Brazil l Nice Olympiad Math Radical Problem l find value of a?

[@TheCktulhu]

√a+√-a=32 √a+√a*i=32 √a(1+i)=32 a(1+2i-1)=+/- 32^2 a*2i= +/- 1024 a= +/- 512/i = +/- 512 * i Вроде так намного проще

[@МиколаДзядук]

Умножим обе части равенства на √а, получим а-а=32√а, или 32√а=0, а=0 Парадокс.

[@Serghey_83]

😂 Ну да. Надо было сразу сказать, что а - мнимое число вида xi

[@TheCktulhu]

Если ты √a+√-a домножишь на √a у тебя получится а - a * i. А не а - а.

[@MorgKev]

Where does "a(1+2i-1)=+/- 32^2" come from? I am referring to the +/-. The original question shows there is pos/neg symmetry but are you saying it follows directly from √a(1+i)=32 to a(1+2i-1)=+/- 32^2? If so, why? Thanks

[@АлександрРоот-п9б]

Конечно, именно такой вариант решения напрашивается в первую очередь.

[@theclown1000]

No need for so much gymnastics. You can just square the original equation on each side and get the solution in less than a min.

[@KUTTTYYOPOPONI.NILNILNIL]

11:11 am, here, particularly here, mainly here, have fully noted that way too many dark-conceited-coakyblots add fully useless imaginary gymnastics++++, ai-blots to nil-up-square-cube, etc but if unable to pay to all vital deals at once, then even all preset solid square deals are always nullified- copying-coakycracked-minds fall the u-turned (-) ai of these xxknowns fall till under roots squared if via all fully ingrained fakes++ preset aiming at any low-fatess later if just as any too remote their too phony dark-black-holes fully unseen upon coaky-minded misusing ?????????????????

[@pavelsanda3149]

Squaring the equation is not an equivalent conversion. Thus the answer must be tested.

[@pascaldechambault8670]

Yes, the function ! f(x) = x2 is not a injective function !

[@MrMLehman]

Yes, I would like to see a follow-up video where the answer is tested.

[@chrisolmsted5678]

The square route of negative 1 is by convention ì. Is there a letter for the square root of ì ? If I assume that letter is ö (in honor of Gödel's incompleteness theorem) then attempts to find the other solution to ö squared (not ì) might show that self referential statements (even in math) can be as paradoxical as Schrodinger's cat. But back to my question, is there a letter for the fourth root of negative 1?

[@RichardPenner]

Sqrt(z) is on the right half of the complex plane, so sqrt(z) + sqrt(-z) is a non-negative real number when z is on the imaginary axis. The equation is symmetric with respect to replacing x with -x everywhere so the two solutions did not represent the introduction of a spurious one.

[@RichardPenner]

@@chrisolmsted5678Gödel has nothing to do with it. The complex numbers are algebraically closed, meaning any polynomial equation with complex coefficients has solutions in complex numbers. x^2 - i = 0 has solutions +/- (1+i)/(2^(1/2)) because (1+i)^2 = (1+i+i-1)=2i All the roots of i exist. [cos(π/(2n)) + i sin(π/(2n))]^n = i

[@mohammednoordesmukh3784]

Better if we square as it is, both sides at the first step: a - a + 2ai = 1024 a = - 512i

[@henishgodwin5643]

Nah bro! That's wrong.

[@henishgodwin5643]

U won't get that 2ai in your 1st step

[@michaelschindler10]

Sure you geht the 2ai in First step. for Dummies: 2*sqrt(a)*sqtr(a)*i But answer musst bei tested for the Same reson th Long stuff in the Video musst bei tested

[@koenraad4618]

@@michaelschindler10 I tested the solution, it is correct, and +512i is also correct.

[@saurabhsrivastava4140]

√a(1+i)=32 Squaring both sides, we get a(1-1+2i)=32×32 a=32×16/i a=512/i a=-512i a=-512i, +512i because during the solution we used squaring both side

[@matthiasroelz]

try to put this result into the equation. You will see, that it will not fit

[@szysztof.buravczek]

@@matthiasroelz √(512i) + √(-512i) = = √512 * (√i + √-i) = = 16√2 * (√i + √-i) = = 16√2 * (√(e^(iπ/2)) + √(e^(-iπ) * e^(iπ/2))) = = 16√2 * ((e^(iπ/2))^(1/2) + (e^(-iπ/2))^(1/2)) = = 16√2 * (e^(iπ/4) + e^(-iπ/4)) = = 16√2 * (cos(π/4) + i*sin(π/4) + cos(-π/4) + i*sin(-π/4)) = = 16√2 * (√2/2 + i*√2/2 + √2/2 - i*√2/2) = = 16√2 * (√2/2 + √2/2) = = 16√2 * 2 * √2/2 = 16√2 *√2 = 16*2 = 32

[@koenraad4618]

@@matthiasroelz Well, I did just that and it fits perfectly. This solution is 100% correct, period.

