Math is Math! Why would they change Math?!

[Brother Oni]

I know exactly how Mr Incredible feels now.

So my daughter has a maths question: "Show every term in this sequence is positive by completing the square or otherwise: n² - 6n + 14"

I don't know completing the square, so I'm thinking 'quadratic equation formula, plug and chug':

Spoiler: I've gone wrong somewhere...

Show

(-b ± SQRT(b² -4ac)) / 2a

(- (-6) ± SQRT(-6² - (4x1x14) ) ) / (2x1)

(6 ± SQRT(36 - 56) ) / 2

(6 ± SQRT(-20) ) / 2

(6 ± 4.46i / 2)

Which means the answers are (6 + 4.46i)/2 and (6 - 4.46i)/2.

TLDR: I get two imaginary roots, which is probably not the answers they're looking for.

I've tried various maths resources online to explain completing the square, and I can't wrap my head around the answer I get from that ( (n - 3)² + 5 = 0 ) as proof that all the terms are positive.

Help please?

[Ninja_Prawn]

I think I remember this from school...

Completing the square means factorising it, and then seeing what is left over. n² - 6n + 14 = (n - 3)(n - 3) + 5

Since the constant on the end is positive, all values in the sequence must be? I can't remember the proof for that, but it must be true, because the question implies it is. According to this, if a quadratic equation is expressed as (x + a)² = b, the x/y co-ordinates of the vertex are (-a,-b). So if b is negative the curve is all above 0.

That said, your approach with the imaginary roots should also be valid, since the question allows "or otherwise".

[Jay R]

First, this is not an equation. The quadratic formula applies to an equation of the form ax² + bx + c = 0.

This is an expression, not an equation. There's no "= 0" part.

The expression n² – 6n + 14 is the same as n² – 6n + 9 + 5.

Since (n – 3)² = n² – 6n + 9, this is the same as (n – 3)² + 5

Any square number is 0 or positive. Add 5 to the square number at you get something that is 5 or more, and therefore positive.

And, finally, they didn't change math. Completing the square is a technique that's been around at least as long as the quadratic formula, and was in fact the tool used to derive and prove the quadratic formula.

[Unavenger]

It's a sequence, not an equation - they're saying "For any value of n, this expression is positive. Prove this by completing the square or otherwise". You could also do so by induction, say.

(Actually, I'm not sure it's possible by induction, and even if it is, you'd only get rational values no matter how meticulous you were about the induction. But that's not the point.)

[Knaight]

First things first - two imaginary roots in the quadratic equation means that the system is either positive at all points or negative at all points, so setting the expression equal to 0 to perform the quadratic equation works fine; you just also need to solve for any individual point (or the limit at positive or negative infinity).

Next up, completing the square is just writing it down differently. Consider the expression ax²+bx+c, where a, b, and c are constants. Now consider that (sqrt(a)x+b/(2*sqrt(a)))² can, through foiling be found to be equal to ax²+bx+b²/4a. Now consider that you can always add 0 to an expression without changing it, and that (b²/4a)-(b²/4a) is definitely 0. So, we add (b²/4a)-(b²/4a) to the original expression of ax²+bx+c and change that to (sqrt(a)x+b/(2*sqrt(a)))²-(b²/4a)+c. The portion that gets squared is obviously at least 0 for all normal numbers, so if -(b²/4a)+c is positive, clearly the whole expression is.

That -(b²/4a)+c term probably looks instinctively familiar to you, so let's do some things that won't change whether it is positive or negative at all. First off, multiply it by 4a/a, for (-b²+4ac)/a. Then, we can multiply it by -1 while now knowing that a negative result means a positive result, and we get (b²-4ac)/a, which is probably looking really familiar. Now we change the understanding slightly. That a in the denominator changes the sign if it's a negative value, but that's about all it can do to a sign, so we can take it out if we reframe getting a negative result as the result being the same sign as a (in graph terms this means the "bottom" of the curve is the same side of 0 as the "top", using non-technical terms), and we get (b²-4ac). That's the very term from the quadratic equation that you took the square root of to get imaginary numbers.

This forum needs an equation editor.

