Kosovo - Math Olympiad Question | A Nice Algebraic Equation

Показать описание

Maths Olympiads are held all around the world to recognise students who excel in maths. The test is offered at many grade levels and provides them with numerous possibilities to win certifications, awards, and even scholarships for higher studies.

LKLogic

Рекомендации по теме

Комментарии

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.

cschaka

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.

imoyal

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

pphguimaraes

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

scmtuk

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

Juice_Wrld_

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...

erautome

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

oussamajaber

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

turtletalks

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

Psykolord

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

lusalalusala

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.

evgtro

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

sauravbisht

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

ayoubkhlifi