[@matthiasroelz]

@@koenraad4618 yes, i made a mistake 😅

[@matthiasroelz]

@@koenraad4618 Just on question that drives me crazy: I can write -512i as (16 - 16i)^2 or as (-16 + 16i)^2 If i put this squares into the equation and directly cut the square and the squareroot i got with the first one 32 but with the last one -32 WTF, try it what i am doing wrong?

[@atul58]

Why show elementary steps. Please skip them. These types of videos are not seen by those who need to be explained - divide both sides by 2 😂

[@RavindraRao]

Very true especially if these are Math Olympiad problems

[@ronan.pellen]

@@RavindraRao some of us are just casual curious viewers, doesn't seem very difficult for anyone familiar with complex numbers so they might just as well skip the whole video

[@E.h.a.b]

√-a = i√a √a + √-a = √a + i√a = √a (1 + i) = 32 a (1+i)^2 = 1024 //Square both sides 2 i a = 1024 a = 512/i = - 512 i

[@NikitaBotnakov]

Great!

[@Doonburn]

This overlooks the fact that -i is also a square root of -1, so you also need to consider sqrt(-a) = (-i)sqrt(a). That will take you to the other solution, a = 512i

[@hanslepoeter5167]

​​​@@DoonburnDoes it ? I mean (-i)^2 = -1,where -(i^2)=1. The latter applies so i do not believe +512i is a valid solution. The original poster squares something and then takes the square root which is not a way to freely convert a positive to a negative number or visa versa. I'm pretty sure my bank would not accept. It's called an extraneous solution and indeed should be checked whenever you square something and use squareroot later.

[@Doonburn]

​@@hanslepoeter5167 Go to the original equation: sqrt(a) + sqrt(-a) = 32. From the symmetry between a and -a, is it not obvious that if a = -x is a solution then so is a = x? So if -512i is a solution then so is 512i.

[@Doonburn]

@hanslepoeter5167 Go to the original equation: sqrt(a) + sqrt(-a) = 32. From the symmetry between a and -a it is obvious that if a= -x is a solution then so is a = x. So if -512i is a solution then so is 512i.

[@felipe3889-p6b]

This result is correct but can be obtained in much fewer lines!🙂

[@olaminiranadetula6683]

Sometimes the steps are needed for better understanding.

[@dextrous3rd485]

Absolutely. Squaring the first line would do it. Then checking to see if the solutions are valid .

[@umitsayman]

It better for math enthusiasts to fully understand each step. Mathematicians surely solve in much less lines but they get this experience in numerous years.

[@jozz2]

Yep. Sqrt(a) + i * Sqrt(a) = 32

[@Farus-dm4tl]

No way. If a is >0 -a is

[@ganeshdas3174]

A short route can be √a(1 + i) = 32 √a = 32/(1 + i) a =[32/(1+i) ]^2 = 1024/2i = 512/i = - 512 i

[@MR..mohamedelsayed]

Trap equation. Notes...Wrong solution

[@levskomorovsky1762]

There is no need to move the root of the minus to the right! If you square both parts of the original expression, the solution is much simpler

[@МалиновыйЗакат-г6к]

You will not record a 6 minutes video doing math in effective way.😂

[@mauriziocernigliaro3177]

per me' e' errato perché se a>0 allora -a

[@ricardomejias3629]

No existe en reales, pero sí en complejos​@@mauriziocernigliaro3177

[@Dreamteeam57]

​@@МалиновыйЗакат-г6к Ooh I get it now 😅😅

[@dougrutledge532]

If you take 10 minutes to do each math olympiad problém, you're not going to pkace very high

[@johnbofarullguix1499]

1. Let x,y be 2 numbers, real or complex, that in turn build the complex a=x+1j*y that satisfies: (x+1j*y)^.5+(-x-1j*y)^.5=-32 2. Expanding and simplyfing : y^2-x^2-2*1j*x*y=512 the only way to carry on is zeroing either x or y 3. when zeroing x one ends up with a contradiction in the start expression : (1j*512^.5)+(-1j*512^.5)=32 So the only way forward is to zero y : -x^2=512 thus x=1j*512^.5

[@rainerzufall42]

Why do you mix integers with square roots? Even for the "long solution", it's more intuitive to combine the square roots: √a + √-a = 32 => √a² + 2 √a √-a + (√-a)² = a + 2 |a i| - a = 2 |a i| = 32² = 1024 => a = ± 1024 / 2 i = ± 512 i.