(Actually, I'm not sure it's possible by induction, and even if it is, you'd only get rational values no matter how meticulous you were about the induction. But that's not the point.)

You could. Moving from the vertex in either direction lets you prove that it is monotonically increasing by induction, a term above 0 that is monotonically increasing can't go below 0, this starts above 0, QED it is always above 0. It's a needlessly painful way of doing things though.

Personally my first instinct would be to note that this is technically a series but that the series will be entirely on a continuous function with the same equation, take the first derivative, solve for 0, solve the equation for that solution, then take the second derivative, note that it's a constant positive, and call it a day. That said if this problem is being put forth I'm pretty sure the kid isn't in a calc class.

I've tried various maths resources online to explain completing the square, and I can't wrap my head around the answer I get from that ( (n - 3)² + 5 = 0 ) as proof that all the terms are positive.

That's because that's not the answer. It's just a change in expression, so there's no =0, just (n - 3)² + 5. (n-3)² is always 0 or positive so the expression is always above 5, and thus positive.

[eggynack]

You could. Moving from the vertex in either direction lets you prove that it is monotonically increasing by induction, a term above 0 that is monotonically increasing can't go below 0, this starts above 0, QED it is always above 0. It's a needlessly painful way of doing things though.

Does this qualify as a proof by induction? You're not drawing any direct relationship between an arbitrary element and a specific subsequent element so much as you are just identifying a quality of the function in a broad sense. Unless I'm missing something, this is taking place in the real numbers. Standard induction requires that you be working with something that can line up with the natural numbers, and even transfinite induction requires that the set be well ordered.

[Unavenger]

You could. Moving from the vertex in either direction lets you prove that it is monotonically increasing by induction, a term above 0 that is monotonically increasing can't go below 0, this starts above 0, QED it is always above 0. It's a needlessly painful way of doing things though.

Ah. I'm used to inductive arguments in the form "This is true for x=1, and also true for x=k+1 wherever it's true for x=k, therefore it is true for x in the natural numbers" and such, so I didn't consider doing it like that.

[DavidSh]

Yes, induction here would be more like: "This is true for n=3. If it is true for n=k, where k >= 3, then it is true for n=k+1 (do the calculation). If it is true for n=k, where k <= 3, then it is true for n=k-1 (again, do the calculation). Therefore it is true for all integer n."

[Caerulea]

Yes, induction here would be more like: "This is true for n=3. If it is true for n=k, where k >= 3, then it is true for n=k+1 (do the calculation). If it is true for n=k, where k <= 3, then it is true for n=k-1 (again, do the calculation). Therefore it is true for all integer n."

But you are not asked to prove it for all n∈Z, you are asked for n∈R. Thus, incrementing by one doesn't work.

Given f(x) = (n-3)² + 5
Knaights' point was that you could start at n=3 (the vertex of the function), and notice that it evaluates to 5. Then, take some epsilon ε > 0 and notice that, for any n = 3 + ε, f(n) > 5, so the function is greater than 5. Also, for any ε, all ζ > ε cause f(3 + ζ) > f(3 + ε), so the function is strictly increasing after n=3.

[Radar]

But you are not asked to prove it for all n∈Z, you are asked for n∈R. Thus, incrementing by one doesn't work.

Given f(x) = (n-3)² + 5
Knaights' point was that you could start at n=3 (the vertex of the function), and notice that it evaluates to 5. Then, take some epsilon ε > 0 and notice that, for any n = 3 + ε, f(n) > 5, so the function is greater than 5. Also, for any ε, all ζ > ε cause f(3 + ζ) > f(3 + ε), so the function is strictly increasing after n=3.

Actually, the question is about a sequence, so it is implied that n∈N. Nevertheless, there are many ways to show that every element of said sequence is positive and every one works.

[Caerulea]

Actually, the question is about a sequence, so it is implied that n∈N. Nevertheless, there are many ways to show that every element of said sequence is positive and every one works.

True. My mistake.

This whole discussion reminds me of a Tom Leher song, titled (appropriately) New Math.

Hooray for ... New Math! New math! It won't do you a bit of good to review math!

