Solving A Nice Rational Equation | Math Olympiads

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[DrQuatsch]

Okay, so I saw it was x(x + 1)(x + 2)(x + 3) = 360. I saw this and thought that if there's an integer solution, then we have a product of 4 consecutive integers equal to 360. I know that 4^4 = 256 and 5^4 = 625, so 3 * 4 * 5 * 6 seemed like a good guess, because the product of four consecutive numbers will be relatively close to the average of the 4 numbers to the fourth power, so in this case 4.5^4, which must lie between 4^4 and 5^4. Turns out it is actually equal to 360. But because it's a product of four terms, the negative values should also work... - 6 * - 5 * - 4 * - 3 is also a solution. Therefore x = - 6 and x = 3 are two solutions to this equation. Now I just expanded the equation; x^4 + 6x^3 + 11x^2 + 6x - 360 = 0. We know two solutions, so we can take out factors of (x - 3) and (x + 6), so (x - 3)(x + 6)(x^2 + 3x + 20) = 0. x^2 + 3x + 20 = 0 gives us x = - 3/2 +- i/2 * sqrt(71).

[Hobbitangle]

Instead of using the substitution y = x^2+3•x in the first method, it's better to use y = x^2+3•x + 1
The equation becomes (y+1)•(y-1) = 360
y^2-1=360
y = ±√361 = ±19

[moeberry]

It is crucial that we state in the beginning that x cannot be 0 or -1 since it makes the right hand side undefined. Also method 2 does not find you the imaginary roots so method 1 is the best.

[draido-dev]

i want to correct a problem in 3:05 it's not 6x^3 it's 6x^2 the cubed one is x^3 recheck if you want :)

[mystychief]

Method 5: write down all the factors of 360 in increasing order. After 18 you can take the rest-factor (20 etc.). The only consecutive series of 4 are 1, 2, 3, 4; 2, 3, 4, 5; and 3, 4, 5, 6. The last one fits: 3*4*5*6 = 360. So x = 3. I did not think of the negative possibility: -6, -5, -4, -3.

[blueshoter]

We have:
x(x+1)(x+2)(x+3)=360
Step 1:
-Multiply x with (x+3), resulting x^2+3x.
-Multiply (x+1) with (x+2), resulting x^2+3x+1.
->We have (x^2+3x)(x^2+3x+2)=360
Step 2:
-Set y=x^2+3x+1
-We have x^2+3x=y-1
-We have x^2+3x+2=y+1
->We have (y-1)(y+1)=360
-Break the brackets out, we have:
y^2-1=360
y^2=361
y=square root of 361, which is 19 or -19
->We have
x^2+3x+1=19
Or
x^2+3x+1=-19
Step 4:
-Solve the quadratic equation:
•With 1st case:
x^2+3x+1=19
x^2+3x=18
x^2+3x+9/4=18+9/4
Using the formula of a^2+2ab+b^2=(a+b)^2:
(x+3/2)^2=81/4
x+3/2=square root of 81/4
->x+3/2=9/2
or
x+3/2=-9/2

->x=3
or
x=-6
•With 2nd case:
x^2+3x+1=-19
x^2+3x=-20
x^2+3x+9/4=-20+9/4
(x+3/2)^2=-71/4
x+3/2=square root of -71/4
x+3/2=(i.square root of 71)/2
x=(i.square root of 71 -3/)2
=>x=3
or
=>x=-6
or
=>x=(i.square root of 71 -3)/2

[stardust-rz]

The same reason why 360 was chosen to be the number of degrees around a circle is that that would make you assume you can factor it to get the solutions: it has a lot of prime factors (which makes it easier to divide angles.)

Anyway, it's always worth the try, but the first method is better because it's not assuming.

[yoshinaokobayashi]

given equation is x(x+1)(x+2)(x+3)=360=3X4X5X6. then x=3 or -6 are answers. then given equation is factorized to be (x-3)(+6)(x^2+3x+20)=0, then other composite answers are (-3 +- isqrt(71))/2. Fin.

[gw]

For the positive solution, after rearranging to give x(x+1)(x+2)(x+3)=360, it can be seen that 360 = 6! / 2 (i.e. 6! / 2!), so x must be > 2 and must also satisfy (x+3) = 6. So x = 3 is an immediate solution.

[nasrullahhusnan]

It is simpler to work it this way. The equation can be written as
x(x+1)(x+2)(x+3)=360
Multiply the two terms in the middle and the remaining two of LHS
[x(x+3)][(x+1)(x+2)=360
(x²+3x)(x²+3x+2)=360

(x²+3x+1)²-1=360
(x²+3x+1)²=361 --> x²+3x+1=19
x²+3x-18=0
(x-3)(x+6)=0
--> x=3 or x=-6

[alextang]

Because x(x+1)(x+2)(x+3)=360 and 360=3*4*5*6=(-6)*(-5)*(-4)*(-3), x=3 or x=-6 are the roots. Then the left 2 roots can be also easily found by long division to quadratic equation. 😊😊😊😊😊😊

[mohamedshafeeq]

x(x+1)(x+2)(x+3) is the product of 4 consicutive integers, if x is an integer.
20 divides 360 and 20 is the product of consecutive integers 4 and 5.360/20=18 where 18= 3x6.Thus the numbers are 3, 4, 5, 6 or -3, -4, -5, -6
Thus x= 3 or -6

[Nameless-qehu]

I suppose I solved it the text book way. I used Descartes’ rule to make a list of potential roots, and then synthetically divided numbers to get the two linear roots and the quadratic.