[@bernysaudino668]

No z² is not equal |z|² for z is complex in effect |z|²=z·conj(z)

[@rainerzufall42]

@@bernysaudino668 I didn't say that, and yes, |z|² = z * conj(z). But here √-a = conj(√a). Thus √a √-a = 512 = 32² / 2. Example: a = 512 i. √a = 16 + 16 i √-a = 16 - 16 i = conj(√a) √a √-a = √a conj(√a) = |√a|² = |a| = 512 = 32² / 2. For a = - 512 i, you have the roles switched.

[@rainerzufall42]

It definitely easier with polar coordinates, and less error prone...

[@rainerzufall42]

The tricky thing here is, that multiplication by i rotates anti-clockwise, while √ goes both ways...

[@sergiofernandez116]

Using Euler's equation (e^ix = cos x + i sin x) yields an infinite number of solutions 512e^(-pi/4*i+(2pi*N)), Where N is an integer from 0 to infinity.

[@nikolaspp5148]

The solution is not full, as (1+i) = sgrt 2× exp(i×(pi/4 + 2N×pi)....-(1+i)= sqrt 2 × exp (i×(5pi/4 +2 N×pi)

[@AndreTewen]

En adoptant la notation polaire des nombres complexes on a x = (|x|,arg(x)+2k.pi), k € Z. Les racines carrées de x sont donc (sqrt(|x|), arg(x)/2 + k.pi), avec k=0 ou k=1. D'où sqrt(a) correspond à 2 valeurs : pour k=0 nommée 0.sqrt(a) et pour k=1 nommée 1.sqrt(a). Ainsi l'equation posée est en fait une quadruple equation : 0.sqrt(a) + 0.sqrt(-a), 1.sqrt(a) + 0.sqrt(-a), 0.sqrt(a) + 1.sqrt(-a), 1.sqrt(a) + 1.sqrt(-a). 2 equations donnent 512i en solution et les 2 autres -512i.

[@Jean-BaptisteOuflè]

Thanks you very much

[@zougaghabdelmajid5449]

(rac(a)+rac(-a))²=2rac(a)rac(-a)=32²=1024. I²rac(a)²=512. Soit rac(a)²=i²522. Ou encore a=+-i512.

[@MrGeralt67]

Just try to use these answers in equation. It doesn't work. sqrt(512i)+sqrt(-512i) = 16*sqrt(2i)+16*sqrt(-2i) = 16*(sqrt(2i)+sqrt(-2i)) - and it doesn't equal to 32

[@koenraad4618]

√i = (1+i)/√2, and √(-i) = (1-i)/√2. Use this, and you will see that it does work. Keep in mind that √(512) = 32/√2. You will see that you get 32.

[@MrMLehman]

@@koenraad4618 Thanks. I was scratching my head until you clarified this.

[@carlosjimenez2848]

El único número real para el que está definida la expresión √a+√-a es a=0 que, obviamente no es solución de la ecuación propuesta. Buscamos entonces una solución en los números complejos. √a+√-a=32 --> √a(1+i)=32 √a=16(1-i) --> a=-512i o también, √a+√-a=32 --> √a(1-i)=32 √a=16(1+i) --> a=512i Al comprobar las soluciones, hay que tener en cuenta que 512i y -512i tienen dos raíces cuadradas (opuesta una de la otra), por lo que sus sumas (miembro izquierdo de la ecuación) ofrecen cuatro posibilidades: 32, 32i, -32 y -32i. Lógicamente, la primera es la que corresponde a la ecuación propuesta: para √512i tomamos 16(1+i), descartando 16(-1-i) y para √-512i tomamos 16(1-i), descartando 16(-1+i)

[@mcwulf25]

Square both sides. The a and (-a) cancel and after factoring the remaining term you get +/- 2ai = 32^2. a = +/- 32*16i = +/- 512i

[@TheWuschelMUC]

No. Let a be > 0. Then -a is

[@tinatin7949]

Да, на западе другой подход к решению: схож с тем что в России, но другой. Меня учили в школе: из отрицательных чисел чётные корни не извлекаются. Да и никаких дополнительных буквы как i и сокращения DBS я не припоминаю. Yes, in the West there is a different approach to the solution: similar to what is in Russia, but different. I was taught at school: even roots are not extracted from negative numbers. And no additional letters like i and abbreviations I don't remember DBS.

