China | Can you solve this ? | A Nice Math Olympiad Algebra Problem

[@Segalmed]

120 = 5! = %*4*3*2*1 = 5*4*3*2 => the trivial solution is x =1. But there must be 4 more solutions since this is a 4th degree polynomial. Another one is obviously -6 since there are four brackets, so if all are negative the result is still positive and we get the same result, if one bracket is -5 the next -4 and so on. My guess is that the the remaining two solutions are complex. So, divide the whole polynomial by (x+6) and (x-1) to get a quadratic equation, that can be solved by the usual formula. Edit: That equation is x^2 + 5x +16 which has the complex solutions -2.5 +/- i*0.5*SQRT(39).

[@ManojkantSamal]

A nice way of elaboration

[@esp-l1n]

That's exactly what I did

[@dr_Sehaq]

Not a math guy. But good explanation

[@Sci-Marvels]

Great breakdown! You've clearly worked through the polynomial to find all the solutions, including the complex ones. I appreciate the detailed explanation and the logical steps you took to solve it. Well done!

[@iitiansaurabhbhu]

2×3×4×5=120 from here by hit and trial method......... X+4=5(bigger with bigger) X=1 and -5×-4×-3×-2 X+1=-5(smaller with smaller) X=-6 Your method is correct but for saving time i analyzed the pattern and hence!!!!!!!!!!! Your method is also appreciable🎉

[@pierrettebalazut9407]

Génial ! J'étais bien partie, mais j'oubliais qu'on pouvait faire un changement de variable !..pour simplifier le calcul

[@bpark10001]

One look at 120, it is 5 factorial. The equation has ascending numbers in it, so X = 1 can be determined by inspection. Because there is even number of factors, negative numbers can give positive result so -6 can also be determined by inspection. These 2 factors (X - 1 & X + 6) can be factored out to give the quadratic equation X² + 5X + 16 = 0. The roots of this are (-5 + i√39)/2 and (-5 - i√39)/2. These are not real numbers so are not part of problem solution, but are the other 2 roots of the 4th order equation for completeness.

[@Sci-Marvels]

Nice observation! Spotting the factorial and using inspection to quickly find the real roots is a great approach. And you’re right about the complex roots-they complete the solution even if they aren’t part of the original problem’s real-number solution. Thanks for sharing your method! 👏😊

[@vacuumcarexpo]

Using x(x-a)(x-2a)(x-3a)+a^4=(x(x-3a)+a^2)^2, (x-1)(x-4)+1=±11

[@cachotrelles4715]

Excelente. Gracias

[@richi2casa127]

import numpy as np # Coefficients of the quartic equation coefficients = [1, 10, 35, 50, -96] # Solve the equation roots = np.roots(coefficients) roots

[@williamBryan-k2e]

here is another one - s^2 + 5x -6 is factorable - ( x +6) (x -1 ) - quadratic equation not needed.

[@Sci-Marvels]

Good point! When the quadratic is factorable like this, it's definitely quicker to just factor it directly. No need for the quadratic formula when the factors are right there! Thanks for the reminder. 😊👍

[@skverma7278]

Sir, we have to find 4 factors of 120 such, these are consecutive numbers. Factors of 120: 2×2×2×3×5 Possible arrangement is 2×3×4×5 Hence value of x is 1. This will be true for negative values of factors i.e. for -2,-3,-4,& -5 for which x will be -6.

[@walterwen2975]

Math Olympiad: (x + 1)(x + 2)(x + 3)(x + 4) = 120, x ϵ R; x =? First method: (x + 1)(x + 2)(x + 3)(x + 4) = 120 = (2)(3)(4)(5) = (1 + 1)(1 + 2)(1 + 3)(1 + 4) = (- 6 + 1)(- 6 + 2)(- 6 + 3)(- 6 + 4) = (- 5)(- 4)(- 3)(- 2); x = 1 or x = - 6 Missing two complex value roots Second method: [(x + 2)(x + 3)][(x + 1)(x + 4)] = (x² + 5x + 6)(x² + 5x + 4) = 0 (x² + 5x + 6)(x² + 5x + 6 - 2) = (x² + 5x + 6)² - 2(x² + 5x + 6) = 120 (x² + 5x + 6)² - 2(x² + 5x + 6) - 120 = (x² + 5x + 6 - 12)(x² + 5x + 6 + 10) = 0 (x² + 5x - 6)(x² + 5x + 16) = (x - 1)(x + 6)(x² + 5x + 16) = 0 x² + 5x + 16 = 0, x = (- 5 ± i√39)/2; Rejected, complex value roots x - 1 = 0; x = 1 or x + 6 = 0; x = - 6 Answer check: (x + 1)(x + 2)(x + 3)(x + 4) = 120; Confirmed as shown in First method Final answer: x = 1 or x = - 6

