The quadratic formula: a tutorial(tigyog.app)
tigyog.app
The quadratic formula: a tutorial
https://tigyog.app/lessons/d1aswpgezqw6/r/the-quadratic-formula-a-tutorial
30 comments
I myself never learned the quadratic formula, I always just used completing the square, i.e. just using the fact that x^2+ax, and that you can split the ax in two put the resulting rectangles on sides of the square, to make a square with side x+a/2, by only adding an a^2/4 which is independent of x. Then you know that you can just remove the little a/2 square from the big square to relate it to x^2+ax.
I think that was the first case where I showed good taste in how to do mathematics.
I think that was the first case where I showed good taste in how to do mathematics.
If you complete the square with generic coefficient, you get the quadratic formula.
Yes, and that's why I've never had to actually learn the formula, and can make do by understand squares.
My algebra II teacher sung the formula to the tune of "pop goes the weasel."
"X equals negative b / Plus or minus square root / B-squared minus four a c / All over 2a"
Decades later, I have never forgotten. Apparently, we should just write parodies!
"X equals negative b / Plus or minus square root / B-squared minus four a c / All over 2a"
Decades later, I have never forgotten. Apparently, we should just write parodies!
My Algebra II teacher used Row, Row, Row Your Boat. She also wore a t-shirt with the quadratic equation on it, under her sweater. As she finished the song, she pulled off the sweater, much to the surprise of the students (the shirt stayed on, so things were still safe for school).
I'm still not sure if it was the song or the shock of seeing your teacher pull her clothes off. While I can still remember the song by heart, I also shudder every time I sing it.
I'm still not sure if it was the song or the shock of seeing your teacher pull her clothes off. While I can still remember the song by heart, I also shudder every time I sing it.
Not exactly "hot for teacher" material from the sounds of it.
Does anyone else find these mnemonics harder to remember than the actual thing itself? I see a lot of people online talk about this and something else for sines and cosines and it always seems really hard.
I’m not saying I’ve got some super memory. Just that remembering the actual expression is easy. The mnemonic is always longer and harder for me to recall.
I can picture the shapeform of the equation itself but cannot muster a single of the rhymes. I wonder if it is a difference in how we were trained as children.
I’m not saying I’ve got some super memory. Just that remembering the actual expression is easy. The mnemonic is always longer and harder for me to recall.
I can picture the shapeform of the equation itself but cannot muster a single of the rhymes. I wonder if it is a difference in how we were trained as children.
My teacher used the following short story:
The negative boy couldn't decide if he wanted to go to the radical party. So he decided to be square, missing out on 4 awesome chicks. It was all over by 2 AM.
The negative boy couldn't decide if he wanted to go to the radical party. So he decided to be square, missing out on 4 awesome chicks. It was all over by 2 AM.
My chemistry teacher had a similar acronym for remembering which of the anode and cathode where positive and negative.
Penises are not cunts.
Penises are not cunts.
My algebra teacher did it to the tune of the Notre Dame fight song, haha. Two decades later and I still remember it!
x equals the opposite of b
plus or minus the square root of
b-squared minus four a c
all over two times a
x equals the opposite of b
plus or minus the square root of
b-squared minus four a c
all over two times a
Same! In fact, I had to sing the entire song in my head to remember what the actual formula was.
I still remember lots of facts about Albania from Cheers: https://www.youtube.com/watch?v=-F_tT-q8EF0
I got my 4 year old to reliably remember our address by just making a simply jingle out of it. The tune is not related to anything. The power of music for memory recall is amazing.
Nice site, but it makes a meal out of the material. The technique of 'completing the square' is about as quick and easy as the quadratic formula (indeed, doing it with generic coefficients is what gives you the quadratic formula) and only calls on a good elementary understanding rather than dredging up a formula from the memory banks.
I think among the first things they taught us when I took physics at university was to forget about the formula, and learn to complete the square. Felt they had a fairly low opinion of the education we had been given thus far, so we had to do it almost as a kata to prepare for the reality of actually needing to rely on algebra abilities.
This was followed by years of professors telling us to learn to derive all sorts of formulas and identities rather than to memorize them.
Learning to derive a formula creates a sort mathematical self-sufficiency that you don't get when you memorize a formula.
This was followed by years of professors telling us to learn to derive all sorts of formulas and identities rather than to memorize them.
Learning to derive a formula creates a sort mathematical self-sufficiency that you don't get when you memorize a formula.
Yeh, "completing the square" is also how I learned it, but I thought this was a nice alternative approach that I hadn't seen before.