[Douglas]

So my daughter has a maths question: "Show every term in this sequence is positive by completing the square or otherwise: n² - 6n + 14"

I don't know completing the square, so I'm thinking 'quadratic equation formula, plug and chug':

Completing the square goes like this:

1. Given an expression of the form an² + bn + c, first divide it all by a. In this case a is 1, so you can skip this step.
2. Divide b by 2 and then square it. I'll call the result d.
3. Form the expression n² + bn + d, and tack on the addition of the value of c - d at the end, making it equivalent to the original expression.
4. Because of how you computed d, n² + bn + d = (n + b/2)² will always be true, so replace the first with the second.
5. You now have an expression consisting of a number squared (which will always be non-negative if you're working in the real numbers) and a constant. The value of this expression obviously cannot be less than the constant.

For your example:
n² - 6n + 14
n² - 6n + 9 + 5
(n - 3)² + 5
This expression is always at least 5, so it's always positive.

TLDR: I get two imaginary roots, which is probably not the answers they're looking for.

That's what happens when the expression is never 0 for real number inputs. That plus evaluating the expression at any arbitrary spot to check whether it's positive or negative is enough to prove that the expression is positive (or negative) for all real number inputs.

[Knaight]

Does this qualify as a proof by induction? You're not drawing any direct relationship between an arbitrary element and a specific subsequent element so much as you are just identifying a quality of the function in a broad sense. Unless I'm missing something, this is taking place in the real numbers. Standard induction requires that you be working with something that can line up with the natural numbers, and even transfinite induction requires that the set be well ordered.

You can draw the direct relationships just fine - you're just splitting it into two sets, and I didn't cover how to actually do it at a detail level.

We'll call the vertex k=0. k=1 would be 1 away in either direction (you might want a quick symmetry proof), k=2 would be 2 away in either direction, so on and so forth.

We can see that the kth term is always k²+5. We also know that the (k+1)th term is k²+2k+1+5, which is the kth term +2k+1. The k=0 term is positive, 2k+1 is always positive, thus the k+1 term is greater than the k term, done.

[gomipile]

I don't understand the title. They taught us this in Algebra 2 at my high school back in the 1990s. I don't see how this is a change to math?

Maybe she's younger than high school, but that would still only imply a reshuffling of curriculum, not a change to its contents?

[eggynack]

You can draw the direct relationships just fine - you're just splitting it into two sets, and I didn't cover how to actually do it at a detail level.

We'll call the vertex k=0. k=1 would be 1 away in either direction (you might want a quick symmetry proof), k=2 would be 2 away in either direction, so on and so forth.

We can see that the kth term is always k²+5. We also know that the (k+1)th term is k²+2k+1+5, which is the kth term +2k+1. The k=0 term is positive, 2k+1 is always positive, thus the k+1 term is greater than the k term, done.

But isn't there also k=.5? Or k=pi? If k can take on non-integer values, then using only integer values isn't sufficient to prove the function is monotonically increasing.

[factotum]

I don't understand the title. They taught us this in Algebra 2 at my high school back in the 1990s. I don't see how this is a change to math?

I don't know how this varies from state to state in the US, but I know I never learned anything about "completing the square" in my algebra lessons here in the UK, and I did maths up to and including A levels (the exams that are set for people who choose to stay on at school for two years after 16 and which are used to determine university entrance requirements).

[Knaight]

But isn't there also k=.5? Or k=pi? If k can take on non-integer values, then using only integer values isn't sufficient to prove the function is monotonically increasing.

This was explicitly defined as a "series", which generally means integer values. That said, k doesn't actually need to be an integer for that to work in any way, so it would work just fine in a continuous case, with the notable exception of needing to prove that the entire k=0 to k=1 range is positive before induction really kicks in in the continuous sense (which is why induction is generally used for discrete math to begin with).

Granted, by the time you're even in position to prove by induction you can already just point to a term that gets squared added to a positive term and call it a day.

