Nice Math Olympiad Simplification Problem

Nice Math Olympiad Simplification Problem | Find "a" and "b"

Місяць тому

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In this Math Olympiad Algebra Problem, you'll learn tips and tricks of solving International Math Olympiad exams quickly. #IMO #matholympiad #algebra #radicalequations #simplify #exponential

КОМЕНТАРІ: 23

@wasimohammad4280 Місяць тому

Consider it a^2=36+b^2,ie a^2=6^2+b^2,the Pythagorean triplets ie 6,8,10. b=8 and a=10.

@jim2376 Місяць тому

"Pythagorean triplets" 👍

@jim2376 Місяць тому

Hint: 100 - 64 = 36.

@superacademy247 Місяць тому

By inspection 🤣💯👌😎🔥

@himo3485 12 днів тому

(a+b)(a-b)=36 a+b=18 a-b=2 2a=20 a=10 , b=8

@ytmiguelar Місяць тому

a^2 - b^2 = 36 a^2 = 6^2 + b^2 a^2 = (2·3)^2 + b^2 a^2 = 2^2 · 3^2 + b^2 ÷ 2^2 => a^2/2^2 = 3^2 + b^2/2^2 (a/2)^2 = 3^2 + (b/2)^2 The best-known Pythagorean triplet is (3,4,5) and it is the only one that has the number 3: 5^2 = 3^2 + 4^2 Then take a/2 = 5 and b/2 = 4, i.e. a = 10 and b = 8 and it’s the only solution (except negative signs).

@superacademy247 Місяць тому

Nice 👍💯 solution

@telemans107 Місяць тому

There is an other solution (6,0)......6^2-0^2=36.

@ytmiguelar Місяць тому

Sí, también está bien. Claro, eso también depende de la posición que tomes en la discusión filosófico matemática sobre si los números naturales inician en 0 o en 1. Yo soy de los que consideran el surgimiento histórico de los números naturales, es decir que comienzan en 1, pues el cero es un concepto abstracto muy posterior al inicio de los números para contar. Igual si consideras las soluciones sobre los números enteros, entonces a y b pueden ser ±10 y ±8, o ±6 y 0. (Y sobre los enteros gaussianos hay soluciones como a = ±8i y b = ±10i, o a = 0 y b = ±6i.)

@dungnguyen-wu8bq Місяць тому

Nghiệm (6,0)

@virnarayansingh7420 Місяць тому

Why restrict (a, b) to natural numbers? (a, b) can also be real numbers. The solutions then include (18.5, 17.5), (10,8), (7.5, 4.5), (6.5, 2.5), (6,0),(6.5, -2.5), (7.5, -4.5), (10,-8) and (18.5, 17.5). All these satisfy a^2 - b^2 = 36. Thus there are 9 solutions to this problem.

@superacademy247 Місяць тому

Restricting to natural numbers is a concept being tested.

@user-nz1fc6ml3s Місяць тому

If a and b can be real numbers than you have infinite solutions: a = +/-sqrt(b^2+36)

@mengzi2009 Місяць тому

The letter b is pronounced p.😮

@superacademy247 Місяць тому

Local dialect interference

@mehmethancicek3372 Місяць тому

why must be (a+b)>(a-b) isn't (a+b)=(a-b) . a,b €N a=6 b=0 0 is a natural number.

@superacademy247 Місяць тому

I restricted natural numbers to counting numbers.

@wasimohammad4280 Місяць тому

0 is not a natural number.

@pianoplayer123able Місяць тому

When you add b to a the result is always bigger. When you subtract b from a the result is smaller. Let's say a=3 b=4. a+b=7 a-b=-1 Try it with several examples.

@mehmethancicek3372 Місяць тому

@@wasimohammad4280 0 is a natural number. a,b€ N it must be a b€N+

@mehmethancicek3372 Місяць тому

@@pianoplayer123able a,b €N and if a=6, b=0 (a+b)=(a-b) but he says a,b € N+ .