Solving x^3-93x-308=0 in Two Ways

[@williamperez-hernandez3968]

The correct name should be Fontana's Method. Nicolo Fontana (1499-1557) was nicknamed Tartaglia (stutter or stammer in English, tartadear in Spanish) because a saber cut to his jaw and palate in 1512 left him with a speech disability.

[@buxeessingh2571]

Thanks for the information!

[@Polpaccio]

Niccolò to be exact

[@SeekingTheLoveThatGodMeans7648]

They sure went through some rough stuff back then, when modern surgical capabilities were not possible (but again maybe Fontana would have gotten shot today). I wonder why the wound. A duel? Service in the military? A robbery?

[@geilesjonboomhaard1414]

@@SeekingTheLoveThatGodMeans7648 he was sliced when he was a child by a barbaric invading french army

[@Icewallocumm]

Good information

[@misterdubity3073]

@8:08 s.b. 3ab=93 ; ab=31

[@jmiki89]

Actually, it was Scipione del Ferro who first found out the method, though Tartaglia found it independently. And while Cardano first learned it from Tartaglia, he didn't published it (as he promised) until he found the late del Ferro's notes and the formula in them. Does this count as cheating on the promise he took to Tartaglia? Maybe. However the cubic formula is named after Cardano not because of solving the depressed cubic (a*x^3 + b*x + c = 0) but for solving the general cubic (a*x^3 + b*x^2 + c*x + d = 0).

[@holyshit922]

but Scipione del Ferro didn't publish his method ad we are not sure are they the same , moreover in these times they dislikes negative numbers so there are many cases (depressed cubic had three "different" cases ) so we are not sure if del Ferro solved all cases (Fior gave to Fontana only one case of cubics) Cardano begged this method from Fontana , so Fontana should be credited Mathematicians knew how to depress cubic before Cardano In Euler's book you can find generalization of Fontana's method to quartics He called it new method

[@rubensramos6458]

Using the Lambert-Tsallis Wq function, one of the roots is: x = sqrt((-93/2)*Wq(2*(308^2)/(-93^3))) = -4 (take the negative value), with q = 1/2.

[@peterbyrne6394]

Thank you This exercise was pretty easy

[@SyberMath]

You're so welcome!

[@pietergeerkens6324]

Nice. I (after a couple of false starts) factored to x * (x^2 - 91) - 308 = 0 and realized that for the sole positive root (Descartes' Rule of Signs) (x^2 - 91) must be a modest positive value. For the factors of 308 only 11 satisfies, and sure enough is a root.

[@skatastic57]

This amounts to the same thing but if you assume it's factorable into (x+a)(x+b)(x+c)=0 then you get x^3+(a+b+c)x^2+(ab+bc+ac)x+abc=0. The rest is as laid out in the video.

[@kuronekonova3698]

Let f be a real function with a single variable x, such that f(x) = x³ - 93 - 308. By inspection, we can see that -4 is indeed a root to the suggested polynominal within this video, which simply means that f(-4) = 0. We may then safely say that (x + 4) is indeed a factor of said polynomial, and from there we may utilize the Polynomial Factorization Theorem to find the rest of all real or possibly complex solutions for the polynominal. This is trivial and will be left as an exercise to the reader.

[@moeberry8226]

The master making the question look way too easy.

[@ghostofsin]

Thank you so much for your videos .

[@HoSza1]

I'd like to suggest a practice: find all the integers n, for which p(n)=n^4+4 has compound values. I wonder in how many different ways can you solve it. Edit: I meant composite here, sorry.

[@chaosredefined3834]

Did not recognise the term "compound values", but I suspect you mean composite. The answer is an infinitely large set: Every even input will result in a value that is even, and it cannot produce 2.

[@HoSza1]

@@chaosredefined3834 That's true. But what about odd numbers? E.g. p(3)=85 is composite. You could also try to find the converse, that is the set, where p has a prime number value. Is that set finite?

