A CRUX MOVE is a trick that comes up over and over when doing math, so it behooves you to remember it. I am slowing posting a list of them on LabKitty.
Complete-the-square is the reason I started Crux Moves. True fact! It's the one thing I simply never think to try. Why? Who knows. Perhaps Donny Kerabotsis' sister wore something slinky on the day we were taught the trick back in 7th grade algebra. Thus distracted, a hole was created in my mathematical psyche. Every time I tried to push off from that toehold, I found only empty. Heartache. Misery. An effect Donny's sister would have for years to come.
No longer leading the list, complete-the-square now checks in at number five. Much like Donny's sister. For I discovered there were bigger fish in the sea who are also utterly unaware I existed (Kari Byron, Clair Danes, Olivia Munn, Sarah Vowell). So, there. As granddaddy used to say, chump don't want no help, chump don't get no help.
None of us ever understood what grandaddy was on about.
It's a tale as old as time.
We have an expression x2 + stuff = blerg. We would like to write this as (x + stuff)2 = blerg. Except the new stuff isn't the same the as the old stuff (it's now squared and there's a cross product with x). And we have to make sure the equation is still true, usually by adding or subtracting something to/from the RHS.
Sometimes we have x2 + y2 + x-stuff + y-stuff = blerg which we would like to write as (x + x-stuff)2 + (y + y-stuff)2 = blerg. This comes up when working with conic sections, as our first example demonstrates.
E X A M P L E 1
Plot this: 4x2 + y2 – 8x + 4y – 8 = 0
Complete the square. For the x part, we have 4x2 – 8x = 4(x2 – 2x) which we can write as 4(x – 1)2 as long as we add 4 to the RHS, because 4(x – 1)2 = 4(x2 – 2x + 1) = 4x2 – 8x + 4 and while we indeed get back all the x-stuff we want, we've picked up a "4" that wasn't there originally. Similarly, in y we have y2 + 4y = (y + 2)2 and we have to add 4 more to the RHS.
Putting it all together:
4x2 + y2 – 8x + 4y – 8 = 0
4(x – 1)2 + (y + 2)2 – 8 = 4 + 4
4(x – 1)2 + (y + 2)2 = 16
(x – 1)2 / 4 + (y + 2)2 / 16 = 1
We've obtained the standard form of an ellipse, which you may recall is:
(x – u)2 / a2 + (y – v)2 / b2 = 1
where (u,v) is the center and 2a and 2b are the major and minor axes, depending on which one is bigger. Thus, the given equation describes a vertically-oriented ellipse centered at (1,–2) with major and minor axes of 8 and 4, respectively. You may now easily plot it, or simply get on with your little life knowing this fascinating tidbit.
E X A M P L E 2
Integrate: ∫ x2 / √ (2x – x2) dx
First, complete the square in the denominator:
∫ x2 / √ (1 – (x – 1)2) dx
Let x – 1 = sin(u), so √ (1 – (x – 1)2) = cos(u) and dx = cos(u) du. The cos(u) cancel and we get:
∫ (1+sin(u))2 du
Expand and apply trig identity sin2(u) = [1–cos(2u)] / 2:
∫ [ 3/2 + 2 sin(u) – 1/2 cos(2u) ] du
Integrate:
3/2 u – 2 cos(u) – 1/4 sin(2u) + C
Unsubstitute, noting if sin(u) = x – 1, then cos(u) = √(2x– x2) (draw a right triangle w/ hypotenuse 1):
3/2 arcsin(x–1) – 2√(2x– x2) – 1/2 (x–1) √(2x– x2) + C
Footnote: Perhaps here you are thinking: oh, come on! Trig substitutions, too? Is one crux move not enough? What can I say? Blood for the blood god and whatnot. The sooner you learn to choke back your impotent rage, the easier life will become. Mathematics, too.
E X A M P L E 3
Prove the sum of any positive number and its reciprocal is greater than 2. We begin by writing down the quantity of interest:
x + 1/x
And now we stare at this for a while. Staring. Still staring (as Curly of the Three Stooges used to say: I try to think, but nothing happens).
When introducing CTS above, I said we want to rewrite x2 + stuff as (x + stuff)2. The problem is that there's no squared anything in this problem. But there can be, if we write x as the square of the square root of x:
√x2+ 1/√x2
Now, what happens if we write this as the completed square (√x + 1/√x)2? Well, the cross product generates an extra 2, so we have to subtract that off. We get:
x + 1/x = (√x + 1/√x)2 – 2
And now we stare some more. The squared term is always non-negative, so if we had a + 2 instead of a – 2 we be done (because we'd have two-plus-something-non-negative. Which is greater than two. Which is what we were asked to prove). But we don't. So we aren't.
Oh, duh. We can also write √x2+ 1/√x2 as (√x – 1/√x)2 (note the minus sign). Now, the cross product generates a negative 2, so we have to add that back in. We get:
x + 1/x = (√x − 1/√x)2 + 2
or, rearranging the RHS to make the logic as obvious as possible:
x + 1/x = 2 + (√x − 1/√x)2
and we're done.
Footnote: I stole this example from the Wikipedia page on completing the square (although to be fair Wikipedia also stole it from someone, probably Evariste Galois). The moral is there's a Wikipedia page on completing the square and they have some nifty examples there.
