An Interesting Quartic Equation

[@timothybohdan7415]

The introduction of a dummy variable k to create two sets of perfect squares is a clever and ingenious approach. Well done, sir.

[@guyhoghton399]

*_x⁴ + 6x³ - 128 = 0_* _x ≠ 0_ so we can divide through by _x³:_ _x + 6 - 2(64/x³) = 0_ Let *_y = 4/x_* ∴ _4/y + 6 - 2y³ = 0_ ⇒ _y⁴ - 3y - 2 = 0_ ... ① With these coefficients it is worth trying a factorisation of: _(y² + ay - 1)(y² - ay + 2) = 0_ Comparing the coefficients of the _y_ terms with ① gives: _3a = -3 ⇒ a = -1_ which works for the _x²_ terms as well, so ① becomes *_(y² - y - 1)(y² + y + 2) = 0_* from which we can solve for _y_ and hence for _x._

[@SyberMath]

Nice!

[@timeonly1401]

Wow!!

[@АндрейПергаев-з4н]

Называется метод неопределённых коэффициентов (х^2+ах+b)*(х^2+сх+d) Надо перемножить х^4+(а+с)х*3+(b+d+ac)x^2+(bc+ad)x+bd Приравняем коэффициенты при одинаковых степенях х Система из четырёх уравнений a+c=6, b+d+ac=0, bc+ad=0, bd=-128 Решаем и получается разложение на произведение двух квадратных трехчленов

[@SyberMath]

Хороший!

[@andreaparma7201]

About the 2nd method: a quartic equation without the x^1 term can also be transformed into a depressed quartic by the substitution x=1/y. In this case we get 1/y^4+6/y^3=128 --> 1+6y=128y^4 --> 128y^4-6y-1=0 --> y^4-3y/64-1/128=0 and we can proceed as usual

[@scottleung9587]

Nice!

[@trojanleo123]

I like that you took your time and did Method 3 fully and did not force yourself to keep the video under 10 minutes.

[@Don-Ensley]

problem x⁴ +6x³ = 128 Put into standard form. x⁴ +6x³ - 128 = 0 Let x = y -3/2 y⁴ - 27 y² / 2 + 27 y - 2291 / 16 = 0 Be rid of fractions by multiplying by 16. 16 y⁴ - 216 y² + 432 y - 2291 = 0 Note that 2291 = 29 • 79 a product of 2 primes. Let's made 1 the leading coefficient. Let y = t/2 t⁴ - 54 t² + 216 t - 29•79 = 0 Factor. There is now no cubic term. 2291 = 29 • 79 (t² - at + 29)(t² + at - 79) = 0 t⁴- (50 + a²) t² + 108 a t - 2291 = 0 a = 2 solves. Factors into (t² - 2t + 29)(t² + 2t - 79) = 0 Quadratic formula yields: Roots in t t ∈ { -1 -4√5, -1 + 4√5, 1 - 2 i √7, 1 + 2 i √ 7 } Back substitute: y = t / 2 Roots in y y ∈ { (-1 -4√5) / 2, (-1 + 4√5) / 2, (1 - 2 i √7) / 2, (1 + 2 i √ 7) / 2 } x = y - 3/2 Roots in x x ∈ { -2 -2√5 , -2 + 2√5 , -1 - i √7 , -1 + i √ 7 } answer x ∈ { -2 -2√5 , -2 + 2√5 , -1 - i √7 , -1 + i √ 7 }

[@ianfowler9340]

I don't know whether you are aware of this or not, but you are using the same technique that was used to develop the general quartic formula. Here, you are applying this technique to a specific case as opposed the general development of the formula. To be clear, this is not a criticism of you or your video. Quite the contrary. Just an interesting observation of the technique. When you use the general quartic formula you always generate a cubic (called the resolvent cubic - I think). In this specific case, this is your cubic in k. As you know this can always be solved by the cubic formula or equivalently the sum of cubes formula. A real MESS in this example. If there is a rational root to the quartic then there will also be a rational root to the resolvent cubic and you can use the factor theorem to find it - as you did for k = -4. However, this all depends on being able to find the -4 "easily" by using the factor theorem. Otherwise you have a very high mountain to climb. It's all good stuff - cheers.

[@Hiro-j5m]

x^4 + 6x^3 = 128 0 = 128 - 6x^3 - x^4 0 = 128/x^4 - 6/x - 1 0 = 256/x^4 - 12/x - 2 0 = (4/x)^4 - 3(4/x) - 2 0 = { (4/x)^2 - (4/x) - 1 }{ (4/x)^2 + (4/x) + 2 } ...

[@snejpu2508]

My method. Trying to estimate 2 roots using Newtons method, figuring out that real roots are symmetric with respect to x=-2, finding that the deviation is 2sqrt(5), evaluation that one of the quadratics is x^2+4x-16 by Viete formulas, finding the values of the other quadratic, verifying if the quadratics are correct and solving for all solutions once more. :-P

[@holyshit922]

You used Newton's method with complex nuumbers or you assumed that there exist real solutions

[@YAWTon]

​@@holyshit922 Since x^4+x^3-128 +oo as x-->+-oo the existence of two real roots is not an assumption. That the difference between the two numerically determined roots is 4 times squareroot of 5 involves some guessing (or intuition), though.

