How to depress a cubic

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Күн бұрын

Пікірлер: 73

@holyshit922 6 ай бұрын

It is possible to generalize this method for quartic (Euler did it) To solve depressed cubic you can substitute x = a+b but to solve depressed quartic substitution would be x = a+b+c Let see what we will get after this substitution x^4+px^2+qx+r x = a+b+c x^2 = a^2+b^2+c^2+2(ab+ac+bc) x^2 - (a^2+b^2+c^2) = 2(ab+ac+bc) (x^2 - (a^2+b^2+c^2))^2 = 4((a^2b^2+2a^2bc+a^2c^2)+2(ab+ac)bc + b^2c^2) x^4 - 2(a^2+b^2+c^2)x^2+(a^2+b^2+c^2)^2 = 4(a^2b^2 + a^2c^2 + b^2c^2 + 2(a^2bc + ab^2c + abc^2)) x^4 - 2(a^2+b^2+c^2)x^2+(a^2+b^2+c^2)^2 = 4(a^2b^2 + a^2c^2 + b^2c^2 + 2abc(a+b+c)) x^4 - 2(a^2+b^2+c^2)x^2+(a^2+b^2+c^2)^2 = 4(a^2b^2 + a^2c^2 + b^2c^2) + 8abc(a+b+c) x^4 - 2(a^2+b^2+c^2)x^2+(a^2+b^2+c^2)^2 = 4(a^2b^2 + a^2c^2 + b^2c^2) + 8abcx x^4 - 2(a^2+b^2+c^2)x^2 - 8abcx + (a^2+b^2+c^2)^2 - 4(a^2b^2 + a^2c^2 + b^2c^2) = 0 So after comparing coefficients we have following system of equations -2(a^2+b^2+c^2) = p -8abc = q (a^2+b^2+c^2)^2 - 4(a^2b^2 + a^2c^2 + b^2c^2) = r This system of equations can be easily transformed into Vieta formulas for cubic with the roots a^2 , b^2 , c^2

@holyshit922 6 ай бұрын

Although such generalization is possible i prefer to solve quartic by factoring into two quadratics For depressed quartic undetermined coefficients (x^2 - ax + b)(x^2 + ax + c) = x^4 + px^2 + qx + r works nicely but when given quartic is not already depressed i prefer to use differece of squares first

@KPunktFurry 6 ай бұрын

@@holyshit922 sounds easyer

@kamra702 6 ай бұрын

Next video on making the cubic cheerful please!!

@GoshISuckAtLife 2 ай бұрын

Yessss.. we need more positivity😂🤩

@xristos.l4259 6 ай бұрын

Why is it sad?

@Samir-zb3xk 6 ай бұрын

It fell into depression after the x² term left 😔

@deimos13710 6 ай бұрын

How to depress a cubic? just show it a reality of life

@Kaptoshka-bs6fj 6 ай бұрын

I recently solved very interesting integral and want you to try it out Integral of sqrt(x^2+a) "a" is a constant

@vedwargantiwar4610 6 ай бұрын

Isnt it done by integration by parts?

@thibautmadjio-xp8vw 6 ай бұрын

I think it will give a hyperbolic inverse, am I right? 🤔

@pandee6923 2 ай бұрын

x/2 *sqrt[x^2 + a] + a/2*log |x + sqrt[x^2 + a] | + C

@caelu.m 6 ай бұрын

person named "Acubic":

@savitatawade2403 6 ай бұрын

why are the views so down nowadays? you are so underrated! hope everything gets better

@baskernatarajan3824 6 ай бұрын

Actually

@Orillians 6 ай бұрын

Woah I never noticed the view counter ever, the videos he makes are SOO SOO SOO good regardless

@iqtrainer 6 ай бұрын

Thats what KZbin algorithm is sp unpredictable. There are still some math youtubers deserving more views including PN!

@iqtrainer 6 ай бұрын

@@OrilliansGood to see you here

@Orillians 6 ай бұрын

@@iqtrainer Do you know me from somewhere?

