Welcome to another an Mathologer video. If I

were to wake you up in the middle of the night, I’m sure that most of you would be

able to rattle off the quadratic formula: minus B plus or minus the square root of B

squared, and so on. But along with all that quadratic fun

you had in school, most of you would have tortured a few carefully cooked up cubic

equations and maybe even a quartic or two. Right? But did you ever wonder why no

one ever taught you the cubic formula or the quartic formula? Did your teachers ever

even mention the higher degree counterparts of the incredibly useful

quadratic formula? Probably not. Why is that? Is it a cultish

mathematical conspiracy? Are they all hiding some deep dark secret? Well

today’s Mathologer is your invitation into the hidden polynomial brotherhood.

Okay, here’s the general cubic equation which we want to solve. And here’s its

solution, the cubic formula in all its glory. Whoa that looks … interesting. So maybe

that’s our answer right there. That’s way too complicated to memorize and use. A

pretty good reason not to teach it, right? Wrong! Sure the full cubic formula

doesn’t exactly roll off the tongue. But it’s not that hard to make sense of. For

starters the two monsters inside the cube roots actually only differ in a

minus sign. But we can also simplify things dramatically by some

straightforward pre-processing. The general cubic equation can be reduced to

solving a much simpler cubic with a genuinely simple cubic formula. What do I

mean by pre-processing? Well, first, if you divide by the leading coefficient a that

gives an equation with the same solutions but with a leading coefficient

of 1. So we may as well turn all the a’s into 1s. That already looks a lot

simpler. The next most frequent coefficient in the formula is b. Now it’d

be good to get rid of all those b’s and it turns out we can. There’s a second

easy pre-processing step which I’ll motivated soon. This pre-processing

allows us to assume b equals zero and so we just have to worry

about solving a simple cubic without the green quadratic term. Pretty simple, right?

Maybe not kindergarten simple but definitely not a big leap from the

quadratic formula. Again, by performing some simple pre-processing, we can reduce

any cubic equation whatsoever to solving one of these simple cubic equations. In

the literature the letters c and d are usually replaced by the letters p and q

and when people say cubic formula they usually have this exact formula in mind. Really very pretty, isn’t it? Okay, let’s

do a quick example to see how it all works.

Let’s choose, well let’s see p equal to – 15 and q equal to – 126.

Sub these numbers everywhere and we get this. The number under the square root

signs pans out to be 3844 which happens to be equal to 62 squared.

Lucky ! 🙂 Now 63-62 that’s 1 and 63 + 62 that’s 125. Cube root of 1 is 1 and cube

root of 125 is 5, and so we finally get mm-hmm so 6 is solution of our cubic and

you can easily check that. Great! Although shouldn’t there be three

solutions? What happened to the others? Don’t worry, we’ll get to them later.

Anyway, all this looks pretty straightforward and very useful for

dealing with cubic conundrums. Also, for people interested in how our mathematics

came to be the discovery of the cubic formula is considered to be one of the

major milestones in the history of mathematics. So, once again, why doesn’t

anybody teach the cubic formula. There are a couple of answers to this puzzling

question, one more surprising than the next. Promise 🙂 Of course I’ll explain it

all in the following. As you’ve probably already guessed this is another master

class Mathologer video and as usual with these long videos we’ll start really

easy and slowly build up to the challenging finale. It’s all broken up

into seven chapters, with truckloads of really interesting cubic maths ahead.

Plus at the very end I’ll throw in solving the quartic equation for free.

Mathematical seatbelts on? Okay here we go. The cubic formula was discovered

independently by two Italian mathematicians: Nicolo Tartaglia and Scipione del Ferro. And then neither del Ferro nor Tartaglia told anyone about their

amazing discovery. Huh? Why not? Well, just like today, the success

of a sixteenth century mathematician depended on the sorts of problems they

could solve. But back then there was no such thing as Publish or Perish.

Instead mathematicians competed for celebrity in mathematical duels. Mathematicians would publicly challenge

others with take-home exams. They would post sets of problems and whoever could

solve more problems was declared to be the better mathematician. And so if you

were the only mathematician with a secret weapon like the cubic formula

then you would have been pretty much invincible. Don’t tell anybody.

Makes sense, right? And also an impressive start to 500 years of NOT teaching the

cubic formula. Anyway after winning a competition featuring nasty cubics it

became clear that Tartaglia had the secret weapon that he knew the cubic

formula. Then the great Gerolamo Cardano started going after him for this formula.