[@snejpu2508]

@@holyshit922 I didn't have to assume, as from the original equation I could see that the left hand side is increasing for positive numbers, so there is certainly a real solution (between 2 and 3). The polynomial is of the 4th degree, so, if there is one real solution, there are two of them. :-)

[@山山-y4q]

x^4+6x^3=128 ζ(4)=π^4 /90 ζ(2)=π^2 /6 If x=π, x^4 + 6x^3=128, π^4 + 6π^3=128, 90ζ(4) +πζ(2)=128, If we let 1≡π, 1+π=128, Also, ζ(2)=π^2 /6 so π=±√6√ζ(2) 1 ±6√6√ζ(2) ζ(2)=128, ±6√6√ζ(2) ζ(2)=127, I tried a math puzzle🧩🤣 The answer may be wrong, but it's fun🤣

[@georget8008]

μεσω τρικαλων μας πηγες.....

[@Mike28a]

first. interesting

[@pwmiles56]

x^4 + 6x^3 - 128 = 0 Integer roots must be x=+/- 2^n, n=0 to 7. None of these work. Try x^4 + 6x^3 - 128 = (x^2 + ax + b)(x^2 + cx + d) Equating coefficients x^3: a + c = 6 x^2: b + d + ac = 0 x^1: ad + bc = 0 x^0: bd = -128 Assuming integer coefficients, note |ac| = |b+d| must be less than or equal to 9 (GM

[@musicsubicandcebu1774]

Would you explain why: "|ac| = |b+d| must be less than or equal to 9 (GM

[@pwmiles56]

@@musicsubicandcebu1774 From the x^3 coefficients, a+c = 6. From the x^2 coefficients, ac = -(b+d). I got a sense from these that |b+d| couldn't be too large. It's a solution strategy, not a proof!

[@pwmiles56]

@@musicsubicandcebu1774 test nm

[@pwmiles56]

test nm

[@NadiehFan]

At 10:45 you have the cubic resolvent k³ + 128k + 576 = 0 When looking for potential _integer_ solutions of this equation, it is immediately clear that we only need to check negative integers, because the left hand side is positive for any positive k. However, it is _also_ clear that k³ + 128k = k(k² + 128) is _odd_ for any odd integer k, so no odd integer can be a solution of this equation since the sum of an odd integer and the even integer 576 cannot be zero. Accordingly, we only need to test _even_ negative integers. Therefore, you did not need to test k = −3, that is just a waste of time. Since k = −2 clearly does not satisfy this equation, the next negative integer to try is k = −4, which is indeed a solution of this equation. At 8:50 you have (x² + 3x + k)² = (9 + 2k)x² + 6kx + 128 + k² and we are looking for a value of k that makes the right hand side a perfect square, i.e. the square of a linear polynomial in x. Most quartic equations with integer coefficients from contests (as well as home made equations) have a nice factorization into two quadratics with integer coefficients. So, before looking at the discriminant of the quadratic in x at the right hand side and setting this equal to zero, the first thing to do is to see if we can find a value of k which makes the right hand side the square of a linear polynomial mx + n with _integer_ coefficients m and n. Since (mx + n)² = m²x + 2mnx + n² both the coefficient of x² as well as the constant term would need to be squares of integers _if_ there exists a value of k which makes the right hand side the square of a linear polynomial in x with integer coefficients. So, we can start by looking for values of k that make the coefficient 9 + 2k of x² the square of an integer (1², 2², 3² ...) and see if the same value of k also makes the constant term 128 + k² the square of an integer. If we want to have 9 + 2k = 1² then we need to have k = −4 and with this value of k we have 128 + k² = 128 + (−4)² = 128 + 16 = 144 = 12², as required. So, without ever having to deal with the cubic resolvent, we select k = −4 and this gives (x² + 3x − 4)² = x² − 24x + 144 which can indeed be written as (x² + 3x − 4)² = (x − 12)² Another approach which can be profitable sometimes (but _not_ in this case) is to work _backwards_ by starting to look for values of k that make the constant term the square of an integer and see if the same value of k also makes the coefficient of x² the square of an integer. To see how this works, let's give this a try anyway. if 128 + k² is the square of some integer n for some integer k, that is, 128 + k² = n², then 128 = n² − k² must be a difference of two squares of integers. Now, any product of two quantities p and q can be rewritten as a difference of squares since pq = ((p + q)/2 + (p − q)/2)((p + q)/2 − (p − q)/2) = ((p + q)/2)² − ((p − q)/2)² Clearly, if (p + q)/2 and (p − q)/2 are to be integers then their sum p and their difference q must be integers and, moreover, p and q must be integers of like parity if (p + q)/2 and (p − q)/2 are to be integers. Since 128 = 2⁷ is even this number evidently cannot be written as the product of two odd integers but we can write 128 as the product of two even integers in several ways, and using the identity above we have 128 = 64·2 = (33 + 31)(33 − 31) = 33² − 31² 128 = 32·4 = (18 + 14)(18 − 14) = 18² − 14² 128 = 16·8 = (12 + 4)(12 − 4) = 12² − 4² which exhausts all possibilities of writing 128 as a difference of two squares of integers. Evidently, if 128 + k² is the square of some integer n for some integer k and therefore 128 = n² − k² then k must be one of the numbers ±4, ±14, ±31, so we need to test at most six integer values of k to see if this will make the coefficient 9 + 2k of x² at the right hand side of our equation the square of an integer as well. For k = 4 we have 9 + 2k = 9 + 8 = 17 which is not the square of an integer, but for k = −4 we have 9 + 2k = 9 − 8 = 1 = 1² which is the square of an integer, so we have again found an integer value of k (in fact, in this case the only integer value of k) which makes the right hand side of our quartic equation the square of a linear polynomial in x with integer coefficients.