@Space_Doge. 6 ай бұрын

what if i want to make it happy

@NadiehFan 6 ай бұрын

Your claim at 17:20 that the other two complex conjugate roots of your depressed cubic t³ + ²⁄₃t + ²⁰⁄₂₇ = 0 are obtained by multiplying the real root t = −²⁄₃ by ω = −¹⁄₂ + i·¹⁄₂√3 and by ω² = −¹⁄₂ − i·¹⁄₂√3 is incorrect, that is _not_ how it works. See my detailed comment on your previous video about solving depressed cubic equations. In fact, what you should do to obtain the other two roots is multiply one of the two cube roots by ω and the other by ω² _in either order._ So, since we have ³√(−10 + 6√3) = −1 + √3 ³√(−10 − 6√3) = −1 − √3 the roots of your cubic in t are t₁ = ¹⁄₃((−1 + √3) + (−1 − √3)) = ¹⁄₃(−2) = −²⁄₃ t₂ = ¹⁄₃((−¹⁄₂ + i·¹⁄₂√3)(−1 + √3) + (−¹⁄₂ − i·¹⁄₂√3)(−1 − √3)) = ¹⁄₃(1 + 3i) = ¹⁄₃ + i t₃ = ¹⁄₃((−¹⁄₂ − i·¹⁄₂√3)(−1 + √3) + (−¹⁄₂ + i·¹⁄₂√3)(−1 − √3)) = ¹⁄₃(1 − 3i) = ¹⁄₃ − i

@NadimShawky 3 ай бұрын

I might just be missing something, but how does cube root(-10 + 6root(3)) end up being -1+root(3)? Please walk me through it

@NadiehFan 3 ай бұрын

@@NadimShawky Well, you could of course verify this by simply cubing −1 + √3. Using the identity (a + b)³ = a³ + 3a²b + 3ab² + b³ we have (−1 + √3)³ = (−1)³ + 3·(−1)²·√3 + 3·(−1)·(√3)² + (√3)³ = −1 + 3√3 − 3·3 + 3√3 = −1 + 3√3 − 9 + 3√3 = −10 + 6√3 But there are systematic ways to denest cube roots of quadratic surds such as ³√(−10 + 6√3) and ³√(−10 − 6√3) provided these nested roots can indeed be denested (which is not always so). As an example, let's denest ∛(2 + √5). Assume there exist _rational_ numbers x and y with √y _irrational_ such that (1) ∛(2 + √5) = x + √y then we must also have (2) ∛(2 − √5) = x − √y From (1) and (2) we have x² − y = (x + √y)(x − √y) = ∛(2 + √5)·∛(2 − √5) = ∛((2 + √5)(2 − √5)) = ∛(2² − (√5)²) = ∛(4 − 5) = ∛(−1) = −1, so (3) x² − y = −1 Also from (1) the cube of x + √y must equal 2 + √5 so we have (x + √y)³ = x³ + 3x²√y + 3xy + y√y = (x³ + 3xy) + (3x² + y)√y = 2 + √5 which implies (4) x³ + 3xy = 2 From (3) we have y = x² + 1 and substituting this in (4) we get x³ + 3x(x² + 1) = 2 which gives (5) 4x³ + 3x − 2 = 0 Since x must be rational, we are looking for rational solutions of this cubic equation. According to the rational root theorem, for any potential rational solution m/n or −m/n of (5), m must divide the absolute value of constant term, which is 2, and n must divide (the absolute value of) the coefficient of the cubic term, which is 4, so m can be 1 or 2 and n can be 1 or 2 or 4. However, we can also see that (a) the polynomial on the left hand side of (5) is strictly increasing on the real number line and (b) the left hand side of (5) is negative for any negative x. This means, first of all, that (5) can only have a single real solution and, secondly, that any real solution and _eo ipso_ any rational solution must be positive. So, we only need to test ¼, ½, 1 and then we quickly find that x = ½ is a solution and consequently the only real solution of (5). With x = ½ we have y = x² + 1 = (½)² + 1 = ¼ + 1 = ⁵⁄₄ and so we have ∛(2 + √5) = ½ + √(⁵⁄₄) which can also be written as ∛(2 + √5) = ½ + ½√5 Of course (1) and (2) imply that we must also have ∛(2 − √5) = ½ − ½√5 Try this method yourself for ³√(−10 + 6√3) and ³√(−10 − 6√3), then you will find that ³√(−10 + 6√3) = −1 + √3 and ³√(−10 − 6√3) = −1 − √3. Reference: Kaidy Tan, Finding the Cube Root of Binomial Quadratic Surds. _Mathematics Magazine_ Vol. 39, No. 4 (Sep., 1966), pp. 212-214 (JSTOR 2688084)