[@AlfredoMirandaSanchez]

I've mentally solved this looking for two factors of 120 that at the same time had two factors which difference is one.: (2x3)x(5x4)=120. X=1 is the value that match the equation.

[@nadiapianist]

1

[@williamBryan-k2e]

I got x =1 from trial and error before seeing the snaswer.. but as someone said - should be 4 solutions as 4th order poly. so I will watch

[@Евгений-ю2р]

Напишу кратко (x^2+5x+3+1)(x^2+5x+3+3)=120 обозначим (x^2+5x+3)- b тогда имеем (b+1)(b+3)=120 далее b^2+4b-117=0 отсюда b1=9 ,а b2=-13 т.е x^2+5x+3=9 и x^2+5x+3=-13 как видим x^2+5x-6=0 и второе x^2+5x+16=0 далее всё остальное подробно на видео.Ответ: x1=1; x2=-6

[@Sci-Marvels]

Отличное разложение задачи! Вы чётко показали, как перейти от замены переменной к решению уравнений и нахождению корней. Прекрасная работа!

[@hakanerci4372]

X=-6 (x+1).(x+2).(x+3).(x+4)=120

[@稲次将人]

2・3・4・5=120だからx=1 x=-6のとき(-5)(-4)(-3)(-2)=120 ∴x=1,-6

[@m.s.9023]

Using a(a+b)(a+2b)(a+3b)+b^4 = (a^2+3ab+b^2)^2, {(x+1)^2 + 3(x+1)+1}^2 = 11^2

[@Sci-Marvels]

Impressive use of identities! It’s always fascinating to see such elegant algebraic manipulations that simplify the problem so effectively. Thanks for sharing this approach! 😊👏

[@manjeetjoon7]

-6

[@สมเจียวเจริญโสภา]

ตอบ X=1

[@sankariraman6453]

X=1

[@lioneljohnpilcher204]

I don't have first one. Please help thanks ❤ l mean it please.

[@TeachChulbul]

putting the value of 1 =x

[@TeachChulbul]

X=1then multiply

[@JeromeTui]

1:20

[@YoussoufSOULAIMANA]

X = 1 is the answer

[@ahmetmete8366]

Böyle matematik düşman başına. O ne öyle ortaokul çocuğuna anlatırmış gibi.

[@Sci-Marvels]

Matematik bazen karmaşık görünebilir, ama basit bir şekilde anlatıldığında herkesin anlaması daha kolay oluyor. Herkesin seviyesi farklı olabilir, ama amaç hepimizin öğrenmesi. 😊

[@SuperAnangs]

1 sec, x=1

[@haitrieu0051]

a+b+c=0, x=1 ,x=c/a , x=1 ; x=-6

[@lioneljohnpilcher204]

Incorrect. A vector can amount it's velocity. I need someone to look just pass proxima B please

[@AmédéeGerest]

I think I found a quicker way this time : y = x“ + 5x + 4 y(y+2) = 10.12 y = 10 x" + 5x + 4 = 10 x" + 5x - 6 = 0 (x-1)(x+6) = 0 S = {-6;1} Is this correct?

[@iitiansaurabhbhu]

​@@AmédéeGerestyes

[@822Adi]

Such a fool You can simply factorise 120 so you get 2 3 4 5 continuous numbers so in order to get those simply add 1 So x = 1

[@neeldesilva8088]

1

[@ramakrishnanwarrier6662]

X= 1

[@lakshmikutty2255]

x=1

[@sakirhussain665]

X=1

[@yordanyordanov7257]

x=1