Thread: Confusion on completing the square

1. Confusion on completing the square

Hi, I've learned two methods of completing the square. One from my math teacher and from a website. The website is: (Completing the Square: Solving Quadratic Equations)

Anyways, I'll post an example problem and try to solve it via both methods, the problem is the answers come out differently.

Teachers method:

0 = x²-4x+3
0 = (x²-4x+4)+3-4
0 = (x-2)² -1

x = 2, -1

Website method:

x²-4x+3 = 0
x²-4x = -3
(x²-4x+4) = -3 + 4
(x-2)² = 1
√(x-2)² = +-√1
x-2 = +-1
x = 2 +- 1

x= 2+1 = 3
x= 2-1 = 1

So x = 3, 1

Maybe I'm missing something important here? If not, it seems the teachers method is more credible because they are a teacher, also, the second method requires me to square root both sides near the end, and if the right side happens to be negative the problem is botched, no?

Any help?

2. Hello,
Originally Posted by cb220
Hi, I've learned two methods of completing the square. One from my math teacher and from a website. The website is: (Completing the Square: Solving Quadratic Equations)

Anyways, I'll post an example problem and try to solve it via both methods, the problem is the answers come out differently.

Teachers method:

0 = x²-4x+3
0 = (x²-4x+4)+3-4
0 = (x-2)² -1

x = 3, -1

Website method:

x²-4x+3 = 0
x²-4x = -3
(x²-4x+4) = -3 + 4
(x-2)² = 1
√(x-2)² = +-√1
x-2 = +-1
x = 2 +- 1

x= 2+1 = 3
x= 2-1 = 1

So x = 3, 1

Maybe I'm missing something important here? If not, it seems the teachers method is more credible because they are a teacher, also, the second method requires me to square root both sides near the end, and if the right side happens to be negative the problem is botched, no?

Any help?
If you continue your teachers' working :
$0=(x-2)^2-1 \implies 1=(x-2)^2$

And square root the equation, this will give the same thing. No matter on which side 1 is, it's the same, you choose what you do by intuition.

Now, a slight difference. Your teachers' method can apply to any case, that is to say if you have for example $\frac{x^2-4x+3}{x-1}$, your teachers' method will better fit in it.

Or another *slight* difference, is if you're asked to factorize x²-4x+3, you'll use your teachers' steps.

3. I can see the similarities now but still unclear on a couple of things.

Take this with a grain of salt, being that it was the first day of class and I may have mistook what he said, but my understanding (and his answers on the board) reflect that if I take the final product of the Teachers Method answer: 0 = (x-2)² -1 I can take the -2 out of the brackets, which will make it x= +2 and the -1 will remain the same making one x = -1. Does it help that this equation is to find the vertex of a parabola? Or have you never seen something like that done before? The square rooting makes more sense to me logically but the answer that he arrived at on the board for the example of his method was V(2, -1).

4. $0 = (x-2)^2-1$

This refers to the x-intercepts of the parabola $y = (x-2)^2 - 1$. You will see that x = 2 will not work since: $(2-2)^2 - 1 = -1 \: \neq \: 0$

I'm not sure what you mean by taking the 2 out of the brackets but you are just trying to solve for x:

$\begin{array}{rcl} 0 & = & (x-2)^2 - 1 \\ 1 & = & (x-2)^2 \qquad(\text{Moved 1 to the other side}) \\ \pm 1 & = & x - 2 \qquad (\text{Took the square root of both sides}) \\ x & = & 2 \pm 1 \qquad (\text{Moved 2 to the other side}) \\ x & = & 3, 1\end{array}$

5. Originally Posted by cb220
I can see the similarities now but still unclear on a couple of things.

Take this with a grain of salt, being that it was the first day of class and I may have mistook what he said, but my understanding (and his answers on the board) reflect that if I take the final product of the Teachers Method answer: 0 = (x-2)² -1 I can take the -2 out of the brackets, which will make it x= +2 and the -1 will remain the same making one x = -1. Does it help that this equation is to find the vertex of a parabola? Or have you never seen something like that done before? The square rooting makes more sense to me logically but the answer that he arrived at on the board for the example of his method was V(2, -1).
Umm, so the "completing the square" was done to determine the vertex of the parabola.
Then your teacher's method is okay.
But you used 0 = x^2 -4x +3.
That is not a parabola. Where is the y or f(x) there?
The parabola is y or f(x) = x^2 -4x +3.

So your teacher "completed the square" of the RHS to arrive at
y = (x-2)^2 -1

A standard form of the equation of a vertical parabola ....where the x is squared....is
(y -k) = a(x -h)^2
where (h,k) is the vertex.

So, in the parabola y = (x-2)^2 -1,
rearranging it,
(y +1) = (x -2)^2
the h is 2, and the k is -1.
Hence, the vertex is (2,-1)

6. Originally Posted by ticbol
Umm, so the "completing the square" was done to determine the vertex of the parabola.
Then your teacher's method is okay.
But you used 0 = x^2 -4x +3.
That is not a parabola. Where is the y or f(x) there?
The parabola is y or f(x) = x^2 -4x +3.

So your teacher "completed the square" of the RHS to arrive at
y = (x-2)^2 -1

A standard form of the equation of a vertical parabola ....where the x is squared....is
(y -k) = a(x -h)^2
where (h,k) is the vertex.

So, in the parabola y = (x-2)^2 -1,
rearranging it,
(y +1) = (x -2)^2
the h is 2, and the k is -1.
Hence, the vertex is (2,-1)

Ah, now I understand. So I guess I just misunderstood the point of the question and that's why I was confused by the method ><. Thanks for clearing that up!