# Algebra function form- completing the square

Discussion in 'Math' started by upand_at_them, May 23, 2015.

1. ### upand_at_them Thread Starter Active Member

May 15, 2010
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29
I'm reviewing some calculus. This example is given in Chapter 1 of the book I have:

I see that both equations are equal, but the last sentence confuses me. I don't see how one jumps to that form. Is there a trick?

Aug 21, 2008
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3. ### upand_at_them Thread Starter Active Member

May 15, 2010
246
29
Great! Thanks. I like reading proofs.

I think this might be something that used to be taught, but no longer is. I never learned that in school.

4. ### djsfantasi AAC Fanatic!

Apr 11, 2010
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I think the trick is in recognizing that to original equation is close to another square, and figuring out what is required to complete the equation for a square. If we start with the next obvious equation to be derived from the original by factoring out -3, I got the following:

$-3(-6x+x^2)$
or
$-3(x^2-6x)$

Now, the square can be completed by adding in a constant for the last term. To do so without changing the equation, we must also subtract the same value (or subtract and add). When we do this, we get:

$-3(x^2-6x+9)+27$

(It's +27, because of the -3. (-3 times 9 is -27))

So the "Trick" you asked about is recognizing that the problem is about completing the square of which $\small{18x-3x^2}$ is part.

Last edited: May 23, 2015
upand_at_them and DickCappels like this.
5. ### djsfantasi AAC Fanatic!

Apr 11, 2010
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@DickCappels, great link. Posted it while I was typing. I just did the same thing by brute force, recognizing the numbers... It was interesting to read the background of how it works formally.

6. ### upand_at_them Thread Starter Active Member

May 15, 2010
246
29
Yeah, that's pretty awesome. There's lots of cool things you can do once you put it into geometric terms.

Nov 9, 2007
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8. ### koyex1 New Member

May 1, 2015
5
0
Rearranging the formula to take the form of its general equation(ax^2 + bx + c) you have -3x^2 + 18x. Obviously a= -3 b= 18 and c= 0. Therefore to make the equation a perfect square c should be a number other than 0. Now to find that number use the formula b^2=4ac. Make c the subject formula. Input a and b into the formula. The value u get for c (-27) input into the original equation without letting the equation lose its meaning(that means u will need to add a +27 into the equation because -27+27=0) so your equation looks like this -3x^2 + 18x -27 +27 which is the same thing as -3x^2 +18x. Then you simply separate out (-3x^2 +18x -27) +27. Factorize whats in the bracket and you get your answer

9. ### WBahn Moderator

Mar 31, 2012
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I don't understand what is being asked for, so it's hard to tell whether completing the square even makes sense.

If you are being asked for the range of values that x can take on, I don't see that completing the square accomplishes anything.

You have a rectangle that is 'x' long by 'w' wide and for which the area, A, obeys the relationship given:

$
A \; = \; xw \; = \; 18x \; -\; 3x^2 \; = \; 3x \(6 \; -\; x \)
$

This is more than enough to answer most of the obvious questions:

1) What is the width of the rectangle, w, in terms of its length, x?
2) What is the range of values that 'x' can take on?
3) What is the range of values that 'w' can take on?
4) Given a width w, what is the length x?

What question is served by completing the square?

10. ### djsfantasi AAC Fanatic!

Apr 11, 2010
2,900
874
The question that is served is how the problem gets from the first equation to the second equation. That's what the TS asked. As I see it, his question didn't address how the equations were going to be used.

11. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Fair enough. Though I am still curious what the text book was getting at when choosing to go this route.

I'll offer a slightly different approach (which is equivalent to what koyex1 presented, just coming at it from a different angle).

A = 18x - 3x^2

After completing the square, we know we will have something of the form

A = a(x + b)^2 + c
A = a·x^2 + 2ab·x + a·b^2 + c

Matching terms, we see that

a = -3
2ab = 18 => b = 18/(2·-3) = -3
a·b^2 + c = 0 => c = -a·b^2 = -(-3)(-3)^2 = 27

Thus

A = 18x - 3x^2
A = -3(x - 3)^2 + 27

12. ### amilton542 Active Member

Nov 13, 2010
494
64
I always complete the square. I could find the roots whether real or complex quicker than you could use a calculator