[@romanitartsssati4670]

Congs 👏👏👏The secret is proceeding step by step, needing no hurry👍

[@abhinitkumar6641]

√ab = √a√b, for a≥0 and b≥0 √a+√-a=32 (square both sides) a -a + 2√a√-a = 1024 √a√-a = 512 (square both sides again, do not write √(a)(-a) = 512) -a² = 512*512 a = ± 512 i

[@Katarzyna240]

Pięknie 😊z zastosowaniem wzorów skróconego mnożenia i przekształcaniem,skracaniem i redukcją 🧐podoba mi się 😊

[@Yogut3k]

pewnie taki jest klucz rozwiązania, ale wynik można dostać w bardziej uporządkowany sposób, jak pokazano wyżej.

[@dumatel]

(1+і)vā=32 Vā=16(1-i)=16 sqrt(2) exp(-pi/4)

[@grandcrappy]

High-level math peole are lucky af, can earn what they wish.

[@bandarusatyanandachary1181]

Explanation is good. But competitive examination steps canbe reduced in view the good standard of olympiad students there by reduce the length of the video.

[@robfrohwein2986]

Nice and more basic solution , very well explained with all steps!! I did it with complex numbers which has more pitfalls... fortunately i arrived at the correct result.

[@akhtarabbasabbas4641]

It is simple.taking both square result will be +- 512 taking both side square equation will be a+a=(32)² 2a=+-1054 a=+-512

[@kimutaiboit8516]

I prefer substitution. Let a=b² b+ib=32 b²+2ib-b²=1024 2ib²=1024 ib²=512 b²=512/i b²=-512i Since a=b² then b=±√a Therefore a=±512i

[@kateknowles8055]

(a^½)(1-i) =32 (a^½)(1+i)=32 a^½= 32/(1-i) = 32(1+i) / (1-i)(1+i) = 32 (1+i)/2 = 16(1+i) a^½= 32 / (1+i) =32 (1-i) / (1+i)(1-i)= 32(1-i) /2 = 16((1-i) Two complex solutions with magnitude 2^9 and arguments pi/2, 3pi/2 So these are a = 512i and a= - 512i ( I found the earlier comments helped me to notice my first error which was halving arguments instead of doubling them)

[@kateknowles8055]

A correction : two imaginary solutions (not complex)

[@zakiater]

la racine de a+la racine de-a=32 la racine dea+t laracine dea=32 a=32 au carre/(1+t )au carre

[@davidbrown8763]

I did it in less lines. However this is the correct solution - tested by substitution.

[@alvinrajs4668]

On transferring --1024a to the other side the sign changes to +. Hence the answer to be checked.

[@joaowilsonvieira6196]

Socorro curió!😂

[@georgech7914]

Почему бы сразу не предупредить, что a - это мнимое число? Тогда многие не стали бы возиться, пытаясь решить эту задачу. Ведь вся "красота" любой задачи заключается в том, чтобы результат оказался вещественным(действительным) числом, которые можно представить на числовой прямой.😀 Что является аналогом "мнимый"? Первый: "воображаемый", второй: "фальшивый", "ложный".

[@ΒασιληςΑρετακης-μ2ψ]

Условии надо поставить:решить во множестве комплексных чисел.,т.к.во множестве действительных чисел решением является пустое множество.

[@rcnayak_58]

√a + √ (-a) = 32, => √a + √ ((-1)(a)) = 32, => √a (1 + i) = 32, => [√ a (1 + i)]² = 32², => a (1 + 2i + i²) =32², => a (1 + 2i -1) = 32², => a (2i) = 32² = 1024. Therefore a = 1024/2i = 512/i = 512i/i² = - 512i

[@omar.ma7]

Too long sqrt (a)+sqrt(-a)=sqrt a +i sqrt a sqrt a (1+i)=32 sqrt(a)=32/(1+i) a=(32/(1+i))^2 a=1024/2i= -512 i

[@birgerschnoor4419]

that was my approach as well. much more intuitive and faster.

[@dariuszb.9778]

Or geometrically on imaginary coordinate system knowing that sqrt(a) norm is equal to sqrt(-a) norm = 16 and phase angle sqrt(a) is Pi- phase angle of sqrt(-a) and these angles are halves of phase angles of a and -a where a and -a have phase angles differing by Pi.

[@cetamon4115]

A²+2AB+B²=1024 -> a+-a+2√a√-a=1024 -> 2√-a²=1024 -> √-a²=512 -> -a²=262144 -> a²=-262144 -> a=√-262144 -> a=±512i.