Cardano was a fascinating character. He was a polymath with his fingers in everything:

he was a famous doctor, a top mathematician, a successful gambler, and

so on. Anyway, eventually Tartaglia gave in. Interesting tidbit: Tartaglia gave

Cardano his formula dressed up as a poem. Quando chel cubo con le cose appresso. Se agguaglia a qualche numero discreto Truouan dui altri differenti in esso. I

put a link to the translation and discussion in the video description.

Ok Tartaglia also made Cardano’s swear a solemn oath not to tell anybody else.

Which of course Cardano broke. After learning the formula from Tartaglia

Cardano kept digging and eventually also found the cubic formula in a

notebook of del Ferro that other discoverer. It became apparent that del

Ferro discovered the formula before Tartaglia. In Cardano’s mind this meant

he was now off the hook. He could now tell people about del Ferro’s solution

and he actually subsequently published a book about it. In a way Cardano’s book

was like a Renaissance WikiLeaks and similar to today and not surprisingly

Tartaglia was incredibly annoyed by this wiki leaking and all hell broke loose.

I’ll link to some sources that have the complete story which, apart from all the

maths also feature a murder, a beheading, and a cameo appearance of the

Inquisition. An absolute must-read for all you maths history buffs. Some great

history, right? Now I’d like you to imagine that we’ve all been transported

back to the time of Cardano and his friends. Try not to be distracted by all

the beheading and inquisitioning. And armed with our knowledge of modern high

school mathswe want to challenge one of these big shots to a mathematical duel.

But since they did not tell us about the cubic formula in school we first have to

rediscover it for ourselves. Time for a bit of a revision

of quadratics to prepare ourselves for the cubic monsters to come. How do you

find a solution to this quadratic equation. Sure, that’s a no-brainer, just

plug the coefficients 2, 8 and -6 into the quadratic formula. Another way of

finding the solutions taught in school since time immemorial is “completing the

square”. This actually translates into the way of

deriving the quadratic formula from scratch. Let me show you a Mathologerized

version of completing the square for this particular quadratic. Ok, start by

dividing through by the leading coefficient. Alright getting there.

Next, interpret x squared as the area of an actual physical square with side

lengths x. Similarly, we’ll think of 4x as the area of a rectangle with sides x and 4.

Chop the rectangle in two like so and rearranged like this. Adding a little

square in the top right corner completes the left side into a new larger square.

But of course to keep our equality we also have to add the little square to

the right side like that. The left side is not just square-shaped it’s also a

perfect square namely x + 2 all squared. It’s now easy to solve the new

equation. Right? Piece of cake. Of course, there’s also that second minus

root 7 solution which we get if we’re not dealing with physical squares. Anyway

here’s an animated derivation of the quadratic formula for the general

quadratic equation by completing the square.

Enjoy the animated algebra and the music. So we know how to complete the square

geometrically in terms of real squares. This is actually similar to how Cardano

and his friends, and probably also the Babylonians thousands of years ago, would

have thought about the process. Algebraically completing the square

boils down to starting with this other super famous identity here, hammering the

quadratic, to look like the right hand side and then using the squareness of

the left hand side to solve the equation. You’ve all done some version of this a

million times, right? To continue our preparation the

cubic puzzling to come we need a second visual derivation of the quadratic

formula. This very nice derivation also illustrates the geometric meaning of the

quadratic formula. Ready? Okay, to be able to quickly refer back to it, let’s write

the quadratic formula in this form and tuck it away up there. Start again with

the pre-processed form of our equation down there. Then the quadratic

corresponds to a parabola like this. Here’s a nice idea. The parabola has a

special point that turning point down there. Let’s shift the parabola

horizontally until the turning point rests on the y-axis. Shift, shift, shift.

The two blue points, the x-intercepts, are still exactly the same distance apart.

