The most fun way of solving this quartic equation
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- Опубликовано: 1 окт 2024
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I created this quartic equation x^4-x^2-2x-1=0 and we will solve it by using the quadratic formula three times! Check out how we can factor x^4-x^2-2x-1 by using the double-cross method: • Solving x^4-x^2-2x-1=0...
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the words you're looking for is "roots of unity" at 5:26
Hey man can you please solve the question on your channel banner
Lots of love ❤
What's funny is that you can actually rearrange this equation into x^4 = (x+1)^2. Then you take the square root of both sides and consider the two quadratics x^2 = ±(x+1), which are the same if you had used the quadratic formula. Idk if something like this is possible whenever the discriminant of the first "quadratic" simplifies nicely, but would be interesting to investigate.
Edit: I just realized you could also write the equation as x^4 - (x+1)^2 = 0 and apply difference of squares to get (x^2+x+1)(x^2-x-1) = 0 which again gives the same quadratics.
Noticed the same but bprp's method was to match the theme
Quadratic formula but
a, b, C are f(x)
@@aaryavbhardwaj6967 Yes I know that but bprp also mentioned that the equation was arranged so that the sqrt(discriminant) term would simplify nicely. I was wondering if alternate solutions like the ones I mentioned always exist for these prearranged polynomials.
The second method is how I just solved it.
you just completed the square
I too thought of the same when I first saw this video.
x⁴ - x² - 2x - 1 = 0
x⁴ - (x² + 2x + 1) = 0
x⁴ - (x + 1)² = 0
(x²)² - (x + 1)² = 0
(x² + x + 1)(x² - x - 1) = 0
Then solve with quadratic formulae to get x = ½(1 ± √5) and ½(1 ± i√3).
sameee
Tbat is the logic and math is logic.great job.
Yes, this solution comes at first glance, but of course you can play
with it.
5:25 The cube roots of unity :)
It's funny sometimes we forget the everyday thing
Did you know it’s possible to prove sqrt(4) is irrational? Assume sqrt(4)=a/b w/ no common factors. Square both sides, 4b^2=a^2, meaning (2b)^2=a^2. So a=2b. Thus our original fraction square root of 4 = (2b)/b which has a common factor of b. Contradiction. Thus sqrt of 4 is irrational
@@thenew3dworldfan
When defining a rational number R, we say R=a/b where a and b are coprime integers. In other words, a and b have no common factors other than 1.
Your proof does not account for 1. The supposed common factor "b" you have found is the number one
@@thenew3dworldfan "So a=2b." - And thus sqrt(4), which you defined as being equal to a/b, equals 2b/b = 2. This is because all integers have at least two common factors: the so-called "units", which are 1 and -1, so your premise of "no common factors" cannot be true. If we restrict ourselves to the study of positive integers only (not all integers), then the only "unit" is 1.
This use of the word "unit" appears in ring theory, where it doesn't actually mean "an element whose magnitude is 1", but rather it means "an element of the ring (algebraic structure) we're considering whose multiplicative inverse is also an element of the ring". A fun, weird quirk of this definition is that if the ring we're considering is the rational numbers, then all elements of the ring are units, making "unit" a bizarre name in this context. But usually/traditionally, number theorists exploring ring theory are/were studying integer rings like the positive and negative integers, or Gaussian integers (where the units are 1, -1, i, and -i), or other polynomial rings of Z.
Can you do sin(cos(x)) = cos(sin(x)) in the next video?
the golden ratio as well as cube roots of unity, this is fantastic stuff
Beautiful!
This quartic is a difference of squares and very easily solved. The method of "abusing" the quadratic formula almost always makes the original equation more difficult to solve, but in this case it happens to work out neatly.
In fact, it only works out at 2:30 because 2^2-4(1)(1)=0. More generally, it can only simplify for ax^2+bx+c-x^4=0 because b^2-4ac=0, meaning that the original equation has to be a difference of squares for this to work.
