Algebra function form- completing the square

Thread Starter

upand_at_them

Joined May 15, 2010
940
I'm reviewing some calculus. This example is given in Chapter 1 of the book I have:

The area A of a certain rectangle, one of whose sides has length x, is given by \(A = 18x - 3x^2\). Here, both x and A must be positive. By completing the square, we obtain \(A = -3(x-3)^2 + 27\).
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?
 

Thread Starter

upand_at_them

Joined May 15, 2010
940
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.
 

djsfantasi

Joined Apr 11, 2010
9,163
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:

djsfantasi

Joined Apr 11, 2010
9,163
@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.
 

koyex1

Joined May 1, 2015
6
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
 

WBahn

Joined Mar 31, 2012
30,077
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?
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?
 

djsfantasi

Joined Apr 11, 2010
9,163
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.
 

WBahn

Joined Mar 31, 2012
30,077
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
 
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