Kosovo - Math Olympiad Question | A Nice Algebraic Equation

[@cschaka1]

You must be kidding, this can't have been proposed in a Math Olympiad contest. This is 5th grade normal day-to-day stuff. The Math Olympiad questions I remember from Romania in the 90s didn't even look like Maths, that's how hard they were. You didn't even know what that was.

[@lucsas9277]

as a fellow romanian i agree tho this stuff is more 7th grade level cuz binomials arent taught till 7th grafe

[@oficialmundodiverso]

Que dices esto es imposible

[@ANIMX_FLAME]

Bro quadratics aren't in 5th grade 😂

[@hive123kbps4]

​@@ANIMX_FLAMEyes they are ..in asia

[@imoyal]

1) Combination of any value for X and 0 for y is a solution. 2) Combination of any value for x and the same value for y is a solution.

[@pphguimaraes]

Using some Basic factorization, we can rewrite this as (x+y)(x-y)=(x-y)^2. This can be further simplified to (x-y)[(x+y)-(x-y)]=0. After some algebra this reduces to y(x-y)=0. This equation has two distinct solutions: y=0 for all x And x=y, for all x,y in Reals

[@劉志仁-r3l]

y=0 for all x, [or not and] x=y, for all x,y

[@scmtuk3662]

These are the actual solutions, according to symbolab. {x = y; y ≠ 0} - this means that if y is not 0, then x is equal to y. For example, if y = 25, then x must also be 25. {y =x; y = 0} - this means that if y = 0, then x = 0, however, the true answer is that if y = 0, then x can be _anything_ However, for some reason, there's no such symbol that means "literally everything ever". There are symbols that mean "real numbers", "natural numbers", etc, but there's no single symbol in mathematics that means "x can be anything you want, ever, no exceptions. The reason is, if the values are the same, then x² - x² = 0, obviously, and then (x - x)² = 0² = 0 If however, the value of y = 0, but the value of x is "anything", then x² - 0² = x², and (x - 0)² = x². In other words: The following solutions will work: If x and y are the same value, then it will work. If y is 0, but x is any value, then it will work. For example: Let x = 25, and y = x, thus y = 25 25² - 25² = 625 - 625 = 0 (25 - 25)² = 0² = 0 Now let x = 25, and y = 0 25² - 0² = 625 - 0 = 625 (25 - 0)² = 25² = 625

[@paulcolburn3855]

correct. y MUST be 0. But x can be anything. She was good in her solution until the last 30 seconds. Stop talking then.

[@scmtuk3662]

@@paulcolburn3855 Actually, y doesn't need to be 0. It can either be 0, or whatever x is. I.e y = x or 0, but x = y, only if y =/= 0

[@paulcolburn3855]

@@scmtuk3662 I agree

[@scmtuk3662]

​@@paulcolburn3855 It just makes sense, even if you don't actually use values for the variables. If we let x = y, then: x² - y² = x² - x² = 0 (x - y)² = (x - x)² = 0² = 0 If however, we let y = 0, and x = x, then: x² - y² = x² - 0² = x² (x - y)² = (x - 0)² = x² The difference between using y = 0, and y = x, is that if y = x, then the solution to both sides is 0, if y = 0, then the solution to both sides is x².

[@realtorn1do]

i did the same

[@erautome]

En passant par les produits remarquables, c'est plus facile. Dans le premier terme de l'équation, on doit considérer que X²-y² = (x-y)×(x+y). Aprés, on fait tout passer à gauche de l'équation, une mise en évidence de (x-y) etc...

[@Juice_Wrld_9]

A much simpler method can be applied. Just convert x²-y² to (x+y)(x-y) and decrease (x-y)² to x-y.

[@neodro4831]

This is a common mistake because we do not know the conditions on x and y, hence when you are dividing by (x-y) to cancel both sides if x=y you are dividing by zero, which is incorrect. hence this cannot be done and you have the actual solution(S) of y=0 and x=y

[@rogeraffleck8677]

@@neodro4831 This is NOT a mistake. You can see just by looking that x=y is a solution. So then you have to ask yourself if there are any other solutions. So assume x not= y and then you are free to divide by x-y. Like most teachers you are trying to turn something simple into something complicated.

[@bilalfilali6334]

​@@rogeraffleck8677you can't make a decision in math juste by looking

[@WattSnehp]

​@@rogeraffleck8677 yess. Simple solution.

[@joon_neel]

Exactly.....acc to me decreaae x-y from both side, is the easy way to understanding

[@oussamajaber3512]

The solution must be {x=y such that y belongs to R}

[@Problemsolver434]

No. 2+2 is not equal to 2-2 The answer is 0 for both

[@themaster9041]

​@@Problemsolver434You are wrong. For example: x = y = 3 3^2 - 3^2 = (3 - 3)^2 9 - 9 = 0^2 0 = 0

[@Zerusa766]

@@Problemsolver434 2+2? You wrong

[@Problemsolver434]

​@hadidmarwan9780 I am not. Can you read at all. I just gave an example of a solution that fits exactly what the commenter said and is obviously wrong. I was proving a point. The comment basically implies that all Real numbers are a solution to the problem

[@suptibhattacharjee9811]

​@@Problemsolver434 Nope. x = y , where x belongs to Real numbers is one of the solution. Just put a random real value of x in the equation and you will see that x² - x² = 0 (for left side) And (x - x)² = 0 (for right side) So the equation holds true for x = y where x belongs to R.