If you like, you can try writing a lesson using the "completing the square" approach! :-) https://tigyog.app/lessons/new
If you like, you can try writing a lesson using the "completing the square" approach! :-) https://tigyog.app/lessons/new
I must confess somewhere around 45% I got bored, simply clicked a lot of wrong answers just to get to the bottom. For me memorizing the formula is easier than deriving it using this lesson. Nevertheless this is a very good resource, it made it to my bookmarks collection
Thanks for the honesty! What was it that made you drop off? E.g. was it too long, or too much tedious algebra, or too much basics at the start?
I dropped off earlier than 45%, somewhere around 25%. It is about deriving the quadratic formula, not about 'what's a root'.
What's a root could be a link you click on to give more information than a special x value when both side of the equation line up (a root is when the graph of a function crosses the x-axis).
I like imagery and the clicking to progress to the next portion. What would be fantastic in my opinion is to take the step-by-step approach of something like WolframAlpha and apply it to defined topics with a set of defined examples rather than hey provide us an answer and if you get it wrong we will take you through a rudiment branch that will take us back to the main branch.
What's a root could be a link you click on to give more information than a special x value when both side of the equation line up (a root is when the graph of a function crosses the x-axis).
I like imagery and the clicking to progress to the next portion. What would be fantastic in my opinion is to take the step-by-step approach of something like WolframAlpha and apply it to defined topics with a set of defined examples rather than hey provide us an answer and if you get it wrong we will take you through a rudiment branch that will take us back to the main branch.
Yeh, I think I tried to pack too much into the lesson. "What is a root" could be more like a prereq lesson.
WA's step-by-step feature is amazing dark magic. I'd love to replicate that somehow ...
WA's step-by-step feature is amazing dark magic. I'd love to replicate that somehow ...
I dropped off at about 70%. For me, it was a few things
1) Too much intro material - The explanation of what a root is was too long. By the time I got to the start of the derivation, my attention span was already running out, due to:
2) Too much tedious algebra - I felt annoyed that I needed to do a bunch of basic arithmetic to advance to the next section.
3) Complicated derivation - Like another commenter mentioned, I think the completing the square method would have been shorter and easier than this method.
Side remark: The completing the square trick can be easily generalised to completing the cube. A few additional straightforward manipulations (substitutions) result in something called a reduced cubic. You can then express the reduced cubic equation as cosh(3 acosh(x)) = C for some constant C (or you can use the function "cos" instead of "cosh") and solve for x.
The quadratic formula is a nice toy exercise in learning some of the pitfalls of floating point numbers.
I have a very distinct memory of the quadratic formula: when my high school geometry teacher just assumed we all knew it, but most of the class had never seen it before.
This sticks in my memory many, many years later because of the hours several of us spent (without being asked to do so) memorizing the thing, that being much easier than the version of the factorization method that we'd been taught previously. I did later learn about other methods (being curious about math, and having access to a much better Internet than during high school). Still, I don't think I'll ever forget that formula.
This sticks in my memory many, many years later because of the hours several of us spent (without being asked to do so) memorizing the thing, that being much easier than the version of the factorization method that we'd been taught previously. I did later learn about other methods (being curious about math, and having access to a much better Internet than during high school). Still, I don't think I'll ever forget that formula.
Great format! I liked the halting problem lesson too. Are you planning to build a company out of this platform? (It's really nice to use)
That's the plan, and thanks for the compliment! :-) If you try writing lessons on the app and have any feedback/requests -- just let me know, [email protected]
3Blue1Brown Episode 1 Lockdown Math, A simpler version of the quadratic formula.: https://www.youtube.com/watch?v=MHXO86wKeDY
It's a good lesson, thanks to him I can derive the whole from my mind. It did take some time with the second part (finding deviation d), but I got there :)
I recommend to watch the video, for people who prefer to read, I'll derive the whole thing from my mind (the video makes it quite intuitive) and write it down.
___DEMONSTRATING THE 3BLUE1BROWN AWESOMENESS OUT OF MY OWN MIND___
I wasn't doing my best to memorize the intuition when he gave the lecture, otherwise it wouldn't have taken me 60 minutes to write it down ;-)
It starts with a sketch of a parabola that has 2 roots. Something like: x^2-4, see a sketch: https://www.desmos.com/calculator/q8mkfjmylc
It then shows that the intuition of the quadratic formula is that it's basically the midpoint m plus or minus the deviation d.
So for root left and root right you get:
l, r = m ± d
Sketch a parabola yourself, take the midpoint and draw arrow -d and arrow +d to the roots to visually verify that this is true.
