrt%2837%29%29%2F3
and
y = %282+-+sqrt%2837%29%29%2F3
:
There are a lot of steps here. I hope you will take the time to study each one
If you have any questions, you can email me. Ankor