[@Psykolord1989]

Before watching: x^2 - y^2 = (x+y)(x-y), thus: -> x^2-y^2 = (x-y)^2 -> (x+y)(x-y) = (x-y)(x-y) PRESUMING that x-y =/= 0, divide by x-y -> x+y = x-y 2y = 0 y=0 If y=0 then x^2-y^2 = x^2 - 2xy -y^2 -> x^2 = x^2. First set of solutions: y=0, x = any real number. SECOND set: Multiply out RHS: x^2 - y^2 = x^2 - 2xy + y^2. Subtract x^2 and add y^2 on both sides: 0 = -2xy + 2y^2 -> 2y^2 - 2xy = 0 -> 2y(y-x) = 0. This actually gets us to our solutions faster, but I'm working this out as I type it, so my mistake. This means either 2y = 0, y-x = 0, or both. 2y=0 -> y=0, x = anything y-x =0 -> y = x. Our solutions are thus: Y=0, X = any real number And Y=X

[@turtletalks1]

Would it not be faster to... (x+y) (x-y) = (x-y) (x-y) x+y = x-y 2y=0 so y=0

[@kee1zhang769]

You are right it is quicker to do it from (x+y) (x-y) = (x-y) (x-y). Obviously if x=y for any x, the equation is true. If x≠y then we can divide both sides by (x-y) to get x+y = x-y and y=0 from your argument, and x can be any number in this case.

[@turtletalks1]

@@kee1zhang769 and I would argue that is the correct answer because plugging back y=0 into the equation, X could = 0, but could equal anything else. So X has an infinite number of possibilities

[@Fernandez218]

this is the way i did it from the thumbnail in my head lol factorization so much easier.

[@user-hi1kl2ej6x]

I thought the same 🤔

[@bakura9683]

Neglecting (x-y) on both sides destroys one of the solution which is x=y

[@alcionidas649]

I really love your videos. Thank you so much!!!

[@Dalesmanable]

I read Maths Olympiad then spent more time looking for a catch than getting the solution that was so easy I did it in my head.

[@AmitYellin]

x²-y²=(x-y)² (x-y)(x+y)=(x-y)(x-y) If (x-y) is not 0 then x+y=x-y 2y=0 y=0 If x-y is 0 then x=y Did it in my head in ten seconds...

[@ayoubkhlifi4903]

x^2-y^2=(x-y)^2 imply (x-y)(x+y)=(x-y)^2 for x different from y: imply x+y=x-y imply y=-y imply y=0 ,and replace the y by 0 we will have x^2=x^2,so x has infinite solutions. for x=y: infinite solutions for x and y

[@lusalalusala2966]

Your final answer should be: (x, 0) and (x,x), for any value of x in the domain of the equation.

[@sauravbisht2019]

x²-y²=(x-y)² (x+y)(x-y)-(x-y)(x-y)=0 How take (x-y) common then (x-y)(x+y-x+y)=0 (x-y)(2y)=0 Thus , (x-y)=0 or (2y)=0 Which means x=y or y=0

[@samriddhisingh7783]

Can you make video for science Olympiad??

[@jim2376]

By inspection the following work: x = 1 or -1 and y = 0.

[@amireallythatgrumpy6508]

Any inspection check should show that y = 0 works regardless of x.

[@davethomson5921]

One equation with 2 unknowns has no unique solution. X=0 for all Y, Y=0 for all X, Y=X for all Y and X.

[@sushreesibanimohapatra7405]

It is not right right The formula is x2-y2= (x+y) (x-y)

[@sudhakaranpillai9623]

(x+y)(x-y)=(x-y)(x-y) (x+y)=(x-y) x=0,y=0 and x=any real number

[@abhinavmehrauwu]

Ma'am but equation with two variables have infinite solutions right?

[@piyushpandey2480]

Very well explained

[@user-hi1kl2ej6x]

(x-y)×(x+y)=(x-y)×(x-y) x+y=x-y 2y=0 y=0 x= infinite solutions

[@evgtro8727]

The equation contains two variables x and y. Therefore its solution must be given in the form of a pair (x,y). You cannot just say y=0 is a solution.

[@paulcolburn3855]

y must be 0. In order for this equation to work, y must ALWAYS be 0. x can be anything.

[@evgtro8727]

@@paulcolburn3855 Try to write your statement as an answer in the proper mathematical form.

[@ashokkukalekar1214]

Thanks!

[@economics01]

thank you so much

[@Mardouck]

Why this questions so easy?

