Philippines Maths Olympiad 2020 Problem | Algebra | 2 Different Methods

[@honestadministrator]

Let m, n be roots of this equation m + n = Sum of the roots = - b Hereby b is an integer Again m n = product of the roots = - 3 b (m - n ) ^2 = ( m + n) ^2 - 4 m n = b^2 + 12 b = ( b + 6 ) ^2 - 36 Hereby (b + 6 ) ^2 - ( m -n) ^2 = 36 = sum of consecutive odd integer 36 = 17 + 19 = (10)^2 - 8^2 b + 6 = 10, -10 i.e. m + n = -b = - 4, 16 at m + n = 16 and m - n = 8 m = 12 n = 4 , b = -16 at m + n = -4 and m - n = 8 m = 2 n = -6 , b = 4 m = 12 n = 4 , b = -16

[@MathOrient]

A simple equation, yet deep problem. Thanks for sharing :) It combines algebra and number theory.

[@benjaminvatovez8823]

Thank you. I propose a 3rd method: Note that (x not= 3), thus (-b)=x²/(x-3)=x+3+(9/(x-3)) -b=alpha+beta in Z, thus (x-3) divides 9 then (x-3) in {-9;-3;-1;1;3;9} Put it in (-b)=(x-3)+6+(9/(x-3)), we get b in {-16;-12;0;4} and we get the answer.

[@alexey.c]

The first solution has a flaw. D may not be a perfect square for an integer x. Example: b = 1/4, D = 49/16, x = {-1, 3/4}

[@MathBooster]

3/4 is not integer. (x is integer means both the values of x is integer)

[@marianne-wt8it]

Please bring more questions on quadratic equation it will help a lot 🤧

[@peterkrenn9051]

Dear! Don't cry Honey, don't cry. In the second method you have to use, only the Viete formulas and the number 9 you have to factorise in all kind of methods. This will drive you to the solution. ❤❤🙂🙂🙂

[@sevencube3]

The quadratic formula is the one that everyone learns, and thus belongs in Tier 2, which is Over Used, the cubic is Rarely Used (Tier 5), and the quartic formula, for all intents and purposes, is Never Used (Tier 6).

[@labzioui1]

…if x is an integer , then D must be perfect square and (yes and) -b ±sqrtD will be divisible by 2a .

[@JabirKarachiwala]

One assumption. Is it necessary for discriminant of the quadratic equation to be a perfect square? It could be in decimal , so its square root also being in decimal with the value of b being in decimal would make the root as integer.

[@MathBooster]

If x is integer that means both the values of x is integer. So, now you can try to put decimal values, you will not get both the values of x as integer (if square root of D is not integer)

[@mathsjourney6608]

Great job bro 🙏

[@ejrupp9555]

Just here to figure out what the question is ... and why you would want to know it.

[@vijayannair2316]

Verynice

[@MathBooster]

Thank you 🙂

[@piman9280]

If x is an integer, then D must be a perfect square and D *and* b must be even.

[@MathBooster]

b can be odd also, if square root of D is odd.

[@alexey.c]

Since b may not be an integer, then D may not be a perfect square for an integer x: b = 1/4, D = 49/16, x = -1

[@MathBooster]

@Alex Chuvilskiy X is integer means both the values of X is integer. (So if b is not integer then both the values of X can't be integer. So, b will be integer)

[@piman9280]

@@MathBooster True.

[@TrebolManger]

Other method! x² + b x - 3 b = 0 Formula Quadratic √(b (b + 12)) - b x = ⎯⎯⎯⎯⎯⎯⎯----------- 2 Condition: b even replace b=2n √((2 n) (2 n + 12)) - 2 n x = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯- 2 Algebra x = √(n (n + 6)) - n Condition it's (+) n (n + 6) > 0 n < -6 ∨ n > 0 Summary Table n b = 2 n x² + bx - 3 b = 0 X1 X2 -8 -16 x² - 16 x + 48 = 0 12 4 -6 -12 x² - 12 x + 36 = 0 6 Only 0 0 x² = 0 0 Only 2 4 x² + 4 x - 12 = 0 -6 2 Sum -24 -.- -.- -.-

[@sumit180288]

You are just using hit and trial and how did you ensure that there are no other integer solutions for more values of n, since n < -6, n>0 there are infinite possibilities of n

[@TrebolManger]

@@sumit180288 Hai risolto e confermato o pensi che ci siano più soluzioni? quali trovo? O semplicemente criticare e minare il lavoro degli altri?

[@sirak_s_nt]

Can m be -ve? As √D = m?

[@MathBooster]

No, m can be any positive integer.

[@sirak_s_nt]

Then in 6:42 case m is -ve? I doubt is also if we right √D = m thn m can be +ve but whn we write m²=D, thn m cn be both -ve n +ve so which case we tke for solving? Plz tell sir

[@MarieAnne.]

By convention, √D ≥ 0. This is why we put ± in front of √D: x = (−b ± √D)/(2a) For example, if we had D = 16, then √D = 4 and we get: x = (−b ± 4)/(2a) But if we allow √D = −4, then we get: x = (−b ± (−4))/(2a) = (−b ∓ 4)/(2a) So if you allow √D = m to be either positive or negative, you get 4 results, but only 2 answers. So what's the point? Furthermore, since we often use exact numerical values in math (π instead of 3.14 or √2 instead of 1.41), it's important that we have rules that we must stick to. For example, tan(60°) = √3, while tan(120) = −√3. If √3 were allowed to be either positive or negative, then we'd have two values for each of tan(60°) and tan(120°), which is not possible. That being said, at 5:25, when finding factors of 36 so that (b+6+m)(b+6−m) = 36, we can assume that b+6+m ≥ b+6−m. This will eliminate the duplicate solutions. Also, in the second method where we denote the root as α and β, we could also assume that α ≤ β without loss of generality. And again, this will eliminate duplicate solutions for b.

[@hazalouldi7130]

Always on top

[@kinno1837]

First