However, the quadratic for this new parabola is now in this simpler form,

super simpler form. Right? It’s just the archetypal y=x^2

parabola just shifted up or down. it’s easy to solve this equation which

amounts to locate in the blue intercepts and now to find the solutions we are

really interested in we simply have to undo the horizontal shift we executed

before and keep track of where the blue points go, there great,

Ok now let’s see this scheme in practice First, how far do we have to shift? Well, that amounts to figuring out the x- coordinate of the green turning point

right? There that one. This would normally be done in school with a little algebra

but we’re on a mission here. So let’s take the express lane and use a little

easy calculus to make short work of this job. If you haven’t seen calculus yet,

doesn’t matter, just run with it or substitute the school algebra. OKay, so

form the derivative of the quadratic, set this derivative equal to zero and solve

for x. There, solve, solve, solve. That’s the x-

coordinate of the turning point and more than that: it’s also the familiar – b/2a, the first orange term in our quadratic formula. So that’s just the shift,

right? Now we can shift our parabola. To obtain

the quadratic equation of the new shifted parabola we replace every x by

x-b/2a. Maybe ponder that for a second. All good? Great!

Now use algebra autopilot to expand and collect like terms. Neat. Of course you

also recognize the fraction the messy constant term, right? There, it’s in the

green bit. Now solve to find the shifted blue points there and finally undo the

shift. Tada! Also very nice, isn’t it? Okay having reexpertized ourselves in

quadratics, we are ready to confront our cubic nightmare. A very natural idea is

to try and mimic completing the square to somehow complete the cube using this

perfect cube formula. Hmm, well the way things line up sort of feels it might

work, right? I think most people would play around with this for a while before

trying anything else. Okay so we really go for it for a day, trying to make

completing the cube pan out. And we come up with … zilch! It turns out it’s not that

easy and maybe it’s time to try something else.

So let’s look to mimic our parabola shifting. Okay, start the pre-processing by

dividing through by the leading coefficient. Graphically this kind of cubic will look

roughly like one of these curves. A cubic equation always has at least one

solution but may also have two or three. As well, every cubic has one special

point, the inflection point. This is the point where the cubic switches from

curving one way to the other. Finally, and this may come as a surprise to many of

you, a cubic graph always features a half-turn symmetry about this special

inflection point. Now let’s shift the inflection point to

the y-axis. For that we need the x-coordinate of this point. I’ll again use

some baby calculus. The inflection point here is characterized by the second

derivative being zero. So differentiate once. Okay. Twice. Set equal to zero and

solve. Solve, solve, solve, solved! That’s it, that’s the x-coordinate. This fraction

-b/3a actually appears in the original monster cubic formula I showed

you. Let me show it to you. There that’s the shift again. Now replace x by x – b/3a

and let the algebra auto-pilot perform our shift. That finishes the

pre-processing. The new coefficients are a messy combo of the original a, b, c, and d

but the fundamental form is one of those simple cubics I mentioned in the intro. So if we can figure out how to solve

this special simple cubic equation, then we can undo the shift to solve the

original equation. All very natural and very pretty.

I just love this stuff. Now to analyze our special cubic, what do p and q stand

for in the picture. Well, q is just the y-intercept.

And p is the slope of the curve at the inflection point. So p is the slope of

this green line. Of course, changing q just shifts the cubic up or down. What if

you vary p? Well here’s an animation of how the shape of the curve changes. So

what we’ve got here is that for negative p we’ve got this rollercoaster

shaped curve and for positive p it’s kind of stretched out like that, all

right? So, again, going into the negative – rollercoaster.

In a little while it will be important to know how many solutions our equation

has. Again, for non negative p the graph is stretched out like this and there can

only be one solution. But for negative p we get a rollercoaster graph like this

and 1, 2 or 3 solutions are possible. To figure out how many

solutions we get for negative p, we need to determine the difference in height

between the inflection point and the two extrema. Okay, because of the half-turn

symmetry, the two green height differences will be equal. Now, how many

solutions will we have? Okay, you got it? Well, if the yellow segment is longer

than the green, as pictured here, then there will be just one solution. It’s

also straightforward to check algebraically when this happens. How? Well

the extrema of the cubic occur when the derivative is zero. And the derivative of

the cubic is a quadratic and so we just have to solve a quadratic equation to

pinpoint the coordinates of the extrema. Again, piece of cake, right? And with the

coordinates of the extrema and the turning point, it is then also

easy to translate the yellow-green inequality into algebra. I leave that

as an exercise for you to complete in the comments. In the end the algebraic

counterpart to our yellow-green inequality pans out like this. Okay, in

exactly the same way the cubic having two solutions corresponds to equality

here and three solutions correspond to the reverse inequality. And so the sign

of the expression (q/2)^2+(p/3)^3 tells us how many

solutions our equation has. And this does not only work for negative p

roller-coaster cubics that we’ve been considering so far but for all cubics.