[@Statistica101]

It could also be 2y^2 -2xy=0 2y^2 = 2xy Y^2= xy Y=x y=x=1

[@amireallythatgrumpy6508]

y^2 = xy should be written as y^2 - xy = 0 and factored to y(y-x) = 0. You lost the y=0 solution.

[@zakariazaki7513]

Thanks for video keep going 🤠 greeting from Morocco

[@pawelskiba5169]

x^2 - y^2 = (x-y)^2 (x-y)(x+y) = (x-y)(x-y) | (x-y) => x y first solution x+y = x -y | -x y= -y => y =0 second solution check y=0 x^2 = x^2 check x=y x^2 - x^2 = 0^2 0=0

[@jmlfa]

Not correct: Y=0 is indeed a solution, but X then can take any value, not just X=Y.

[@amireallythatgrumpy6508]

There are two different solutions. It's true when y = 0, or when y is the same as x.

[@rahulp9352]

Find f(s) where f(s)=1+1/2^s+1/3^s+.....=0

[@Call_ZZ]

very easy but good practice i guess

[@नेाटटिवदिललीआळेधीप्रग]

Ques. find value of x and y. ans. after solving, y = 0 ... i x = y ... ii from i and ii, we get, x = y = 0 so, x = 0 and y = 0 ... final answer.

[@mauri46100]

What if y=0 and x may have any value you want?

[@chakrig6080]

Do any of you know this pen brand?

[@properbeatz]

x=1, y=0 is anther soln, unless I'm mistaken.

[@amireallythatgrumpy6508]

That would be covered by y = 0. It is true if y = 0 no matter what x is.

[@edumathhub]

Cool video

[@enes3573]

Bu soru Türkler için çok basit. This question is very simple for Turks.

[@economics01]

thanks

[@jesuspuhwillreturn]

What really used in life . I mean real time where applicable?

[@hmunjam]

Why not x = y = 1?

[@MabelGyamfi-v9g]

I think this a basic math for binomial is too easy

[@gatorpuppy]

infinite solutions 🤯🤯🤯 as long as x = y

[@ABG__GARMERZZ]

mam the answer you got x=y know mam we know that y=0 so that we can substitute the value of y in the equation of x so that we get x=0 is that correct mam ?

[@ijeomachigozieejefobiri9911]

Think about a much simpler and faster method to teach us later

[@soheilabarahimi-sw3bj]

X=1. Y=1

[@mathcem138]

No what is this X=y Any value

[@saurabhgarg9935]

This is a very simple question. Can't be an Olympiad question 😅

[@lifeatph]

I'm sorry but I don't understand it. Make it simple to understand..... it's more complicated dear ....too vague

[@lawrenceliebman9079]

y = 1 also works

[@pawelskiba5169]

x=2, y=1 4-1 (2-1)^2

[@scmtuk3662]

There's no real way to write the correct solution in symbols, so I'm going to describe it in words. The value of x can be anything you want it to be. And, the value of y can either be the same value of x, _OR_ 0. If we let y = 0, x² - 0² = (x - 0)² x² = x² However, if we let y = x, then: x² - x² = (x - x)² 0 = 0² 0 = 0

[@amirah23]

@@scmtuk3662so y is dependant on x’s value, but not vice versa?

[@scmtuk3662]

​@@amirah23 Basically yes. x can be anything you want. But y must be either the same value of x, or 0.

[@vijayannair2316]

Nice

[@AASHUKUMAR-bo6si]

1by 2 is answer 😅

[@VijayaRamya-k9t]

X^2-y^2=(X-y)^2 (X+y)(X-y)=(X-y)(X-y) X+y=x-y 2y=0 y=0 (1) X^2-y^2=(X-y)^2 X^2-y^2=X^2+y^2-2xy 2xy=2y^2 X=y (2) From (1)&(2) X=y=0

[@안녕하세요-b6w9t]

Too easy for Korean

[@mariakos756]

1:31 or , not and

[@chandanmeher2897]

This equation makes no sense

[@aamirahmed898]

Not good enough

[@djhumantorch]

Kosovo 😀😀😀😀 Serbia, you mean

[@l_h_2]

She means Albania🇽🇰🇦🇱

[@djhumantorch]

@@l_h_2 In your sweetest dreams, pal, Kosovo will be free once more

[@fsponj]

​​@@djhumantorch Kosovars & Bosnians were oppressed enough by Serbians

[@劉志仁-r3l]

Shame on you. That wrong. The answer is y=0 or y=x

[@Dosadniste2000]

Kosovo is Serbia, thank you.

[@harshakalikiri6587]

😂😂😂😂😂😂😂😂

[@noanmee]

Is this Olympiad for dumbest kids? if x² - y² = x² - 2xy + y², so -2xy = 0. So => x = 0 or y = 0

[@oksanalysenko671]

Жах.Не має відповіді і якщо взялися щось пояснювати то не дуоіть людям голову,а пояснюйте до кінця правильно ,а не просто коментуєте розв'язання ,як для сліпих.

[@juancarlosvergaramartinez8234]

This demo is wrong, error