SOLUTION: I am working on an assignment for "completing the square" in quadratic functions. The problem I'm working on is
9x^2-12x+4=-3
I am struggling with putting this into vertex f
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Question 805484: I am working on an assignment for "completing the square" in quadratic functions. The problem I'm working on is
9x^2-12x+4=-3
I am struggling with putting this into vertex form.
Thank you very much!
Found 2 solutions by josgarithmetic, Alan3354:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
Factor the 9 from :
Form the missing square term, ;
You must add to both sides of the equation, being careful to account for what is inside and what is outside the factored grouping from your previous factoring step.
Watch this step carefully:
Start simplifying.
If you want to solve for x, you can from that. "Vertex" form suggests you have a function but you did not show a given function, but a quadratic equation. Maybe ?
The vertex is (2/3, 3).
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I am working on an assignment for "completing the square" in quadratic functions. The problem I'm working on is
9x^2-12x+4=-3
------------------
This is a tougher than necessary equation to learn comleting the square, but...
9x^2-12x+4=-3
Step 1, get the coefficient of x^2 to be 1. Divide by 9
x^2 - (4/3)x + 4/9 = -1/3
Put the coefficients terms on the right
x^2 - (4/3)x = -1/3 - 4/9 = -7/9
Find the c term, where it's ax^2 + bx + c = ...
c = (b/2a)^2
Since a = 1, it's just (b/2)^2
b = -4/3
b/2 = -2/3
(b/2)^2 = 4/9
------------
Add c (= 4/9) to both sides
x^2 - (4/3)x = -7/9
x^2 - (4/3)x + 4/9 = -7/9 + 4/9 = -3/9
x^2 - (4/3)x + 4/9 = -3/9 = -1/3
Now the left side is a "perfect square."
Take the sqrt of both sides.
---------------
===========
Not a good teaching example, imo.
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