Right? For example, if p is positive, we have only one solution. But, also, with p

positive (q/2)^2+(p/3)^3 is positive. Works: one solution.

But there’s one tiny little annoying exception. Can you spot it? Well can you?

Well, when both p and q are zero, we still have

only one solution however in this case Q (q/2)^2+(p/3)^3 is

equal to 0 and so our table wrongly suggests there are two solutions. A minor glitch in the fabric of the

universe, which is easily fixed by talking for example about single and

repeated solutions. Anyway, the take away message from all this is that (q/2)^2+(p/3)^3 plays the same role for cubics as the

discriminant b^2-4ac plays for quadratics and just like b^2-4ac features prominently in the quadratic formula, will see that this

cubic discriminant takes center stage in the cubic formula. Remember when we wasted that day trying

to complete the cube? It turns out that time wasn’t

totally wasted. While playing with that perfect cube identity for hours we

noticed something very promising: the two middle terms have a common factor 3 u v.

If we pull it out, we get this. And now we can line up things like this. What

this shows us is that if we somehow find u and v so that all the lined up boxes

are equalities, then our cubic is effectively a perfect cube after all and

we will have solved our equation. What this amounts to is solving for u and v

in the green and yellow boxes. And then having done that we obtain our solution

x by just adding u and v. Pretty easy, actually. Let me just show you one way of

doing this. Well, first cube both sides of the green equation. Next, multiply the

yellow equation by v^3. Now we’ve got the same expression here and here and we

can sub like this. Shuffle everything to the left, as usual.

Looks bad at first glance, right? But it’s just a quadratic equation in

disguise. Think of the v^3 as the unknown.

There, blue squared plus something times blue plus something equals 0. Now

just use the quadratic formula to solve for the blue v^3 and cube rooting on

both sides gives this. Almost there. Remember that in our original equations u and v are indistinguishable and so if we solve for u, we get exactly the same

result. And now because of the plus/minus it would seem we can make two choices

each for both u and v, giving a total of three different ways to add them to get

solutions to our cubic equation: plus plus, minus minus, plus minus, and what

gives the same sum, minus plus. So here we have

three candidates for solutions AND cubic equations have up to three solutions.

That looks very promising, doesn’t it? Yes, it does. However, if we double-check

by substituting our candidates in the original equation, we find that only one,

the plus minus or minus plus gives us a genuine solution of the cubic. Why don’t

the other candidates for solutions work out? Can you figure out what the reason

might be? Let us know in the comments. Also, if all this only gives one of the

solutions, where are the other solutions hiding? Well, we’ll get to that. Why don’t they teach this? Well, there’s a

few related reasons that have to do with the way the cubic formula spits out

solutions. Remember that really nice example that I showed you at the

beginning. In that case, the solution given by the cubic formula easily

simplified down to a final answer of 6. Lucky with all those integer squares

and cubes materializing in just the right spots, wasn’t it? Of course it

wasn’t luck. One has to work damn hard to find an

example where the roots simplify that well. But things in general are even

worse than you might expect and some really weird stuff can happen. Let’s have

a look at another example. It’s pretty easy to guess a solution of this

equation. Can you see it. Ok, too late 🙂 Yep x=4 and just to check 4^3 that’s 64, 6 times 4 that’s 24, 64 minus 24 that’s 40, minus 40 is 0. Bingo!!

But what does our formula tell us? Plugging in -6 and -40 we get

this. So the cubic also has a monster rooty solution. But remember that 392

under the square root sign is our cubic discriminant and since 392 is positive

that tells us the cubic equation has only one solution. But that means that

the rooty monster in front of us must equal 4, the solution we guessed earlier.

Can you see at a glance that the two sides of this equation are really equal?

No, not obvious at all. It’s a challenge for you, can you do this from scratch? But

it can be even weirder. Have a look at this third example. Ok, plugging in -6 and -4 we get, well, under the square root sign our cubic discriminant

is 4 minus 8 that’s -4, which is negative. Hmmm, I hear you think. But let’s

first just focus on a negative discriminant which tells us that our

equation has three solutions. Let me just show you those solutions.

But now have a closer look at the expression spat out by our cubic formula.