[FrogworfKnight]

Done before watching the video. Apologies for how the synthetic division portion of this is formatted, youtube doesn't like extra spaces it seems. Quotation marks are so things are spaced out nicely.
(x+2)(x+3)=360/(x^2+x)

(x^2+x)(x+2)(x+3)=360
x(x+1)(x+2)(x+3)=360
3*4*5*6=360 and -6*-5*-4*-3=360 //saw that it was just a sequential number and made an educated guess (partially done with knowledge of factorials)
x=3, -6

(x^2+x)(x^2+5x+6)=360
x^4+5x^3+6x^2+x^3+5x^2+6x=360
x^4+6x^3+11x^2+6x-360=0

3|1 6 11 "" 6 -360
"| " 3 27 114 360

" 1 9 38 120| 0
-6| 1 9 38 120
"| " -6 -18 -120

" 1 3 20| 0
x^2+3x+20=0
x=(-3+-sqrt(9-4*20))/2
x=(-3+-sqrt(-71))/2
x=(-3+-isqrt(71))/2

x=3, -6, (-3+isqrt(71))/2, (-3-isqrt(71))/2

[RexxSchneider]

x(x+1)(x+2)(x+3) = 360. If there are integer solutions, they are four consecutive integers, much smaller than 360.
That motivates us to look at the fourth root of 360 ≈ ± 4.36, which strongly suggests the consecutive integers are arranged two below ± 4.3 and two above ± 4.3.
Those would be 3, 4, 5, 6 or -6, -5, -4, -3. Check: 3 * 4 * 5 * 6 = 360.
So x = -6 or x = 3 are the real solutions. There are two other complex solutions in the vicinity of ±4.3i, but I guess we're not interested in those.

[something]

I watch all your vidoes can you make vidoes teaching Calculus
pls i would love to watch you teach Calculus

[goldfing]

Definition set D = C \ {0, -1}
Multiply:
x(x+1)(x+2)(x+3) = 360
Rather than expanding and collecting and Solling
x^4 + 6x^3 + 11x^2 + 6x - 360 = 0
and testing all divisors of 360 (good luck!), I take

Method 2: Substitute t = x + 1.5 = x + 3/2
(which is the average of x, x+1, x+2, x+3), so we get

(t - 3/2)(t - 1/2)(t + 1/2)(t + 3/2) = 360
We rearrange this to
(t + 1/2)(t - 1/2)(t + 3/2)(t - 3/2) = 360
Use the 3rd binomic formula:
(t^2 - 1/4)(t^2 - 9/4) = 360
Multiply with 4 and 4:
(4t^2 - 1)(4t^2 - 9) = 360*16
16t^4 - 36t^2 - 4t^2 + 9 = 5760
Subtract 9:
16t^4 - 40t^2 = 5751
Add 25:
16t^4 - 40t^2 + 25 = 5776
Use the 2nd binomic formula:
(4t^2 - 5)^2 = 76^2
4t^2 - 5 = +-76
4t^2 = 5 +- 76
t^2 = (5 +- 76)/4
Either t^2 = 81/4, thus t = +-9/2
Or t^2 = -71/4, thus t = +-sqrt(71)/2*i
We resubstitute: x = t - 3/2.
For x, we get the four solutions (two real, two conjugate complex):
x1 = +9/2 - 3/2 = +6/2 = +3
x2 = -9/2 - 3/2 = -12/2 = -6
x3, 4 = -3/2 +- sqrt(71)/2*i = (-3 +- sqrt(71)*i)/2

[andylee]

Instead of y and y-2 I chose z-1 and z+1 for my substation so the product would be a convenient difference of squares. Just a slightly different path to the same solution.

I really need to learn my two-digit squares, I did not recognize 361.

[bqaez]

370 est choisi car c'est le ratio de 2 factorielles donc pour trouver des solutions sur N.
(x+3)! / (x-1)! = 360 2! = 2 6! = 720 6!/2! = 360 (x+3) = 6 x = 3 puis en reprenant les factorielles développées et en changeant pour des valeurs négatives dans l'écriture des factorielles en vérifiant les signes -1.-2.-3.-4.-5.-6 = 720 -1.-2 = 2 donc les facteurs du polynôme sont -6 -5 -4 -3 soit x=-6 et .... x+3=-3
Pour les racines complexes je n'ai pas de solution "originale".

[pedrovargas]

Tried checking the quartic formula. My phone browser collapsed and had to reboot. 🤣