I believe he used this method in a previous video
Next, solve polynomial with degree 6 with cubic equation 😎
It's 5am and I'm watching a math video lol rip.
x^3 term: allow me to introduce myself
actually you can always get rid of the x^3 term with a substitution
@@Adrian-cq2tt not always. The coeffecients for x⁴ and constant should be same. And also coeffecient of x³ and x should be same. These are not rules given by a book just a logic. There is no other way to make substitution if not for tjese.
@@ThorfinnBus It’s always possible through the substitution y = x - a/4, where a is the coefficient of the x^3 term. Ferrari himself only considered a depressed quartic for his method of solving a general quartic because it’s always possible to go from one to the other
If I am not wrong, this works only if b^2 = 4*a*n, where n is the constant in c (n+x^4). This way, we can get rid of the square root.
Pour raccourcir la rédaction remarquons que cette équation peut s'écrire:
x⁴ = (x+1)² ce qui donne de nouvelles équations du second degré: x² = x+1 ou x² = -x-1. Etc.
Une question se pose: ne serait-ce pas cette particularité qui permet d'utiliser ta méthode ?
J'adore toujours autant tes tutos 👍👍👍
Just put x= omega(w) and boom.
This is my 1/0 timeth seeing brilliant on a black pen red pen video💀
Which 1/0? +∞, -∞, ∞i.. what's the argument
@@fsponjyes
My -1/12 th
Using this, we can derive an ULTIMATE Fibonacci sequence defined as:
a0 = 1
a1 = 1
a2 = 1
a3 = 1
a(n+4) = a(n+2)+2*a(n+1)+an
So, the sequence will be
1, 1, 1, 1, 4, 4, 7, 13, 19, 31, 52, 82...
compared to the original Fibonacci sequence of 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 231, 375, 606, 981...
where the ratio of a(n+1)/an as n goes to infinity also approaches the Golden Ratio.
5:19 the phrase you were thinking of was "the cube roots of unity" or possibly "the non real cube roots of unity"
OMG!!!? PHI, 2nd n 3rd cbrt of Unity all 3 as answers????????
What are those ?
before watching: convert to difference of two squares (x^2)^2 - (x+1)^2 then two quadratic factors [x^2 - x - 1][x^2 + x + 1] = 0 . Real roots are phi, Complex roots (-1+- iRoot3)/2
Please anyone solve 4+n=2^(n-1)
i enjoy your videos as you methodically slay these problems. In light of this possible "discovery" here's a merch idea for a shirt( with the appropriate graphics) : Quad-Form Man: Crushing Unruly Quartics Since 2024
Why keep this one unlisted?
Foarte interesant acest mod de a rezolva această ecuație, nu l-am mai întâlnit până acum. Felicitări. Succes in continuare.
Am i the only one who noticed at 3:50 he said "Quadratic Formula aa gaya"
now i can't unhear it 😭😭
he said 'again' but it does sound like 'aa gaya'
All indians want to know you location
x^4-(x^2+2x+1)
=x^4-(x+1)^2
=(x^2-(x+1))(x^2+(x+1))
=(x^2-x-1)(x^2+x+1)
Perfect square trinomial and difference of squares. I wonder which quartic equations are solvable via your method.
That means the original equation allows a pretty simple factorization. Either notice:
x^2 + 2x + 1 = (x+1)^2, so x^4 = (x+1)^2
or
(x^2 + x + 1) (x^2 - x - 1) = x^4 - x^2 - 2 x - 1
which has a simple solution in terms of quadratics. There's 2 real and 2 complex roots.
The person who made the quadratic formula would not have even though of this lol
No they certainly wouldn't because it was discovered over 4000 years ago during a time when math was linked to geometry and the modern algebraic notation that we use did not exist.
You’d think the inverse of f(x) = x(x+1) would be √(x+√(x+√(x+…))). If you solve it it would be (√(1+4x)-1)/2, but the closest recursive formula is √(x+√(x+√(x+…))) - 1; this doesn’t apply at x = 0 (or x ≤ 0? Idk), but where does the -1 come from???