Not only does this expression not look anything like those three solutions. In

addition, it contains square roots of negative numbers. This means that even

though our three solutions are real numbers, our formula expresses these

solutions in terms of complex numbers. How crazy weird is that. Not surprisingly

Cardano and his buddies were completely weirded out by this. They barely had an

understanding of negative numbers and complex numbers were way beyond their

imagination. And as it’s clear from the formula, the case of three roots always

leads to these weird complex number infested outputs. This horror case came

to be referred to as the casus irreduciblis. In fact, the struggle to

overcome the casus irreducibilis led to the discovery of complex numbers by

Rafael Bombelli not long after Cardano published the cubic formula.

I’ll make sense of the casus irreducibilis in the next chapter. First, let’s

just get all the cubic formula weirdness out in the open. How much weirder can it

get? Well, let’s see. When you’re only dealing with real

numbers, as we have until now, in the first instance it makes sense to speak

of THE cube root of this number, the one real cube root of this number.

However, when we’re dealing with non-real complex numbers,

there is no distinguished cube root anymore. Just as every nonzero complex

number has two square roots, every nonzero complex number has three cube

roots and none is distinguished in any way that makes it qualify for the job of

THE cube root. As we shall see this means that the expression up there does not

just stand for one of the real solutions down there, but for all three of them. One

final bit of weirdness. So Cardano’s formula outputs tame

integer, rational, etc. solutions in weird complex ways. That prompts the question

whether there is another totally different cubic formula that avoids this

weirdness. Right, there’s really no reason why there should not be several cubic

formulas. The answer to this question is both `no’ and `yes’. If by formula we mean a

combination of the coefficients using just basic arithmetic and square roots

and cube roots and the like, as an our cubic formula, the answer is `no’. Any such

formula cannot totally avoid the complex weirdness. This can be proved using a

heavy weapon called Galois theory. We skirted Galois theory in our previous

video on impossible constructions and it will also be the topic of a completely

insane, promise, completely insane Mathologer video in the near future. On the

other hand, if we allow ourselves to also use a bit of trigonometry, then we can

make up cubic formulas that always give real solutions in terms of real

expressions, avoiding the detour into the complex numbers. Anyway, to summarize,

except for the overcomplicated output of simple solutions, our cubic formula is

well-suited to dealing with cubic equations that have positive or zero

discriminant. That is, for equations that have exactly one or two solutions we

don’t need to enter the complex wilderness. Okay,

and that just leaves us with making sense of complex infested outputs in the case

of negative discriminates. Here’s one of the complex monsters

we created earlier. Let’s try to make sense of it. To begin we’ll just cross

our fingers and go for it without worrying too much about any worrying

details. Again, if you’ve never seen complex numbers, please just run with it

as best as you can. So we can always rewrite the square root in terms of the

square root of -1, like this. Now the square root of -1 is usually abbreviated

by the letter i. Marty behind the camera is shaping to throw something at

me. Okay, yes there’s a large nit you can pick here but please save your nitpicks

and projectiles for the comments in the end. It all works out ok. The two complex

numbers under the cube root only differ by the plus or minus sign in the middle.

That means they are complex conjugate numbers and so in the complex number

plane we picture them on opposite sides of the real axis. These two complex

numbers are also pinned down by this distance and this angle. That means we

can rewrite our complex numbers in terms of sine and cosine like this. There,

beautiful ! Ah, so nice. This way of writing complex

numbers is called the polar form. To find the roots of complex numbers written in

this form is very easy. First take the cube root of the distance like so.

Actually it simplifies quite nicely here. And then divide the angle by 3. Ok, there

roots coming up. Now, when we add up the two complex numbers up there, the

plus/minus imaginary parts cancel out, leaving us with a real number. That’s one

of our real solutions. But that’s not all. So far we’ve only used one cube root

each for the two terms in Cardano’s formula. There are two more each. All

these roots together from the corners of two equilateral triangles with centers

the origin. There’s one and there’s the second triangle. In total there are three

conjugate pairs among our six cube roots. Add up these conjugate pairs to get all

three solutions of our cubic equation. And just in case you’re wondering, and

this is a little trig challenge for you. Not `trick’, `trig’ 🙂 these numbers are

equal to these nice expressions. Great stuff, don’t you agree? I have to

admit that these videos are always a killer to put together but they are also a

blessing. While preparing them I always end up stumbling over lots of great

maths and amazing history of which before I only had the barest inkling,

always enough for another video – which I know I’ll never live long enough to

get to. So, in lieu of a second video here are just three fun facts

I stumbled across while working on this video, without explanation. Feel free to

give proving them a go in the comments. Fun fact number one. Here is a cubic with

three zeros. Draw in the tangents at those zeros. The tangents intersect the

graph in one more point each. Then those three points lie in a straight line.

