A Cubic Equation with Double Roots

[@Popcorn31416]

I wish to memorize the cubic formula one day. That will be the day I reign supreme over the world

[@SyberMath]

Let's go!!!

[@anshumanagrawal346]

LoL

[@aryaanpetiwalla2655]

What about the quartic formula?

[@Popcorn31416]

@@aryaanpetiwalla2655 Cubic is enough, I'll need to upgrade my brain for quartic

[@aryaanpetiwalla2655]

@@Popcorn31416 Agreed. I remember I freaked out the first time I saw the quadratic formula two years ago.

[@mathunt1130]

Third method: divide through by a and then write the polynomial as (x-b)^2(x-c)=0 Expand this polynomial and compare coefficients, you get some very interesting results.

[@kamarinelson]

That's what I did, but I def got slightly different values for a which points to arithmetic error. I didn't even think to use the first method tho which was more efficient.

[@anshumanagrawal346]

I don't know, I tried that first but it didn't seem to give anything helpful. Which was why I used an approach similar to the one in the video

[@chandrasekhar2187]

My favourite part - Discriminant of a cubic polynomial (7:25).

[@engjayah]

@2:15 without calculus we could obtain both the Eqns as follows: as r is a double root; then (x-r)^2 = 0 from which we get x^2 = 2rx - r^2 Plug in this result in f(x) wherever x^2 appears; as f(x) is zero we obtain (3ar^2 - 2r - 1)x - (2ar^3 - r^2 - 2) = 0 from which we get 3ar^2 - 2r-1 = 0 and 2ar^3 - r^2 - 2 = 0

[@242math]

great job bro, thanks for sharing this solution

[@SyberMath]

No problem 👍

[@ThAlEdison]

I used vieta's formulas. If ax^3-x^2-x+2=(ax-u)(x-r)(x-r)=ax^3-(u+2ar)x^2+(2ur+ar^2)x-ur^2=0 Then you can eliminate u and get the same equations as with using the derivative. I was actually worried because with how problems are made, I was expecting a to have integer values so I thought I might've made a mistake, but no.

[@SyberMath]

Nice!

[@WahranRai]

You are not using Vieta formula, you are using factorisation saying that you have a double root and one single root. Vieta formula (formules de Viete en français), with double root r and single t : 2r+t = 1/a r^2 + 2rt = -1/ a r^2*t = -2/a Solve the system of 3 equations with 3 unknows

[@anshumanagrawal346]

@@WahranRai Potato Potahto

[@WahranRai]

@@anshumanagrawal346 What is the meaning of your comment

[@anshumanagrawal346]

@@WahranRai That you are being Nitpicky, there's nothing wrong with the OP calling their method Vieta's Formula

[@samosavaglio2141]

The cubic formula isnt hard at all! U Just have to memorize it and use it using p and q: When u subsitute (x+b/3a) you eliminate the x^2 coefficient and the equation becomes f(x)=x^3 +px+q. The discriminant at this point it's just (q/2)^2 + (p/3)^3 wich is kinda nice

[@srijanbhowmick9570]

Nice my answer matched with yours

[@SyberMath]

Cool!

[@Qermaq]

That's always a good sign :D

[@srijanbhowmick9570]

@@Qermaq Yep

[@srijanbhowmick9570]

@@Qermaq Haven't seen you in a long time

[@Qermaq]

@@srijanbhowmick9570 Been here ;)

[@TheShacharZiv]

Thanks! I did it in a long way and it worked this time but I'm not sure it would always work out... I assumed (ax+b)(x+c)^2=0

[@anshumanagrawal346]

Nice!

[@yoav613]

Cool problem and your methods are even cooler! Using derivative nice,i did not think of that

[@SyberMath]

Thanks! 😃

[@jinhuiliao1137]

We can let x=1/y and get a new cubic equation 2y^3-y^2-y+a=0. Then solve with vieta.

[@oussamalaaroussi5467]

Thanks for the video , but i didn t understand why the derivative of r is 0

[@oussamalaaroussi5467]

@@anonmath much appreciated

[@asutosh2429]

Can you provide the proof for the Discriminant of Cubic equation?

[@SyberMath]

That would be complicated but there's a video if you want to check it out: kzbin.info/www/bejne/fqHCeKmgp7KapZI

[@physicist55]

2.denklemi r ile çarpmak bir şeyi değiştirmiyor mu? Bazen fazladan kök geliyor da.

[@SyberMath]

r=0 getirir yani problem yok.

[@physicist55]

Teşekkürler hocam

[@gemeni0]

U did not prove that if r is a double root as u call it, than f'(r)=O. Where do you get this from? It's not apparent.

[@SyberMath]

If f(x)=(x-m)(x-r)^2 then x=r is a double root f'(x)=(x-r)^2+2(x-r)(x-m) f'(x)=(x-r)[x-r+2(x-m)] f'(r)=0

[@jimmykitty]

Cool 😎

[@SyberMath]

Thanks!

[@jimmykitty]

@@SyberMath ❤❤❤

[@SyberMath]

@@jimmykitty 😊🥰

[@jimmykitty]

@@SyberMath 💖😸

[@ajiwibowo8736]

Why you can do tht system

[@carloshuertas4734]

Another great explanation, SyberMath!

[@SyberMath]

Thank you!

[@राजनगोंगल]

👍👍👍

[@ahmadmazbouh]

I didn't know about the discriminant for the cubic equation, this will be helpful for me, thank you ❤❤❤❤

[@SyberMath]

Glad to hear that!

[@ahmadmazbouh]

@@SyberMath ❤❤

[@georget8008]

Couldn't we use de vetta's formulas? We would end up with a system of 3 equations with 3 unknowns: a, x1, x2. What do you think?

[@SyberMath]

We could 😊

[@YossiSirote]

What about using a third method, using Vieta's formulas. (Which I learned from watching this channel).

[@SyberMath]

Absolutely! Thanks for sharing! 😊

[@mcig98]

a = 0 ;)

[@SyberMath]

😁

[@aashsyed1277]

1st

[@jfcrow1]

a=-.1578 and a=.53