I always try the problems before watching the videos. After watching your last video, I had a clue how this would work, so I factored out x^2-1 from the quartic and quadratic terms and used the quadratic formula with a=x^2-1, b=-2, and c=-1. It works just the same as far as resulting in the 2 same quadratics, x^2-x-1=0 and x^2+x+1=0. I love this new to me use of the quadratic formula and I would not call it ABUSE. Did you invent this? If so, congratulations!
You can do more you can take x=2 and get the answer. That is Lapo method. If i remember. YOU CAN EVEN TAKE X=1 AND IGNORE THE OTHER X'S AND GET THE SOUTIONS. THATBIS CALLED THE SLEEPING DOG IF I REMEMBER
IT DOES NOT WORK
Nice eq. I suggest this solve
x^4 - x^2 - 2x - 1 = 0
x^4 - (x+1)^2 = 0
(x^2-x-1)(x^2+x+1)=0
x(1,2) = (1+-5^0.5)/2
x(3,4) = (-1+-i3^0.5)/2
Does this method works for others cases?
Because this case is easier by addition by diference
x^4 - x^2 - 2x - 1 = 0
=> (x^2)^2 -(x^2 + 2x + 1) = 0
=> (x^2)^2 -(x+1)^2 = 0
=> (x^2 + x + 1)(x^2 - x - 1)=0
And resolve two equations of second grade
Hmm...I don't know what that would mean, lol, to abuse the quadratic formula, but maybe one can rewrite it as x⁴- (x²+2x+1)=0 => x⁴- (x+1)²=0, so, the difference of two squares, that would be (x²-x-1)(x²+x+1)=0, etc...just set each factor equal to 0, regular quadratics, etc...hopefully I didn't make some silly mistake...
This equation is scary, but we can see, that x2+2x+1 = (x+1)2 =>
(x+1)2-(x2)2 = 0 =>
(x+1-x2)(x+1+x2) = 0
Second equation has no solution in real numbers. First equation has two solutions.
It is easier, than abusing the quadratic formula.
Just a small doubt on my part. while we derive the quadratic formula, it is usually established that a is not zero(since it wouldn't be a quadratic if it was). However if a, b and c can be functions in x(How does this generalize to all functions? Piecewise functions too?) zero(or any root of a) can be a perfectly legible root for the equation but the denominator for the quadratic formula can become undefined. Eg. (x-2)*x^2 + x - 2. I am really curious about this idea, but I don't think we will be able to derive powerful conclusions without rigorously answering these questions. I highly enjoyed your video nonetheless. Thank You.
DISCLAIMER : This is a repost of my comment on the previous video posted, but I am also putting this here because I believe it would be more likely that you read it here.
Good evening sir from India , i have recently discovered a new method for finding 1st root of a quadratic equation ,, i would like you
where can i contact you (i dony use insta or facebook )😊
i'm in class 8 or grade 8 or standard 8. so I don't know derivatives, differentials, integrations. but I know complex numbers, some trigonometries, logarithms and other things of grade 9-10. blackpenredpen helped me a lot to understand these. still I don't know any calculus topic. i have invented some formulas and theorems like 3d trigonometry with 2 angles and 12 ratios, product of some infinite sums, relation between all types of means/averages like quintic>quartic>cubic>quadratic>arithmetic>harmonic etc. recently I'm trying to make quintic formula with radicals, because I don't know calculus and some special functions. many people thinks , it's not possible and they already have given the proof. but I think there's still a probability that we can make quintic formula with only radicals. blackpenredpen, can you help me by explaining without calculus that why this is'nt possible? please make a video on it.
You could have more easily done it as the difference of 2 squares. (X squared all squared and x+1 squared. Gives the product of 2 quadratics, solved as per original
x^4-(x+1)^2=0, then we use the square difference formula and get the roots
SIR, I AARUSH YOUR BIG FAN FROM INDIA. SIR, I LIKE CALCULUS THE MOST. SO, COULD YOU PLEASE UPLOAD A VIDEO ON A TOUGH INDEFINITE INTEGRAL.
Can anyone please explain me the relation between the eqn, and golden ratio. I mean its amazing that they are linked together, but I want a visual explanation, or possible graph.