Always! Nice, huh? Fun fact two. This is again about cubics with three zeros. Highlight

the inflection point. Draw an equilateral triangle whose centre is hovering

somewhere above the inflection point. Then it is always possible to rotate and

scale this triangle so that the three corners end up above the zeros. Pretty

neat. You can prove this easily based on what I said in the previous chapter. But

there’s more. Inscribe a circle into the triangle. Then the two extrema line up

with the left- and rightmost points of the circle like this. Final related fun

fact and this one is a real killer: Start again with a cubic polynomial but this

time the coefficients can be any complex numbers. So in general the three zeros of

such a polynomial will be three points in the complex plane, forming a triangle.

Highlight the midpoints of the edges of this triangle. It turns out there’s

exactly one ellipse that touches the edges in those midpoints. And here’s the

killer fun fact three: the two zeros of the derivative of our polynomial are the

focal points of this ellipse. This amazing result is called Marden’s theorem.

And one more thing is true. The zero of the second derivative, which is also the

z-coordinate of the “complex inflection point” is the center of the ellipse. And, phew, that’s just about it for today,

to finish I’ll just show you an animation of a method for solving quartic

equations. This solution is due to Ludovico Ferrari Cadano’s assistant. One of

the super surprising and famous results in algebra is that Ferrari’s method is as

far as we can go when we are asking for solutions in radicals.

There is no quintic formula of this type or a formula for anything beyond. As I

said proving this will be the mission of a future Mathologer video. For now enjoy

Ferrari and the animations and the music. Until next time 🙂

What font is he using for the equations?

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Today I hardly remember all this stuff but actually they teached me these formulas with demostrations at school in Italy in the eighties.

By lining up that cubic formula you assumed that -3uv is equal to p. This means that you can't choose u at will if you've already solved v. You have to divide p by -3v to find it. As for the other 2 roots, you wrote a quadratic equation to solve for v3. Well the cube root of the roots isn't the only solution for v, you can rotate v 120 or 240 degrees around the origin in the complex plane to find the other 2 solutions.

Could you please make a video about Martin's Axiom and why it fails for "aleph 1"? I find no good videos about Martin's Axiom in YouTube.

The cubes are really headache that's the reason I avoid that in the history of me. Anyway your video made me to think of rewriting my history. Thats why I tell people that your head is the best head on YouTube…….

到底是在学数学还是在学英语(ฅ•﹏•ฅ)

In my school in Brazil they teached the cubic formula, but didnt force us to remember everything on the test

I have an equation l came up with and I can't go beyond the 4rth or the 7th step of it I suppose there are infinite steps to it .

x.♾+0.1 – 0.1 = x+1×(-1^2) .

Where were you when I was studying this is college?

Just wonderful. Thanks a lot. Looking forward on your Galois video.

Mr. Mathologer. Thank you for your good work. I understand every single thought, every single word in every single second in this video and it is an extreme pleasure to follow the steps of these old math geniuses. The perfectness and beauty of math amazes me for years now. And it never stops. In math we see … god.

Wasn't Omar Khayyam the one who came up with the solution for the cubic equation?

The cubic and quartic formulae are tedious to apply, and very tedious to memorise. Computer algorithms find decimal approximations to the zeroes of polynomials very quickly, to as money decimal places as are desired. This capability applies to polynomials of arbitrary degree. Even Excel (via GoalSeek) will find the real roots of polynomials.

What is marvelous here are the deep reasons (Galois theory etc.) why no formulae exist for the zeroes of polynomials of degree higher than 4.

For the homework at 18:38 (figuring out the discriminant for the cubic):

cubic equation -> x^3 + px + q = 0

—

first derivative -> 3x^2 + p = 0

3x^2 = -p

x = +-sqrt(-p/3)

These are the extrema (x values)

—

[+-sqrt(-p/3)]^3 + p[+-sqrt(-p/3)] + q = 0

This gives us the y values for the extrema

—

[sqrt(-p/3)]^3 + p[sqrt(-p/3)] = green (difference in height between the inflection point and the 2 extrema)