@NadimShawky 3 ай бұрын

@@NadiehFan thanks, that explains it really nicely

@KPunktFurry 6 ай бұрын

hello first title sound intersting but maby the cubic needs mental help? xD 0:47 but if i can depress a cubic it has to be posible to get back the original form so in terms of the last video wouldn´t that be easyer ? 1:12 what is not posible?? how ever. 2:22 i think if you use the normal ruls of equantions it is unimportent wich number you choose it will work anyway! 3:19 ok i understand how it works but not whatfor you need it! 3:58 that way of using t i have seen it somewere befor and why you choose t and not z or r or what ever? 7:10 ok yes thats quite simple at this point 8:34 i think i schould learn the solution for (a+b)³ 9:53 thanks for simplify it for me! 10:35 wouldn´t it be easyer to write (t/3) ? 11:28 i think i would had simplifyed the sum in the brackets before instead of writing it out but allright! 11:51 ok yes now it´s fine! 13:51 yes i remeber that! but anyway less you if you wan´t to use that 15:55 ok much fun to calculate that with out an calculator xD 16:34 oh ok now it looks posible 17:34 allright 17:41 an could you please explain me why do you add quotes at the end? 17:42 and please also explain me the quote it self! yours sincerly K.Furry

@holyshit922 6 ай бұрын

There is Tschirnhaus transformation which allows to get rid of terms x^{n-1} , x^{n-2},x^{n-3} from polynomial

@xotlxoltyna 26 күн бұрын

Depress a cubic by showing it the cubic formula - we've got you all figured out bud

@boguslawszostak1784 6 ай бұрын

I prefer x=a+b , I find it easier to memorize. (a+b)^3+(a+b)^2+(a+b)+1= a^3+a^2 * (3b+1)+a*(3b^2+2b+1)+(b3+b2+b+1) We need the coefficient of a^2 to be zero, which means (3b+1)=0 b=-1/3 (a + b)^3 + (a + b)^2 + a + b + 1 = a^3 + (2 a)/3 + 20/27

@benhassineiem 4 ай бұрын

I always enjoy the intro, every time👍👍

@SYN_synonym 6 ай бұрын

Misread as ‘how to make a child depressed’ and still clicked

@Nishchaya.01stha 6 ай бұрын

Can this method be still used if all the roots are real? For example I have a cubic equation x³-10x²+31x-30=0 Then t³-7/3t-20/27=0 after this I am having trouble for computing t. Can anyone help me out?

@jendf654 5 ай бұрын

Just divide original by x-5, no depresing

@davidbrisbane7206 3 ай бұрын

You can use the quadratic formula to solve for t. It will be a real value, but you might find the t is the sum of two complex numbers, which simply to be a real number.

@aashsyed1277 Ай бұрын

@@davidbrisbane7206 cubic*

@Th3OneWhoWaits 6 ай бұрын

Question: he said to multiply by the cube root of unity to get from -1 to the other two answers, what does he mean?

@NadiehFan 6 ай бұрын

Don't bother, what Prime Newtons claimed is wrong. See my main comment on this video for an explanation how to do this correctly.

@Th3OneWhoWaits 6 ай бұрын

@@NadiehFan thank you kind viewer.

@PsYcHoCI2usHeI2 6 ай бұрын

How in the world did you get minus 2

@NadiehFan 6 ай бұрын

We have ³√(−10 + 6√3) = −1 + √3 ³√(−10 − 6√3) = −1 − √3 so we get ³√(−10 + 6√3) + ³√(−10 − 6√3) = (−1 + √3) + (−1 − √3) = −2

@dirklutz2818 6 ай бұрын

if 6sqrt(3)-10 = (sqrt(a)-b)³ then you get 6sqrt(3) -10= (a+3b²)sqrt(a) - (3ab+b²). You see immediately that a=3 (sqrt!) and after solving (a+3b²)=6 you get b=1. So, (sqrt(a)-b)³ = (sqrt(3)-1)³ = 6sqrt(3) -10 and therfore sqrt(3)-1 = (6sqrt(3) -10)^(1/3) The solution of (-6sqrt(3) -10) yields -sqrt(3)-1. The sum (sqrt(3)-1) + (-sqrt(3)-1) = -2

@MyUploads-f3v 5 ай бұрын

multiplying (x-1) on both sides taking x ≠0 ⇒ x= e^(2πi/4) x= e^(πi/2) i=1,2,3 x=i,-1,-i

@lukaskamin755 6 ай бұрын

Still I would like to see the whole process, anв where those formulas for 2 other roots emerge (I've seen at least 2 variantsб maybe more). And, perhaps I missed smth, but i didn't get how we see how many roots there are. Also I'm wondering how Cardano (or Tartaglia) found all the three solutions without using complex numbers.