[@chaosredefined3834]

@@HoSza1 Actually, I missed out on something. Suppose p is of the form 5n. Then, raising it to the 4th power gives you something of the form 5k. Adding 4, we get 5k+4. Suppose p is of the form 5n+1 or 5n+4. Squaring it gives you something of the form 5k+1. Squaring it again gives you something of the form 5m+1. Adding 4 gives you 5m+5, which is divisible by 5. Suppose p is of the form 5n+2 or 5n+3. Squaring it will give you something of the form 5k+4, and squaring that again will give you something of the form 5m+1. adding 4 gives you 5m+5, which is, again, divisible by 5. The only prime number divisible by 5 is 5, and it turns out that p(1) = 5 (and no other input). So, the only possible answers are 1 and multiples of 5.

[@jmiki89]

@@HoSza1 I came up with three method, spoilers incoming in 5 . . 4 . . 3 . . 2 . . 1 . . So the first method: n^4+4 = n^4+4*n^2+4 - 4*n^2 = (n^2+2)^2 - (2*n)^2. Since this is a difference of two squares, we can use the a^2-b^2 = (a-b)*(a+b) formula, and (after rearranging the terms) we get (n^2-2*n+2)*(n^2+2*n+2). As we can factor out the polinomial, it's almost always composite except if one (or both) of the factors is (are) either 0 or unity. Both factors can be written as (something)^2+1, hence they can't be 0 nor -1 (unless we want to take Gauss-integers into account, but in that case both quadratic expression can be factored further so it always will be composite). So it has to be 1, which can only occure if the aformentioned (something)^2 is 0, and that will only be case if n is either -1 or 1 (and p(-1) = p(1) = 5 which is indeed a prime). The second method: n^4+4 = n^4-2*n^2+1 + n^2+1 + n^2+1 + 1 = (n^2-1)^2 + n^2+2*n+1 + n^2-2*n+1 + 1 = [(n-1)*(n+1)]^2 + (n+1)^2 + (n-1)^2 + 1 = (n+1)^2*[(n-1)^2 + 1] + 1*[(n-1)^2 + 1] = [(n+1)^2 + 1]*[(n-1)^2 + 1]. Again, the polinomial is factored now, the analysis of the factors are the same. The third method: n^4+4 = n^4 - (-4) = (n^2)^2 - (2*i)^2 = (n^2 - 2*i) * (n^2 + 2*i). Given that the square roots of 2*i are 1+i and -1-i: n^2 - 2*i = [n - (1+i)] * [n - (-1-i)]. Similarly: n^2 + 2*i = [n - (1-i)] * [n - (-1+i)]. To get back to the real numbers, we need to regroup the factors by pairing the complex roots with their conjugates: [(n-1-i) * (n-1+i)] * [(n+1-i) * (n+1+i)] = (n^2-2*n+2) * (n^2+2*n+2). PS: Bojler eladó XD

[@HoSza1]

@@chaosredefined3834 I don't understand your proof entirely, but consider n=5, which is a multiple of 5. So p(5)=25*25+4=629=17*37, and this is a composite number. Similarly p(10)=50629=257*197 also composite, but p(1)=5, which is prime.

[@mystychief]

calculated the formula in excel for integers x from -20 to + 20. You get three solutions zero. Then you have the solutions because there are not more and not less solutions.

[@hamzalouliditv7227]

Can you solve please this equation:5f(-x) +f(1-x)=2x

[@WahranRai]

A solution would be of the form : f(x) = ax + b etc...

[@Vladimir_Pavlov]

f(x)=-x/2.

[@WahranRai]

@@Vladimir_Pavlov NOOOOOOOOOOOOOOOO, FALSE !!!! When you propose solution, verify the relation !

[@Vladimir_Pavlov]

@@WahranRai f(x)=-x/3+1/18. :)

[@holyshit922]

You could show trigonometric solution to this equation Reduce this equation to triple angle formula for cosine xor sine Here we have casus irreducibilis

[@NadiehFan]

Exactly. Since (308/2)² + (−93/3)³ = −6075 the equation has three distinct real roots, so it is appropriate to solve this cubic trigonometrically right away if you are looking for a formal solution and don't want to test for integer solutions first.