—

| q | = absolute value of q

—

| q | > green -> 1 solution

| q | < green -> 3 solutions

| q | = green -> 2 solutions

—

q^2 > green^2 -> 1 solution

q^2 < green^2 -> 3 solutions

q^2 = green^2 -> 2 solutions

—

q^2 > {[sqrt(-p/3)]^3 + p[sqrt(-p/3)]}^2 -> 1 solution

q^2 < {[sqrt(-p/3)]^3 + p[sqrt(-p/3)]}^2 -> 3 solutions

q^2 = {[sqrt(-p/3)]^3 + p[sqrt(-p/3)]}^2 -> 2 solutions

—

(q/2)^2 + (p/3)^3 > 0 -> 1 solution

(q/2)^2 + (p/3)^3 < 0 -> 3 solutions

(q/2)^2 + (p/3)^3 = 0 -> 2 solutions

Edit: I just realized that "green" can be simplified into {[sqrt(-p/3)]^2 + p}[sqrt(-p/3)] -> [(-p/3) + p] * [sqrt(-p/3)] -> [(-p/3) + (3p/3)] * [sqrt(-p/3)] -> (2p/3) * [sqrt(-p/3)] -> sqrt(-4p^3/27). This is easier to substitute in.

–

q^2 = green^2 -> 2 solutions

q^2 = [sqrt(-4p^3/27)]^2 -> 2 solutions

q^2 = -4p^3/27 -> 2 solutions

q^2 + 4p^3/27 = 0 -> 2 solutions

(q^2)/4 + p^3/27 = 0 -> 2 solutions

(q/2)^2 + (p/3)^3 = 0 -> 2 solutions

–

which gives us the same final answer:

(q/2)^2 + (p/3)^3 > 0 -> 1 solution

(q/2)^2 + (p/3)^3 < 0 -> 3 solutions

(q/2)^2 + (p/3)^3 = 0 -> 2 solutions

😲😲 Why nobody told me that ?

His tshirt is cube root…..nice sending message to student

11:00 – For those watching in low quality..

SIR PLEASE HELP ME…….

I RECENTLY FOUND SOMETHING

Something that can be astonishing!!

Please tell me how can i show it to the world or i may share to u ………

Please help

I studied these on my own (I had some strange curiosity).

It was not easy material (or an easy method was not easily found back then — and I asked some smart math profs).

You made it look very simple. That's a pleasant illusion.

Thank you for covering this — one of the best math videos I've ever encountered.

I have just constructed the first Klein Rubicks cube : https://quantumantigravity.wordpress.com/unzicker/

Notice in English how Card-ano, a gambler and polymath (Renaissance type man … proficient in a number of fields) has had an abbreviated version of his name associated with our 'Card' deck, 'Card-inal' directions on a compass, status of papal prelates in the Vatican, a red bird and baseball team. Being something of a gambler, he predicted his own death; as the story goes, the day came around and it was a nothing-burger…so he committed suicide to be accurate in his prediction ! That's why we Go to the Clubs, to Pick up the Diamonds, and Win some Hearts in hope of Getting it all back in Spades !

Only when u ≠ v does x³ + p x + q ≡ 0 (cancellations by diff.o.sqrs, etc.); u=v gives (-3q ± 8√D) + 2 p ∛(-½ q ± √D), for Mthlgr’s (p,q) discriminant, D, all of which isn’t identically 0. Why not? I don't know, but try (p,q)=(3,2), (I think) it’s ≈10, not zero. I can’t see any more deeply or convincingly than that.

wolframalpha.com/input/?i=6%28%28sqrt%282%29-1%29%5E%281%2F3%29%2B8%2F6*sqrt%282%29-1%29

Please make the video about unsolvability of a quintic and higher soon.

I always wanted to learn Galois theory in a simpler way!

Thanks you and bless your bliss

The square root of a negative number is an imaginary convenience that on occasions turns out to be useful in reaching an acceptable solution. I often wonder why so many accomplished mathematicians are such devout atheists? … I am perfectly happy with the Boar Incarnation of Vishnu, The Supreme personality of Godhead. "Lord Varaha" https://en.wikipedia.org/wiki/Varaha These ancient sages say that Planet Earth was stuck in a messy part of the universe, so this boar selected a suitable Star with a Solar System to park up Planet Earth, about 4.32 Billion years ago. That sounds perfectly reasonable to me. Also Jesus The Anointed Celestial Treasurer, "Lord Isah" As Jesus is named in India.