[jayem]

There's a proof by contradiction for well ordered sets that can be used more or less the opposite way to induction. You assume you have the least element that doesn't fit the bill and then create ANY lower one. This is probably easier to make watertight.
Classical proof by induction you know you get each number, if your elements are even the rationals that goes. I mean it's 'clearly' the case [basically what Knaight says]

This is a case where knowing and understanding the theory is different from just using it. By stopping the factorizing half way, you get a square of a real number (positive) plus a positive which is positive (as you did).
Alternatively if you do go straight for QF you could know that the roots are where the graph cuts the axis, and again it follows that it must be one side or another (assumming continuity and assorted things).

Personally I like completing the square as my first resort for finding the roots anyway, it means I don't need to do a temporary substitution when I may already have used x,a,b,c.

[Radar]

I don't know how this varies from state to state in the US, but I know I never learned anything about "completing the square" in my algebra lessons here in the UK, and I did maths up to and including A levels (the exams that are set for people who choose to stay on at school for two years after 16 and which are used to determine university entrance requirements).

Well, I for example didn't at first understand the phrase "completing the squares" either, but this was due to language difference. Back in my school days however we were tought all three forms to write down a quadratic expression and we did use all three to solve specific problems.

[Lord Torath]

Completing the square goes like this:

1. Given an expression of the form an² + bn + c, first divide it all by a. In this case a is 1, so you can skip this step.
2. Divide b by 2 and then square it. I'll call the result d.
3. Form the expression n² + bn + d, and tack on the addition of the value of c - d at the end, making it equivalent to the original expression.
4. Because of how you computed d, n² + bn + d = (n + b/2)² will always be true, so replace the first with the second.
5. You now have an expression consisting of a number squared (which will always be non-negative if you're working in the real numbers) and a constant. The value of this expression obviously cannot be less than the constant.

I was taught the Completing the Square method (7th grade - age 12-13), but I don't remember the algorithm ever being explicitly laid out like that by my teacher. Thanks, Douglas!

[eggynack]

This was explicitly defined as a "series", which generally means integer values.

Maybe, but I don't think there's anything in the expression being evaluated that only takes on integer values.

That said, k doesn't actually need to be an integer for that to work in any way, so it would work just fine in a continuous case, with the notable exception of needing to prove that the entire k=0 to k=1 range is positive before induction really kicks in in the continuous sense (which is why induction is generally used for discrete math to begin with).

I mean, yeah, I guess you're fine if you can prove this for that interval. That scenario is super bizarre though, where you can prove something for an uncountable infinity of cases without having already proved that thing for the higher numbers where it's also true. I wonder if that's ever occurred in all of math.

[Unavenger]

I don't know how this varies from state to state in the US, but I know I never learned anything about "completing the square" in my algebra lessons here in the UK, and I did maths up to and including A levels (the exams that are set for people who choose to stay on at school for two years after 16 and which are used to determine university entrance requirements).

I did A level maths (it's no longer a choice to do A levels [unless you want to do some other kind of further education], although you can very well decide not to do maths) and learned about completing the square. IIRC, it's the stuff of C1. My mother's a maths tutor, who has been teaching this stuff a long time, and she never mentioned completing the square as being a new thing in any sense.

[jayem]

There was a nice video on youtube that have completing the square done graphically for the case x*x+b*x=(-c) [where x, b, -c is positive]

You have your square + your (same widthed) rectangle.
Divide the rectangle in half (lengthways) and attach it to your square
Complete the square and add the same area to the other side
Reassemble the other side into a square somehow (if necessary by using explicit arithmetic, but if you can go via Pythagoras it shows it's more generally true)
Identify the two sides of the constructed squares as being the same (as the squares are the same area and square) and hence take off the known length to find 'x'

Which makes you identify with the Babylonians, which is kind of cool.
I think deriving a similar picture for negative bx (when b<x) is also just about intuitively understandable. When b>x I didn't like it so much

To me converting the two (eight?) pictorial cases to algebra seems to give a justification for the basic rules of negative numbers and it being much easier to treat them as genuine numbers.

[Brother Oni]

I don't understand the title. They taught us this in Algebra 2 at my high school back in the 1990s. I don't see how this is a change to math?

Because the exact curriculum and methods taught vary from era to era, country to country, school to school and even teacher to teacher.

I was never taught completing the square, but I went to public school in the UK in the 90s. Your experiences may vary.