[@vladislavlukmanov4998]

I suppose ab must equal 31 rather than 93

[@Avighna]

I actually made an application to solve cubics with at least one integer root. Here's how it handled it (everything below this is generated by the program.) x^3 - 93x - 308 = 0 Step #0: Synthetic Division 1 | 1 0 -93 -308 | 1 1 -92 | 1 1 -92 -400 2 | 1 0 -93 -308 | 1 2 -89 | 1 2 -89 -486 4 | 1 0 -93 -308 | 1 4 -77 | 1 4 -77 -616 7 | 1 0 -93 -308 | 1 7 -44 | 1 7 -44 -616 11 | 1 0 -93 -308 | 1 11 28 | 1 11 28 0 Step #1: Factor out (x - 11). (x - 11)(x^2 + 11x + 28) = 0 Step #2: Use the zero-product rule: x - 11 = 0 (equation 1) x^2 + 11x + 28 = 0 (equation 2) Step #3: Solve equation 1 for x: x = 11 Step #4: Solve equation 2 for x: Using the quadratic formula: x = (-b + or - sqrt(b^2 - 4ac))/(2a) x = (-11 + sqrt(9))/(2*1) x = (-11 - sqrt(9))/(2*1) x = -4 or x = -7 x = {11, -4, and -7}

[@christianthomas9863]

very smart!

[@michaelempeigne3519]

incorrect, it is right that ( a + b )^3 - 3ab ( a + b ) = a^3 + b^3 then let x = a + b 3ab = 93 and a^3 + b^3 = 308 but ab = 93 / 3 = 31 so x^3 - 93x - 308 = 0 becomes a quadratic that is solveable with m^2 - 308m + 31^3 = 0 m = ( 1 / 2 ) * [ 308 + or - sqrt ( 308^2 - 4 * 31^3 ) ]

[@voltalimwabbit5492]

But the discriminant is less than zero. Could you explain how to find the 3 roots using this formula? Thank you

[@michaelempeigne3519]

@@voltalimwabbit5492 you can find the values of a^3 = ( 1 /2 ) * [ 308 + sqrt ( ....... ) ] and b^3 = ( 1 / 2 ) * [ 308 - sqrt ( .........) ] find the a and b values

[@voltalimwabbit5492]

ab = -93/-3 = 31

[@NadiehFan]

Unfortunately this video does not make it clear that the second method will not get you the solutions of the equation in any usable form because the expressions for the solutions involve cube roots of complex numbers which cannot be evaluated algebraically: any attempt to do so will result in a cubic equation which is equivalent with the cubic equation you were trying to solve in the first place. This Catch-22 is commonly referred to as the *casus irreducibilis* (Latin for 'irreducible case'). The appropriate way to handle this equation without guessing any solutions is trigonometrical. It can be demonstrated that a cubic equation x³ + px + q = 0 with real p and q has three different real roots if and only if (q/2)² + (p/3)³ < 0 and that the three real roots are then x₁ = 2·√(−p/3)·cos((1/3)·arccos((−q/2)/((−p/3)·√(−p/3)))) x₂ = 2·√(−p/3)·cos((1/3)·arccos((−q/2)/((−p/3)·√(−p/3)))+2π/3) x₃ = 2·√(−p/3)·cos((1/3)·arccos((−q/2)/((−p/3)·√(−p/3)))+4π/3) For the cubic equation in the video we have p = −93, q = −308, so (q/2)² + (p/3)³ = 154² − 31³ = 23716 − 29791 = −6075 which means that the equation does indeed have three different real roots and that the trigonometric expressions for the roots can be used. If we plug in p = −93 and q = −308 we get x₁ = 2·√31·cos((1/3)·arccos(154/(31·√31))) x₂ = 2·√31·cos((1/3)·arccos(154/(31·√31))+2π/3) x₃ = 2·√31·cos((1/3)·arccos(154/(31·√31))+4π/3) However, it is not at all obvious how these expressions can be simplified or evaluated numerically without using some calculating device or trigonometric tables. But you can plug these expressions into WolframAlpha to verify that they indeed simplify to x₁ = 11 x₂ = −7 x₃ = −4

[@КатяРыбакова-ш2д]

Спасибо за оба эти способа, но ab=93:3=21 во во втором способе. Я тоже сама решила это уравнение двумя способами, но другими:1) группировкой, (x^3-16x)-(77x-308)=0. 2) Нашла -11 как делитель свободного члена и получила после деления на x+11 квадратное уравнение x^2-11x+22=0.