Bg 10.23

rudrāṇāṁ śaṅkaraś cāsmi

vitteśo yakṣa-rakṣasām

vasūnāṁ pāvakaś cāsmi

meruḥ śikhariṇām aham

Synonyms:

rudrāṇām — of all the Rudras; śaṅkaraḥ — Lord Śiva; ca — also; asmi — I am; vitta-īśaḥ — the lord of the treasury of the demigods; yakṣa-rakṣasām — of the Yakṣas and Rākṣasas; vasūnām — of the Vasus; pāvakaḥ — ﬁre; ca — also; asmi — I am; meruḥ — Meru; śikhariṇām — of all mountains; aham — I am.

Translation:

Of all the Rudras I am Lord Śiva, of the Yakṣas and Rākṣasas I am the Lord of wealth [Kuvera], of the Vasus I am ﬁre [Agni], and of mountains I am Meru.

The "Rākṣasas" are always creating a hellish situation, so Jesus or "Isah" "Jeshua" and a dozen or more other names. has to come and rescue the Atheist Mathematicians. who use their knowledge to become exceedingly wealthy. That causes a disturbance in society. Rival Mathematicians start Nuclear Wars. Then we end up with a planet full of mutant rats and cockroaches. etc.

https://en.wikipedia.org/wiki/Rakshasa

Cause its Freemasonry.

Are you guys bad at factoring or something? The 1st example could have been solved at a glance by factoring, and no need for trig; 2+2i is (i-1)^3

learned the quadratic formula literally today in math

I call it dumbing down society

hit like on 1:05

way too hard… whyyyyy?

Can some one tell how to solve 12.18 with algebra

More useless number flipping. Just like in school.

Not everything needs to be taught to you. A truly educated person takes enough interest to learn things on their own. School, at best, only gives you a taste of how to begin to self-study. Anytime I hear somebody say, "why don't they teach this in school?' I think, "why don't you have enough interest to go beyond what your teachers tell you?'

Idk why I’m learning this, I know damn well I’m going to get the quadratic formula confused with this somehow 🤷🏾♂️

okay, this is my 11:57pm math class!

Or you can just plot the points of the line and find where it intersects 0…

Who are "they"?

0 is everything

They didn't teach because from a physics perspective its nowhere near as useful.

i guess i am just stupid but i hit a brick wall when he said we might as well just make B equal to zero, I cant for the life of me see how he can do that, i know there is a trick in there somewhere but i cant see it

guess what, i NEVER learned that formula, still i can solve these equations. This formula is useless, just for teachers to test their students.

I enjoyed the 'Cubic Formula' video. Please make a sequel video on these :

quartic, quintic, sextic, septic, octic, nonic, decic, undecic, dodecic, tredecic polynomials, ….

The problem with 100% of all demonstrations of the cubic & quartic is that they are all identical in considering real coefficients only. We need examples with complex-valued coefficients.

Can someone explain to me how to obtein the expression (q/2)^2 + (p/3)^3? thank you

7:18 "..and a cameo appearance of the Inquisition"

NOBODY EXPECTS THE SPANISH INQUISITION!It's not taught in High School because we are required to focus only on what is on Standardized Testing. ugh.

7:23 Well I sure wasn’t expecting it…

I like how he changes his shirt in the video. Subtle. Smooth.

25:18

Pretty simple, bring the numbers to polar form. You're welcome 😉

Hi Mathologer, Can you please write a formula for the below Fibonacci expression.

Basically the fixed number plus 10 percent of itself plus 10 percent of the number above it, equals the number below the number you started with, in a decimal form. Must be in a 5 column sequence. Being able to find a vertical answer, I would like to invert the formula to apply it and assume the same numbers be inverted before zero with a negative expression. Forgive the lack of knowledge, I'm no math expert of such. Not even amateur!

https://www.youtube.com/watch?v=mRaAe6SswSE

I wish these were taught in the Schools.

His shirt has square roots

Thank you very much for another exhaustive treatment. Just like WelchLabs' on the imaginary number.

Actually, everything I've seen has m and n in the reduced cubic and under the radicals, with p and q equal to the cube roots.

No big deal I suppose.

This is the greatest video talking about cubic equation!

cool content love that

Great, even math is glitched now -.-

Is there an algorithm or generalized formula (formulae) that yields any n-dimensional to the exponent n formula? Geometrically I can visualize that it should exist. However, you said that a quintic formula can be proved to not exist.

I simply use sythetic division method to convert the equations into quadratic form and find roots 🙂 no need 2 memorise that formula anyway