To solve 2x−3x2=−8 we will collect terms on one side. I prefer a positive in front of x2, so I'll collect them to make that happen:
2x−3x2=−8
0=3x2−2x−8
Now, of course, this equals that exactly when that equals this, so we can write:
3x2−2x−8=0
There are a couple of ways to fill in the details of completing the square, but discussing all possibilities is more confusing than helpful, so we'll go through it one way.
3x2−2x−8=0
On the left, we want a square like (x−a)2=x2−2ax+a2. Let's get that 8 out of our way:
3x2−2x=8
We don't want that 3 out front, so we'll multiply both sides by 13 (Don't forget to distribute on the left.)
13(3x2−2x)=13(8)
x2−23x=83
Cookbook: now take 12 of the number in front of x, square that and then add that square to both sides. (Why later, in the Note below.)
12⋅23=13
(13)2=19
Add 19 to both sides:
x2−23x+19=83+19
Now we can factor on the left -- remember the 13 we squared? That's what we need now:
(x−13)2=249+19
Notice that I also got a common denominator on the right so I can do the addition on the right:
(x−13)2=259
Now a2=n when a= either √n or −√n. So
x−13=±√259 Simplify the right, to get
x−13=±53 Add 13 to both sides:
x=13±53
Remember that this means there are two solutions.
One of them is 13+53=63=2
and the other is 13−53=−43=−43
Note
When we got
x2−23x=83 Why did we do what we did?
We want
x2−23x+something to be a perfect square like:
x2−2ax+a2
So the number in front of x is 2 times the thing I need to see the square of. That is 23=2a. To fins a, take 12 of the number in front of x.