In order to make this slightly easier, he suggests to immediately take the formula ax^2+bx+c=0 and rescale it to x^2+bx+c=0 and work from there. The bx and c are not the same symbols, you should see it as b-prime and c-prime in the second equation.
___GETTING THE MIDPOINT m___
Now you need to figure out what m is.
m is the midpoint, how do we calculate the midpoint algebraically?
Well, if you'd average out the roots, then you'd get the midpoint, so m = (l+r)/2. "But we don't know the roots!"
That's true, but suppose we would, how would that look like? It would look like:
y = (x-l)*(x-r) for roots l and r
Expanding it makes: x^2 -l*x -r*x + l*r --> x^2 - x(l+r) + l*r
Comparing it to x^2 + bx + c this means that: b = -(l+r) [**] and c = l*r
Remember that m = (l+r)/2. So we can somehow substitute -(l+r) for some version of m. Let's rewrite to see what it is.
m = (l+r)/2, 2*m = l+r, -2*m = -(l+r). -2*m = b (see [**])
therefore b = -2*m. Let's rewrite m in terms of b.
-b = 2*m, -b/2 = m. So m = -b/2.
We have our m in ways we understand! m = -b/2 from our rescaled x^2+bx+c=0 equation.
___GETTING THE DEVIATION d___
How do we get d?
3Blue1Brown shows a very cool thing here about midpoints geometrically and numerically.
3*5 = 15, 4^2 = 16
1337 * 1339 = 1338^2 - 1 (checking if correct: 1790243 = 1790244 - 1, yep)
So the closer the roots are to the midpoint, the closer it's looking as if you're multiplying the midpoint with itself (makes sense). How does it look like you make the deviation big?
Resuming with 3*5 (my own insights, 3B1B probably showed similar things).
l*r = c, m^2 = q (I use c because it's related to the c from x^2+bx+c = 0, I use q because I need a letter)
3*5 = 15, 4^2 = 16
2*6 = 12, 4^2 = 16
1*7 = 7, 4^2 = 16 [*]
The deviations are clear to see, they are:
3,5 = 4 ± 1
2,6 = 4 ± 2
1,7 = 4 ± 3
Observation: multiplying the roots l and r gives c. c is somehow related to the deviation as we can see that in the first case 16 - 15 = 1 which is the deviation we want.
Let's try the following, and see how close we get to the deviation we want:
d1 = 16 - 15 = 1 --> seems to check out
d2 = 16 - 12 = 4 --> does not check out, we need to square root it
d3 = 16 - 7 = 9 --> does not check out, we need to square root it
What did we just calculate?
m^2 - c = d^2
I find it hard to explain why this is the case. 3Blue1Brown might have an explanation for it. I simply feel the following "if you square a midpoint, and you multiply 2 points equally far away from the midpoint in opposite directions, that's kinda like weirdly squaring a midpoint, so there must be some relationship there. The further those 2 points are, the bigger the deviation must be. This deviation cannot be d since we're talking about squares here. Ah m^2 - c = d^2! Yea that seems to look alright."
Square rooting it we get:
d1 = 1, from m = 4, l,r = m ± d --> 3,5 = 4 ± 1
d2 = 2, from m = 4, l,r = m ± d --> 2,6 = 4 ± 2
d3 = 3, from m = 4, l,r = m ± d --> 1,7 = 4 ± 3
So let's put it into a formula.
d^2 = m^2 - c --> e.g. 3^2 = 4^2 - 7 (see [*] )
So d = sqrt(m^2 - c)
___PUTTING IT TOGETHER___
We know the quadratic formula is l, r = m ± d for x^2+bx+c = 0 which was rescaled from ax^2+bx+c=0 (again the bx+c symbols scale with it since I am too lazy to write b-prime and c-prime :D).
We know m. m = -b/2
We know d. d = sqrt(m^2 - c)
So the whole thing for left root l and right root r: l,r = -b/2 ± sqrt(m^2 - c) for x^2+bx+c=0.
It's a good lesson, thanks to him I can derive the whole from my mind. It did take some time with the second part (finding deviation d), but I got there :)
I recommend to watch the video, for people who prefer to read, I'll derive the whole thing from my mind (the video makes it quite intuitive) and write it down.