Finally bprp came up with C as C(x) for quadratic formula
You could probably use the same method to solve certain quintic equations.
I presume quadratic formula lawyers still did not appear just to sue you
Can you do the same with cubic formula?🤔🤔
I have a moral question for you
Before proposing this method, if you put this problem in one of your tests, and one of your students worked out his way through this method, would you have accepted his soluation and granted him the grade?
I would give the student full credits!
NICE
Keralites were shocked at the intro sponsor, until we saw the logo😂
The complex solutions are cube roots of 1.
Math is the way of expression in different ways of terms.
Does this work on cubic aswell?
Can't we just substitute x^2 as t and then solve ?
3:45 shouldn’t it be x^2 - x + 1?
You can do dots to immediately factor it. You have x²+2x+1 - x⁴ → (x+1)² - x⁴ → ( x²+x+1)(-x²+x+1)
One way I like to derive the quartic equation is to split the reduced quartic into a difference of squares of the form (x^2+ax+b)^2 = c(x+d)^2. In order to solve for the coefficients in this equation, you need to solve a cubic 'determinant' equation. Of course in this case, the coefficients are trivial.
Sir please make videos on sieve theory
If you have x^3 in an equation like this, can you say that x is the coefficient of x^2 and use the quadratic formula, substituting a, b, or c as x and then solving for x?
okay i literally thought we getting to the point that we would start getting a quadratic formula inside a quadratic formula smth like that but luckily still not i guess..
Nice method - BTW does it only work for depressed quartics?
UTTPs with childs be like
I need an explanation for why this works in the first place, that is taking x^4 as a constant in a quadratic equation which itself is in terms of x?
What if we solve a quintic equation with the quadratic and cubic formulas?🤔
My kind of ASMR
Can we have the quadratic equation with a equal to x square minus 1?
The line right before the quadratic formula is clearly a difference of squares. Thus we factor immediately to the 2 quadratic you solve.
That's what he said already at about 0:30, followed by "I'm not going to do that".
Proof quartic formula
Proof quartic formula
Proof quartic formula
Proof quartic formula
Proof quartic formula
Proof quartic formula
This was posted on members only right whyhere
x!=i^x+3
That smile as he breaks math 😊😅😂
Sir, How to know the number of rows for DI method
ruclips.net/video/N1KLaLi_LjA/видео.html
just do it until it becomes simple or repeats
it might get more and more complicated instead, which may mean you need to divide what's D and what's I differently, but that comes from experience
but for simplest cases:
x^n needs n+1 rounds of D to become 0
e^x neess one round to repeat
and sin, cos, tg, ctg need 2 steps
You add the rows as you go along??? Or you can predict it. 2-3 is a good start.
Maths is half abusing notation and half cheating notation
The real mystery created by maybe is that so much of the abuse of notation turns out to work anyway...
This becomes even more obvious when you progress onto stuff like simplifying tensors in general relativity. The rules for simplification are so trivial that it's hard to believe they actually give you an answer that matches experiments.
This is sometimes referred to as "the unreasonable applicability of maths to the real world".
Cube roots of unity
And the golden ratio
tq the rock
Ok
Damn! You looked so much younger in previous videos😢
this is so cool
He's onto something🔥🔥🔥🔥
That's at the same level of maths as 64/16 = 4/1 by simplifying the 6... It works on one example without a particular reason and has mathematically no sense ...
Not really, this trick will always give you a valid equation and you are allowed to treat variables as constants. Eg. (Partial derivatives). Bare in mind, most of the time it just makes the equation harder to solve, but it is a neat trick when working with specific equations.
Mathematically it is completely valid. But it needs kind of contrived examples for it to be useful, as the square root complicates things.
@@be-mr3tj Okay I understand why you can do it, but it's not because you can treat x as a constant ... It's just the way it's proven is just factorisation.
Nevertheless I think it's really better to factor it ourselves with
(x²)² - (x+1)² = (x²-x-1)(x²+x+1) = 0
So we see why the simplification works