Anyway, it wasn't a serious title, just that scene from The Incredibles 2 seemed apt.


Nevertheless, thank you all for the replies! I'll try and explain it to my daughter when she stops giggling at her old man running to the forums to complain.

[Knaight]

Maybe, but I don't think there's anything in the expression being evaluated that only takes on integer values.

As it's explicitly defined as a sequence you can assume this holds for x, or more generally any domain term (technically you should explicitly define this, in practice that tends not to happen, similarly whether the first term is n=0 or n=1 should have been specified, though it doesn't actually matter). The very term "sequence" is a discrete math term.

The range here is also all integers as a result, but that's not a requirement at all. It could be octonions with irrational numbers for every term, and that would still be fine, though the very idea of trying proof by induction on an octonion is pretty terrible.

I mean, yeah, I guess you're fine if you can prove this for that interval. That scenario is super bizarre though, where you can prove something for an uncountable infinity of cases without having already proved that thing for the higher numbers where it's also true. I wonder if that's ever occurred in all of math.

It's a pretty textbook example of why proof by induction is a discrete math process generally. If you treat it as a discrete process (and again, sequences) you just need to explicitly show the k = 0 case and the rest takes care of itself.

*Which is what I should have said when I said series; it's just the sequence points on the parabola, not their sum.

[Algeh]

I don't know how this varies from state to state in the US, but I know I never learned anything about "completing the square" in my algebra lessons here in the UK, and I did maths up to and including A levels (the exams that are set for people who choose to stay on at school for two years after 16 and which are used to determine university entrance requirements).

In much, but not all, of the USA, it's currently in the high school Common Core standards as:

A.SSE.3 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression.

b. Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines.

A.REI.4 Solve quadratic equations in one variable.

a. Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x – p)² = q that has the same solutions. Derive the quadratic formula from this form.

and

F.IF.8 Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function.

a. Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context.

Which is, generally speaking, taught in Algebra 1 near the end of the school year (if you actually get through the whole curriculum), touched on again the next year in Geometry since it's relevant to the equation of a circle:

G.GPE.1 Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation.

(which is taught near the end of the year if you actually have time to get through the whole curriculum)

and then quickly gone over again by an aggravated teacher in Algebra 2 early in the year when it becomes clear that most of the class still doesn't know how to do it...

Or maybe that's just the case in districts that I've taught in.

(I currently teach all 3 of those classes, and we do indeed get to the end of the curriculum each year, but completing the square is still one of those things that most of my students don't seem to retain from year to year...the time I most wish they understood it is the geometry case, really, since that's the one I find hardest to end-run with other ways to deal with quadratics that they do understand and remember.)

[snowblizz]

Because the exact curriculum and methods taught vary from era to era, country to country, school to school and even teacher to teacher.

I was never taught completing the square, but I went to public school in the UK in the 90s. Your experiences may vary.

Anyway, it wasn't a serious title, just that scene from The Incredibles 2 seemed apt.


Nevertheless, thank you all for the replies! I'll try and explain it to my daughter when she stops giggling at her old man running to the forums to complain.

I'm told you are by definition old when you try to do your kids homework and realise they've changed all the rules.


My mother was a teacher and every 15 years or so she'd complain how they "changed math". Some of the stuff she showed me to complain before she retired also made me go "huh? why would they try and explain it like that???".

At least it's not the 1970s "New Math".

Which ironically would be superbly adapted for today's computer age.

[Radar]

I can't beleive I only now remembered this.

Enjoy.

[Rockphed]

I can't beleive I only now remembered this.

Enjoy.

I was about to post that exact video. Tom Leher is great.

[OutOfThyme]

Because the exact curriculum and methods taught vary from era to era, country to country, school to school and even teacher to teacher.

I was never taught completing the square, but I went to public school in the UK in the 90s. Your experiences may vary.

Anyway, it wasn't a serious title, just that scene from The Incredibles 2 seemed apt.

It's kind of ironic, because the person responsible for coming up with the idea of completing the square (at least, IIRC, he was the first person to publish about it) is Al-Khwarizmi, who wrote the book that the word "algebra" is derived from.

I don't recall if that text is the one where he discusses completing the square.