Our Olympiad math problems are not so plain. This one would be OK for 7th grade' regular school lesson, definitely not for Olympiad, unless primary school also takes part. I've opened list of problems for recent 2024 contest and you know what? There are NO equations to solve at all there. I can translate for you the very first problem targeted for 9th grade. Try to solve it. Petya and Vasya know only natural numbers not greater than 10^9-4000. Petya counts as "good" ones only such numbers, which can be expressed as abc+ab+bc+ac, where a, b and c - natural numbers, not less that 100. Vasya calls "good" ones only numbers, which can be recorded as xyz-x-y-z, where x, y and z are natural numbers greater than 100. Who of them has more "good" numbers? That's it. THIS is the level of our Olympiad. Of course, you can simplify the task significantly by renaming Petya and Vasya to Peter and Basil respectively.
x = +-sqrt(2i) x = +-(sqrt(2)* sqrt(i)) That is not the end of it, as a complex number should be of the form a+bi where a and be are real numbers. sqrt(2) is, but the rest is not. You have to find out then what sqrt(i) is.
I just solved it in 1 min by putting x = a+ib , BTW I m preparing for iit jee advanced
friend I am not from the United States so that is that 😂
Our Olympiad math problems are not so plain. This one would be OK for 7th grade' regular school lesson, definitely not for Olympiad, unless primary school also takes part.
I've opened list of problems for recent 2024 contest and you know what? There are NO equations to solve at all there. I can translate for you the very first problem targeted for 9th grade. Try to solve it.
Petya and Vasya know only natural numbers not greater than 10^9-4000. Petya counts as "good" ones only such numbers, which can be expressed as abc+ab+bc+ac, where a, b and c - natural numbers, not less that 100. Vasya calls "good" ones only numbers, which can be recorded as xyz-x-y-z, where x, y and z are natural numbers greater than 100. Who of them has more "good" numbers?
That's it. THIS is the level of our Olympiad. Of course, you can simplify the task significantly by renaming Petya and Vasya to Peter and Basil respectively.
Ok
Really nice problem 😂😂😂
hahahaha...😂🤣😂🤣
Really nice math problem.
Thanks for watching and commenting my good friend.
Much love...❤️💖💖
Слишком много лишних вычислений в ролике.
x⁴ = -4
x⁴ = 4∙(cos(π + n∙2∙π) + i∙sin(π + n∙2∙π)) - тригонометрическая форма комплексного числа
x = (4 ** ¼)∙(cos((π + n∙2∙π)/4) + i∙sin((π + n∙2∙π)/4)) - максимально удобна для извлечения корней
x = √2∙(cos(π/4 + n∙π/2) + i∙sin(π/4 + n∙π/2)) - 4 корня для n = 0, 1, 2, 3
x = √2∙(±√2/2 ± i∙√2/2)
x = ±1±i
Thanks my good friend and mathematician
X^2 = plus or minus sqrt(2i)
X^2= plus or minus sqrt(-2i)
Ok
Really maths challenge from you. Thanks
X= + -(√2i) also
Not true. (√2i)^4 + 4 = 8 = (-√2i)^4 + 4.
Simple and tough
Laugh!!! 🤣😂🤣😂🤣
X⁴+4=0> ,(X²)²+ 2²=0,(x²+2)(x²+2)=0,x²+2=0,x²=-2, la somme de deux Carré n'est jamais nul,S={0}.
This approach will give only two roots out of the four roots we are expected to get from here sir.
Thanks for the suggestion sir.
12 минут это решать🤯? Вон x/7=7
Due to detail explanations hence the much time spent solving.
Really nice math problem.
Glad you liked it
Excelente aula.
Thanks for watching and encouraging us sir.
143
143? how do you mean sir?
❤
We love you too sir...❤️💖❤️💖💖
can you read it your self
How do you mean?
Your question, comment or statement not clear sir.
Rephrase your question or comment sir.
Thanks a million sir.
x^4 - (sqrt(2i))^4 = 0 ?
Kindly simply sir.
You are a very good teacher
Thanks and welcome sir.
We respect you sir
Square root -3
How do you sir?
Please kindly rephrase your question for proper understanding.
Thanks sir.
X^4 - (2i)^2 = 0
You are good at math
Good ❤
Thanks a bunch sir.
1
Wrong, kindly resolve the challenge or watch this video tutorial from beginning to the end.
X=2i
correct also
Wait is it?
I think it’s wrong tho
I just took the square root of -4 not the fourth root
Bro its hard,the simple solving way in my opinion:
x⁴=-4
x²=2i
x=±√2i
x = +-sqrt(2i)
x = +-(sqrt(2)* sqrt(i))
That is not the end of it, as a complex number should be of the form
a+bi
where a and be are real numbers.
sqrt(2) is, but the rest is not.
You have to find out then what sqrt(i) is.
ok sir
Дуже захмарно
Really!!!
Nice math solution
Thanks my wonderful friend, hahaha 🤣🤣🤣.
College level
Yes boss!!!