[@jmart474]

I gave up. I don't know how to solve c^2-308c+29791=0 in order to obtain the three solutions.

[@Jha-s-kitchen]

Yes, Solving c^2 -308c +29791, I get x = cbrt(154+45 * sqrt(3) * i) + cbrt(154 - 45*sqrt(3)*i) And if I cube both sides here, I get the original equation 😂 Is this the expected answer ? {-4, -7, 11}

[@jmart474]

@@Jha-s-kitchen Yes. I arrived to this solutions via Ruffini. But I don't know how to obtain them from the depressed cubic formula.

[@voltalimwabbit5492]

Could you explain us how to arrive at the three solutions? The equation has a negative discriminant

[@Rbmukthegreat]

Beautiful!

[@SyberMath]

Thank you!

[@SuperYoonHo]

awesome!

[@lf9361]

pronounced Tartal’a

[@devondevon4366]

x= -4, 11, and -7

[@voltalimwabbit5492]

The discriminant of c^2 - 308c + 29,791 = 0 is less than zero.

[@unonovezero]

It doesn't matter, when you sum the two c values (which are a and b) that are complex conjugates, you get a real number for x.

[@Vladimir_Pavlov]

Bezu's theorem solves the problem.

[@bjornfeuerbacher5514]

Could you please elaborate? (BTW: you probably mean Bezout?)

[@Vladimir_Pavlov]

@@bjornfeuerbacher5514 I was referring to the consequence of Bézout's theorem : "A free term of a polynomial is divisible by any integer root of a polynomial with integer coefficients (if the highest coefficient is 1, then all rational roots are also integers)." In this case , it is necessary to check all the divisors of the number 308 : ±1, ±2, ±4,±7,±11,±77,±154, ±308, are they not the roots of the original equation. Do not be afraid of a large number of options. It is necessary to "feel" how the individual terms work. x^3 gives too small (in absolute magnitude) value compared to 308 for small (in absolute magnitude) divisors ( ≤2) and too large for absolute magnitude divisors ≥77. The multiplier at x is also important because it has a large absolute value. So, the search should be performed among ± 4, ± 7, ± 11. Not so much. If the whole root has a place to be, then it will be found! I found x=-7. After dividing a given polynomial of the third degree by x+7 (in a column)we find a square trinomial whose roots must be found. It's not a problem.

[@Vladimir_Pavlov]

As it turned out, the given equation has three integer roots (which is rare in tasks) and you could initially pick up the root not x =-7, but x=-4 or x=11. Not too detailed?

[@bjornfeuerbacher5514]

@@Vladimir_Pavlov This is known to me under the name "rational root theorem". And he uses that in the video already.

[@Vladimir_Pavlov]

@@bjornfeuerbacher5514 OK. I was sorry to spend nine minutes watching this video, and I watched it for several seconds at minute intervals.) Note that at 3:53 and a little further, the author says that from the equality m +n+k=0 it follows that at least one root has a negative value. Why?!! After all, it is possible that one root is positive, and the other two are complex conjugate, having a negative real part. A hastily invented example x^3-90*x-341=0. But the author knows the result in advance (!) (there will be three integer roots), so he makes such a statement, and uses a method suitable in this case. It's not interesting to me. "Rational root theorem" is a consequence of Bézout's theorem. Enough to find one root, after that, the other two roots, real or complex, are found from the quadratic equation.

[@amaraouassima4154]

so. much fun

[@-basicmaths862]

X=-4

[@giuseppemalaguti435]

11,ma lho fatto semplicemente con la formula della cubica

[@yakupbuyankara5903]

X=11.

[@MrLidless]

*IF* we assume whole number solutions, let’s factorise 308: 2x2x7x11. Can any combination of them sum to zero (the x^2 term)? Yes: -4 (ie 2x2), -7 and +11. Checks…Solved. This is more a case of how to short-cut an answer on a maths test against the clock as opposed to using Cardano’s solution, which is the more rigorous approach.

[@rakenzarnsworld2]

x = 11

[@SyberMath]

That's it?

[@rakenzarnsworld2]

@@SyberMath In algebraic geometry, yeah, actually that's it because length must be a positive real number. Also, I was too lazy to use the quadratic formula to get the leftover roots. 308 is too large :(