Previous Move: Apply a Trig Identity
Next Move: Special Snowflakes
Complete-the-square is the reason I started Crux Moves. True fact! It's the one thing I simply never think to try. Why? Who knows. Perhaps Donny Kerabotsis' sister wore something slinky on the day we were taught the trick back in 7th grade algebra. Thus distracted, a hole was created in my mathematical psyche. Every time I tried to push off from that toehold, I found only empty. Heartache. Misery. An effect Donny's sister would have for years to come.
No longer leading the list, complete-the-square now checks in at number five. Much like Donny's sister. For I discovered there were bigger fish in the sea who are also utterly unaware I existed (Kari Byron, Clair Danes, Olivia Munn, Sarah Vowell). So, there. As granddaddy used to say, chump don't want no help, chump don't get no help.
None of us ever understood what grandaddy was on about.
Crux Move #5
Complete the Square
Complete the Square
It's a tale as old as time.
We have an expression x2 + stuff = blerg. We would like to write this as (x + stuff)2 = blerg. Except the new stuff isn't the same the as the old stuff (it's now squared and there's a cross product with x). And we have to make sure the equation is still true, usually by adding or subtracting something to/from the RHS.
Sometimes we have x2 + y2 + x-stuff + y-stuff = blerg which we would like to write as (x + x-stuff)2 + (y + y-stuff)2 = blerg. This comes up when working with conic sections, as our first example demonstrates.
E X A M P L E 1
Plot this: 4x2 + y2 – 8x + 4y – 8 = 0
Complete the square. For the x part, we have 4x2 – 8x = 4(x2 – 2x) which we can write as 4(x – 1)2 as long as we add 4 to the RHS, because 4(x – 1)2 = 4(x2 – 2x + 1) = 4x2 – 8x + 4 and while we indeed get back all the x-stuff we want, we've picked up a "4" that wasn't there originally. Similarly, in y we have y2 + 4y = (y + 2)2 and we have to add 4 more to the RHS.
Putting it all together:
4x2 + y2 – 8x + 4y – 8 = 0
4(x – 1)2 + (y + 2)2 – 8 = 4 + 4
4(x – 1)2 + (y + 2)2 = 16
(x – 1)2 / 4 + (y + 2)2 / 16 = 1
We've obtained the standard form of an ellipse, which you may recall is:
(x – u)2 / a2 + (y – v)2 / b2 = 1
where (u,v) is the center and 2a and 2b are the major and minor axes, depending on which one is bigger. Thus, the given equation describes a vertically-oriented ellipse centered at (1,–2) with major and minor axes of 8 and 4, respectively. You may now easily plot it, or simply get on with your little life knowing this fascinating tidbit.
E X A M P L E 2
Integrate: ∫ x2 / √ (2x – x2) dx
First, complete the square in the denominator:
∫ x2 / √ (1 – (x – 1)2) dx
Let x – 1 = sin(u), so √ (1 – (x – 1)2) = cos(u) and dx = cos(u) du. The cos(u) cancel and we get:
∫ (1+sin(u))2 du
Expand and apply trig identity sin2(u) = [1–cos(2u)] / 2:
∫ [ 3/2 + 2 sin(u) – 1/2 cos(2u) ] du
Integrate:
3/2 u – 2 cos(u) – 1/4 sin(2u) + C
Unsubstitute, noting if sin(u) = x – 1, then cos(u) = √(2x– x2) (draw a right triangle w/ hypotenuse 1):
3/2 arcsin(x–1) – 2√(2x– x2) – 1/2 (x–1) √(2x– x2) + C
Footnote: Perhaps here you are thinking: oh, come on! Trig substitutions, too? Is one crux move not enough? What can I say? Blood for the blood god and whatnot. The sooner you learn to choke back your impotent rage, the easier life will become. Mathematics, too.
E X A M P L E 3
Prove the sum of any positive number and its reciprocal is greater than 2. We begin by writing down the quantity of interest:
x + 1/x
And now we stare at this for a while. Staring. Still staring (as Curly of the Three Stooges used to say: I try to think, but nothing happens).
When introducing CTS above, I said we want to rewrite x2 + stuff as (x + stuff)2. The problem is that there's no squared anything in this problem. But there can be, if we write x as the square of the square root of x:
√x2+ 1/√x2
Now, what happens if we write this as the completed square (√x + 1/√x)2? Well, the cross product generates an extra 2, so we have to subtract that off. We get:
x + 1/x = (√x + 1/√x)2 – 2
And now we stare some more. The squared term is always non-negative, so if we had a + 2 instead of a – 2 we be done (because we'd have two-plus-something-non-negative. Which is greater than two. Which is what we were asked to prove). But we don't. So we aren't.
Oh, duh. We can also write √x2+ 1/√x2 as (√x – 1/√x)2 (note the minus sign). Now, the cross product generates a negative 2, so we have to add that back in. We get:
x + 1/x = (√x − 1/√x)2 + 2
or, rearranging the RHS to make the logic as obvious as possible:
x + 1/x = 2 + (√x − 1/√x)2
and we're done.
Footnote: I stole this example from the Wikipedia page on completing the square (although to be fair Wikipedia also stole it from someone, probably Evariste Galois). The moral is there's a Wikipedia page on completing the square and they have some nifty examples there.
Previous Move: Apply a Trig Identity
Next Move: Special Snowflakes
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