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May 3rd, 2007, 12:19 AM   #1
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Completing the Square?

I know the quadratic formula is basically the completing the square method expressed algebraically. But does anyone know exactly why this method works? Why would one add (b/2a)²? Also, why does

x²+bx+(b/2a)² = (x+b/2a)² ? Where does the x in the bx go?
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May 3rd, 2007, 02:44 AM   #2
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If you try completing the square with the equation ax^2+bx+c, then you will find that this is the quadratic formula. I don't have time right now to show this, but I will later if you are interested.
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May 3rd, 2007, 03:41 AM   #3
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Quote:
Why would one add (b/2a)²?
I always add (b/2)², and divide the equation out so as to make sure that a=1.


Quote:
Also, why does

x²+bx+(b/2a)² = (x+b/2a)² ? Where does the x in the bx go?
It's still there; try multiplying it back out, assuming that a=1:

(x+b/2)²
= x² + (b/2)x + (b/2)x +(b/2)² = x²+bx+(b/2)²
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May 3rd, 2007, 04:22 AM   #4
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Quote:
Originally Posted by Infinity
It's still there; try multiplying it back out, assuming that a=1:

(x+b/2)²
= x² + (b/2)x + (b/2)x +(b/2)² = x²+bx+(b/2)²
Thank you for clearing that up.

I know how to get the quadratic formula from standard and normal form. My question isn't how but why this works. Is the completing the square method of adding (b/2)² just a trial and error method that proved correct, or is there some other justification for it?
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May 3rd, 2007, 09:48 AM   #5
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Quote:
I know how to get the quadratic formula from standard and normal form. My question isn't how but why this works. Is the completing the square method of adding (b/2)² just a trial and error method that proved correct, or is there some other justification for it?
It makes it possible to factor otherwise un-factorable expressions.

For instance, suppose we have to find x in the equation x² + x + 1 = 0

There is no way to factor the thing, so we must complete the square in order to get it factored. x² + x + 1/4 + 1 = 1/4

That way, we can now write: (x + 1/2)² + 1 = 1/4

The completing of the square just lets us write it in factored form, since all we are doing is getting the expression into an expanded-square form, which can naturally be written as a square of some expression, in this case: (x + 1/2)²

So now, we can easily see that we can solve for x algebraically in our now-factored expression: (x + 1/2)² = -3/4
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May 3rd, 2007, 11:22 AM   #6
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Quote:
Originally Posted by Infinity
Quote:
I know how to get the quadratic formula from standard and normal form. My question isn't how but why this works. Is the completing the square method of adding (b/2)² just a trial and error method that proved correct, or is there some other justification for it?
It makes it possible to factor otherwise un-factorable expressions.

For instance, suppose we have to find x in the equation x² + x + 1 = 0

There is no way to factor the thing, so we must complete the square in order to get it factored. x² + x + 1/4 + 1 = 1/4

That way, we can now write: (x + 1/2)² + 1 = 1/4

The completing of the square just lets us write it in factored form, since all we are doing is getting the expression into an expanded-square form, which can naturally be written as a square of some expression, in this case: (x + 1/2)²

So now, we can easily see that we can solve for x algebraically in our now-factored expression: (x + 1/2)² = -3/4
Thanks for the explanation. I hope I'm not annoying, but my question still was WHY does adding (b/2)² do all that as opposed to some other operation? I know how to solve quadratics and all, but I never found deeper justification as to exactly WHY these methods work. Is it really just by chance that someone discovered adding (b/2)² solves quadratics or is that linked to other mathematical properties that elude me? I'm dying to know.
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May 3rd, 2007, 01:02 PM   #7
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The methods don't work all that well at all. You get to crack a quadratic, but somewhat messily. Most higher-degree polynomials can't be cracked similarly, as there isn't a general method for finding exactly what the factors are.
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May 3rd, 2007, 01:11 PM   #8
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Now that I have some time, I will try to explain the generalized method.

Assume ax² + bx + c = 0.

