So you want to solve a cubic equation?
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- čas přidán 15. 02. 2021
- This cubic equation is from the 2020 Oxford MAT. See my video here: • Solving An Oxford Elli...
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bprp #fast
While watching these kinda fast videos , I randomly pause the video in between so that you can relax .
Thanks!
Btw, cool profile picture.
@@bprpfast oh thanks , I just customized it cause I'm thinking of starting my channel after my exams
@@bprpfast and yeah it's cools but to be honest your beard is more cooler than it.
"Isn't it!"
I just randomly clicked your channel and it's actually scary how much similar our subscribed channels list is, btw which country are you from?
*Everybody gansta till some square roots inside cube roots show up to town*
just to confirm, you did not actually need to solve this equation, you only had to answer questions about it - obviously they would not expect you to remember and use Cardano's formula on such a weird example lol
What
I love the frequent uploads on this channel
you suggested him in his second last video to make video on cubic equation. am i right?
you commented on that video 4hours ago and he made a video 1 hour ago
@@sarvesh_soni yes
@@sarvesh_soni cool
very relaxing.
How to remember :
You have (E) x^3=px+q
Solve (F): y^2-qy+p^3/27=0
You get two solutions A and B,
x=cbrt(A)+cbrt(B)
Is it the Cardan's method ?
@bprp fast.. Isn't this Cardan's formula for solving cubic equations??
Yes sorry.. Typing mistake.. Thanks anyways.
My video be buffering to just slow him down lulz
Hi, can you solve that 7 grade math problem? I tried to solve, but this is too hard for me.
If (n^3 + 1 + n!) is prime number, n is natural, n > 2. Proof that n^2 + 2 = p1 + p2 where p1 and p2 are primes.
That’s 7th grade? You’re russian, right?
@@randomblueguy yes, this is Olympic problem
@@user-pg5ks4ww1u Oh yes, that explains it.
Is so satisfyng
X^2 term ?
It's always possible to write an equivalent cubic without an x^2 term using a substitution.
You r allowed a calculator right?
Comm' on dude, He is an asian
For this exam, nope
@@5dots297 me too😂
We didn’t have to solve it, just answer questions revolving it
I am also asian 😂😎
You forgot to demonstrate the cardan's formula.now power for,Ferrari, forza italia
was the formula given or were the students taking this test forced to suffer...
We didn’t have to solve it, it was just questions about it
Can u proof the cubic formula?
I make math/calculus videos please check them out and subscribe
Have you seen the whole formula before (not this one)? Its too complex to be proofed.
@@ntth74 yes I am just curious to know
I can, do you have discord ?
Fun fact: That answer is real
Yup!
Ofc it's real
I trust Bprp
He wont give us a fake answer
How?
@@mrhatman675 Both cube roots are complex numbers with opposite imaginary parts, thus the imaginary part gets cancelled when both are added, resulting in a real number.
Ok
but isn't there 2 more solutions?
You can get the other two with polynomial division
Multiply the first cubic Root by 1/2+sqrt(3)*i/2, and the second cubic Root by 1/2-sqrt(3)/2.
Swap those factors for the third solution.
@@kuchenzwiebel7147 sorry but who the hell wants to do polinomial division with such a monster that x is
Would you believe me if i told you i know how to solve a cubic equation the algebraic way?
#CubicEquation
And if there's a ...x^2?
You'd say, "Welp, it's algebra time"
There s an updated cardanos formula for that case too but it s twice as long as this one lol
I wonder what is the cube root of i
(sqrt(3)+i)/2
@@blazedinfernape886 how do you calculate that ?
@2C (02) Chan Kwan Yu no, I just ask Blazed, I understand your calculation and just liked it.
@@ntth74 i = e^(iπ/2)
cbrt(i) = (i)^(1/3)
= ((e^(iπ/2))^(1/3)
= e^(iπ/6)
= cos(π/6) + isin(π/6)
= (sqrt(3) + i)/2
This is only one answer. There are two more answers and we get them by adding multiples of 2nπi.
i = e^(iπ/2) = e^(5iπ/2) = e^(9iπ/2)
So both of you are correct ?
😵
this is insanity
Shouldnt there be 3 solutions because of fundamental theorem of algebra? how do you produce the other two
Complex solutions, 2 of them because complex solutions always occur in conjugates
@@shivam5105 they could be two complex solutions but not always tho... either 3 real zero complex or 1 real 2 complex
@@TheEndernal but here with the cubic formula if I’m not wrong, he’s only getting one real solution, so the 3 real 0 complex is ruled out I think
@@shivam5105 3x^3-4x graph that on desmos for proof that has 3 real roots which are -2 0 and 2 or you could factor
@@shivam5105 I think it involves you having to multiply by the cube root of unity? Like the complex one. I think all you need to do is just find one and then just switch the sign of the i part bc of fundamental theory of roots
I still don't get how all the answers are real...
I don’t get it either lol but they are!
@@bprpfast a + bi + a - bi = a real number.
@@michaelempeigne3519 isn't it supposed to be (a + ib)(a-ib) which is real ?
@@thecrazzxz3383 cubic have at leat one real root.
@@michaelempeigne3519 then what if p^2/4 + q^3/27 = -1 ?
Please solve questions on board! Please this is a request
The solution is imaginary lol
Dude i don’t understand this?
How does this make any sense? (except the cube root part that’s definitely right
I think Oxford is just a stupid university, what were they thinking with these equations?
He is very slow.