@darcash1738 6 ай бұрын

I legitimately memorized both the cubic and quartic before but I think I memorized the quartic wrong bc the font of the thing I was looking at was too small

@lawrencejelsma8118 6 ай бұрын

Another KZbinr shows a cubic formula derivation of one complex conjugate from a pair root found in ax^3 + bx^2 + cx + d = 0 whether or not it was a real root from the more than one real roots possibility. Because a cubic polynomial is at least from a real root (x + r) multiplied to a quadratic polynomial, the cubic root formula is related to if the resulting quadratic polynomial had real two roots or two imaginary numbers. The more versatile cubic formula is that of ax^3 + bc^2 + cx + d root formula.

@moeberry8226 6 ай бұрын

Great video brother but I want to point out that you can complete a cube it’s analogous to completing the square. This was done over 500 years ago. Scipione Del Ferro and others have done it.

@NadiehFan 6 ай бұрын

Yes. Read my comment on Prime Newton's video about solving the general depressed cubic equation x³ + px + q = 0.

@childrenofkoris 4 ай бұрын

if all the cubic roots have 3 or 2 with different root values, how can we apply this formula?.. OR we assume that all the roots for the cubic equation will have the same value?.. because what i can think of, the other options to find the cube root is by factoring the cubic equation

@childrenofkoris 4 ай бұрын

i gone through on minute 13.. and yes u already stated its only for one single root on a cubic equation with real numbers

@prakashlakhapate1598 6 ай бұрын

Can you give proof of formula for root of cubical equation?

@justgold1 Ай бұрын

p = (3ac - b^2) / 3a^" q = (2b^3 - 9abc + 27(a^2)d) / 27a^3

@holyshit922 6 ай бұрын

How to use cube root of unity to find all solutions ? kzbin.info/www/bejne/onnIZ5yfmtateNk Suppose omega is principal cube root of unity to get solutions you multiply one cubic radical by omega and the other one by omega^2 If you look at system which you get after solvig depressed cubic p = 3ab , -q = a^3 - b^3 it would be clear What about casus irreducibilis in my opinion it is worth considering this case because after using de Moivre theorem you will get trigonometric solution from complex cubic radicals

@lubiemuze6368 6 ай бұрын

I don't wanna cubic to be sad MONSTER😢

@moeberry8226 6 ай бұрын

This can be solved with factoring by grouping no need for depressing this cubic. The answers are i and minus i and -1.

@shasbucks 6 ай бұрын

I must be a cubic.

@AlexanderofMiletus Ай бұрын

Just tell it no one loves it, no need to overcomplicate things.

@cremath 6 ай бұрын

depress?? 😂

@pedropiata648 6 ай бұрын

😂

@afuyeas9914 6 ай бұрын

One way I never see is to set x = (t-b)/3a, which also eliminates the quadratic term but scales the cubic to only have integers as coefficients and a monic leading coefficient. The numbers are bigger but working with integers is much nicer than working with fractions.

@robertveith6383 6 ай бұрын

You wrote that incorrectly. The denominator would be inside grouping symbols: x = (t - b)/(3a).

@user-gr5tx6rd4h 3 ай бұрын

Should it not be x = t - (b/(3a))? If the original coefficients were integer, the new ones will still be? For 2x^3 - 6x^2 + x + 6 = 0 we get x = t + 1 and 2 t^3 - 5t + 3 = 0. (One solution for x will be 2, with t = 1.)

@orionfacts 6 ай бұрын

Why are you making it sad? 😭😭

@mmfpv4411 6 ай бұрын

I commend you on not making a depressed cubic joke in this video. If humor is a low hanging fruit most people can't resist picking a fruit like that. I enjoyed the video, this was a new topic for me. Thank you!

@AlexisCledat 6 ай бұрын

Hum... Why ? 😭 I mean that's interesting but it's very complicated and there's an evident solution that you can find faster.

@AlexisCledat 6 ай бұрын

The Cardano's formula is really awful...

@kevinmadden1645 Ай бұрын

Quite right! Increase each root by 1/3 and write the cubic equation with the increased roots .Also let y be any of the roots of the transformed equation and let x be any of the roots of the original equation. This x equals y-1/3. Therefore f(y-1/3) equals 0.. The result is the same.