First we subtract c from both sides, to obtain the expresion
ax² + bx = -c

Factoring out an a from the left side yields
a[x² + (b/a)x] = -c

Now, we want to add something inside the brackets that will turn it into a perfect square. Now (m+n)²=m² + 2mn +n² by the distributive property. Now, in our equation m=x, and 2mn=bx/a=bm/a (*). Dividing the left and right sides of equation (*) by 2m yields n=b/2a. Now, going back to our formula, if we add m² inside the brackets, we will have a perfect square, as demonstrated.
a[x² + 2(b/2a)x + (b/2a)² - (b/2a)²] = -c

Anything we add to one side of an equation, we must also subtract from the same side of that equation (multiplying and dividing by 2 was just to make the relationship clear). Now, by the distributive property,
a[x² + 2(b/2a)x + (b/2a)²] - a(b/2a)² = -c

Next, we move the term outside the brackets to the right side of the equation and factor the expression inside the brackets.
a[x + b/2a]² = a(b²/4a²) - c

dividing both sides by a yields
(x + b/2a)² = b²/4a² - c/a

Now, we want the right side of the equation to be a single fraction. To do this, we multiply both the numerator and the denominator of the right fraction by the quotient of the two denominators, which is 4a²/a=4a. We now have
(x + b/2a)² = b²/4a² - 4ac/4a²

Adding the fractions gives
(x + b/2a)² = (b² - 4ac)/4a²

Is some of this starting to look familiar? We now use the fact that k = ±√(k²) to reduce the above equation to
x + b/2a = ±√[(b² - 4ac)/4a²]
x + b/2a = ±[√(b² - 4ac)]/2a

To isolate x, we subtract b/2a from both sides and add the fractions
x = -b/2a ±[√(b² - 4ac)]/2a
x = [-b ± √(b² - 4ac)]/2a
which is the familiar quadratic formula.

I hope this gave you some insight into why completing the square and the quadratic formula work.

This has already been a lengthy post, and the rest of it is not essential. You may prefer to not read it (or just go ahead, but don't say you weren't warned).










In my opinion, the one critical flaw with math education through high school is that so much of it is teaching students a bag of tricks to memorize without teaching why those tricks work (I am still in high school, so my memory is very fresh). I think that it is much more important to teach students how to solve problems that they have not seen, rather than just memorizing how to solve specific types of problems (and then forgetting many of them after the test). If students were taught problem solving (as opposed to having specific methods beat into their heads), mathematics would be more worthwhile and enjoyable for students.
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May 3rd, 2007, 01:14 PM   #9
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Quote:
Originally Posted by skipjack
The methods don't work all that well at all. You get to crack a quadratic, but somewhat messily. Most higher-degree polynomials can't be cracked similarly, as there isn't a general method for finding exactly what the factors are.
I will agree with you on that one; however, dispite how messy cracking a quadratic is, it is emphasized rather a lot (read: too much) in American High Schools (this also goes along with my rather lengthy paragraph above about math education).
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May 3rd, 2007, 06:18 PM   #10
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Quote:
If students were taught problem solving (as opposed to having specific methods beat into their heads), mathematics would be more worthwhile and enjoyable for students.
That's the problem I've seen in a lot of the math classes I've taken so far.
The students all know that 1 + 2 = 3, but they can't figure out that because of that fact, 3 - 1 = 2. That's why I work hard to understand things at a basic level, so I can build from scratch or re-structure any formula I want. It's a very effective method, and works great. As a result, it is now a widely held belief among many of the students that I am a born genius, and that nobody could ever become as smart as they imagine I am unless they have some sort of 'magical touch'. This excuse allows the rest of the students to handily escape any guilt they feel over not having studied hard enough, because if I soundly beat them on some test or other, they just blame it on my 'unattainable genius' and still don't bother to work any harder. Of course, a few of the students (the very good ones) know better, but the average student just keeps blindly stumbling on, assuming a defeatist attitude and assuming that they'll never do any better than they are already doing. This attitude is so prevalent among many students that is shocks me to see how many of them really never learn anything new. Oh, they get more knowledge stuffed into their heads ( most of which they promptly forget), but they don't actually ever learn anything new at a basic level, nor do they ever change their method of learning, which is typically badly flawed. Moreover, it doesn't matter if they realize it's badly flawed, because they are convinced that they just aren't bright enough to fix it anyway. You know, it's the old story of the goats who won't go past the line in the field where the old electric fence stood, because nothing will induce them to think that the electric fence was removed 3 years ago. And if anyone does get over to the other side of the old fence line, the goats just convince themselves that he must have been born there, because how else could he have wound up on the other side?
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