___DEMONSTRATING THE 3BLUE1BROWN AWESOMENESS OUT OF MY OWN MIND___
I wasn't doing my best to memorize the intuition when he gave the lecture, otherwise it wouldn't have taken me 60 minutes to write it down ;-)
It starts with a sketch of a parabola that has 2 roots. Something like: x^2-4, see a sketch: https://www.desmos.com/calculator/q8mkfjmylc
It then shows that the intuition of the quadratic formula is that it's basically the midpoint m plus or minus the deviation d.
So for root left and root right you get:
l, r = m ± d
Sketch a parabola yourself, take the midpoint and draw arrow -d and arrow +d to the roots to visually verify that this is true.
In order to make this slightly easier, he suggests to immediately take the formula ax^2+bx+c=0 and rescale it to x^2+bx+c=0 and work from there. The bx and c are not the same symbols, you should see it as b-prime and c-prime in the second equation.
___GETTING THE MIDPOINT m___
Now you need to figure out what m is.
m is the midpoint, how do we calculate the midpoint algebraically?
Well, if you'd average out the roots, then you'd get the midpoint, so m = (l+r)/2. "But we don't know the roots!"
That's true, but suppose we would, how would that look like? It would look like:
y = (x-l)*(x-r) for roots l and r
Expanding it makes: x^2 -l*x -r*x + l*r --> x^2 - x(l+r) + l*r
Comparing it to x^2 + bx + c this means that: b = -(l+r) [**] and c = l*r
Remember that m = (l+r)/2. So we can somehow substitute -(l+r) for some version of m. Let's rewrite to see what it is.
m = (l+r)/2, 2*m = l+r, -2*m = -(l+r). -2*m = b (see [**])
therefore b = -2*m. Let's rewrite m in terms of b.
-b = 2*m, -b/2 = m. So m = -b/2.
We have our m in ways we understand! m = -b/2 from our rescaled x^2+bx+c=0 equation.
___GETTING THE DEVIATION d___
How do we get d?
3Blue1Brown shows a very cool thing here about midpoints geometrically and numerically.
3*5 = 15, 4^2 = 16
1337 * 1339 = 1338^2 - 1 (checking if correct: 1790243 = 1790244 - 1, yep)
So the closer the roots are to the midpoint, the closer it's looking as if you're multiplying the midpoint with itself (makes sense). How does it look like you make the deviation big?
Resuming with 3*5 (my own insights, 3B1B probably showed similar things).
l*r = c, m^2 = q (I use c because it's related to the c from x^2+bx+c = 0, I use q because I need a letter)
3*5 = 15, 4^2 = 16
2*6 = 12, 4^2 = 16
1*7 = 7, 4^2 = 16 [*]
The deviations are clear to see, they are:
3,5 = 4 ± 1
2,6 = 4 ± 2
1,7 = 4 ± 3
Observation: multiplying the roots l and r gives c. c is somehow related to the deviation as we can see that in the first case 16 - 15 = 1 which is the deviation we want.
Let's try the following, and see how close we get to the deviation we want:
d1 = 16 - 15 = 1 --> seems to check out
d2 = 16 - 12 = 4 --> does not check out, we need to square root it
d3 = 16 - 7 = 9 --> does not check out, we need to square root it
What did we just calculate?
m^2 - c = d^2
I find it hard to explain why this is the case. 3Blue1Brown might have an explanation for it. I simply feel the following "if you square a midpoint, and you multiply 2 points equally far away from the midpoint in opposite directions, that's kinda like weirdly squaring a midpoint, so there must be some relationship there. The further those 2 points are, the bigger the deviation must be. This deviation cannot be d since we're talking about squares here. Ah m^2 - c = d^2! Yea that seems to look alright."
Square rooting it we get:
d1 = 1, from m = 4, l,r = m ± d --> 3,5 = 4 ± 1
d2 = 2, from m = 4, l,r = m ± d --> 2,6 = 4 ± 2
d3 = 3, from m = 4, l,r = m ± d --> 1,7 = 4 ± 3
So let's put it into a formula.
d^2 = m^2 - c --> e.g. 3^2 = 4^2 - 7 (see [*] )
So d = sqrt(m^2 - c)
___PUTTING IT TOGETHER___
We know the quadratic formula is l, r = m ± d for x^2+bx+c = 0 which was rescaled from ax^2+bx+c=0 (again the bx+c symbols scale with it since I am too lazy to write b-prime and c-prime :D).
We know m. m = -b/2
We know d. d = sqrt(m^2 - c)
So the whole thing for left root l and right root r: l,r = -b/2 ± sqrt(m^2 - c) for x^2+bx+c=0.
That was a big step in simultaneously realizing that mathematics could be beautiful but also that mathematics could be ruined by schools.