[@AntonioVazquez-wf5qi]

By squring system 4=5

[@AntonioVazquez-wf5qi]

4-9/2=5-9/2 (4-9/2)^2=(5-9/2)^2 16+81/4-72/2=25+81/4-90/2 16+81/4-36=25+81/4-45 -20=-20....so (4-9/2)^2=(5-9/2)^2 Squring √(4-9/2)^2=√(5-9/2)^2 4-9/2=5-9/2 4=5 So squring is false

[@VVG-tq2nj]

С точки зрения советской математической школы это бред!

[@David_Lloyd-Jones]

The Soviet Union has gone where the bindweed twines, so maybe you'd better come up with a piece of actual mathematical reasoning...

[@VVG-tq2nj]

​​@@David_Lloyd-Jones, прежде чем мастурбировать на Советский союз, раскажите откуда взялся минус под √ ? И прежде чем глумиться на советской школой, сможете рассказать о чем теорема Пуанкаре, которую доказал советский математик Григорий Перельман?

[@VVG-tq2nj]

​@@David_Lloyd-Jones, квадрат числа не может быть отрицательным. И математика не может быть политизированной. Нравится вам или нет, советская наука опередила время. То что изучают в западных унивеоситетах в России изучают в школах.

[@MrGeralt67]

@@David_Lloyd-Jones Just try to use these answers in equation. It doesn't work. sqrt(512i)+sqrt(-512i) doesn't equal 32

[@tarcisiobomfim91]

Squaring the lefter member we can obtain de answer faster.

[@solneeman518]

Definitely

[@maxime_0912]

No. I stopped at 1'10" because you made a mistake. (√a) ² not equal a, but | a |.. Also, you should know that to solve an equation, one must know the definition set.. 🙄

[@eli37co]

Correct result, long way. It is simple with complex numbers from the beginning.

[@JohnPodesta-yu4tf]

Factor i out of 2nd term and solve for sqrt a in one line. I solved it in 30 sec!

[@SylviaVanderlay]

Thanks for this video

[@SPetrosyan1960]

Test the answers, prove it

[@fsantosneto]

Well. Do now replacing 32 by i.

[@Serghey_83]

Вот что я сразу увидел, так это каллиграфический почерк!!! Браво!

[@ВладимирГречишкин-в7х]

А где проверка этого решения, подстановкой ответов в уравнение?

[@RobertGabor]

I think it is everything starts wrong. We should present square of a and -a as R+iJ and -R-iJ because roots of C numbers could have multiple answers.

[@erichendriks2807]

√a(1+i)=32 => √a=32/(1+i) With this as the first step it takes very little to finish the calculation.

[@oo7521]

(a)^(1/2)+(-a)^(1/2)=32 ((a)^(1/2)+(-a)^(1/2))^2=a+2*(a)^(1/2)*(-a)^(1/2)-a =2ai=1024 ai=512 (a)^2=-(512)^2 a=512i or a=-512i

[@荻野憲一-p7o]

More easily, just squaring both sides, 32^2 = (√a + √-a)^2 = a +- 2√(-a^2) + (-a) = +- 2√(-a^2). Squaring again, 32^4 = 4(-a^2). So, a = +- √( 32^4 / (-4) ) = +- 512i. Be aware. √a √-a not= √(-a^2), √a √-a = +- √(-a^2). √(-a) not= (√a)i, √(-a) = +- (√a)i.

[@fisti1208]

√a+√-a=2^5 √a+i√a=2^5 (√a+i√a)^2 = 2^10 a + 2 * √a * i√a - a = 2^10 2 * i * a = 2^10 i * a = 2^9 a = 2^9/i a^2 = 2^18/-1 = -(2^18) a = 512i and because i = -i: a = +- 512i

[@lips7111]

Nice. But the music is too rich.

[@ПавелГерманов-ь4я]

Насколько нас хорошо учили в СССР! Хватило половины действий для ответа... Зачем так всё разжёвывать?

[@paulvangastel8665]

Square both sides a + 2✓(-a^2) - a = 2^10 2ia = +/- 2^10 a = /+ i2^9 = +/ 512i

[@mrdad1858]

How did "b" come into the equation?

[@antgonsu]

Por favor, para obtener ese resultado se pueden dar bastantes menos vueltas.😮

[@RobertRoth-oj6zz]

How did ever make it through algebra Seeing problems and solutions like this.?

[@akoeng1979]

Solve it easier: root(a)+root(a)*i=32 Root(a)(1+i)=32 a=32^2/(1+i)^2= 32^2/2i

[@dac374]

Always check your answer because this one is incorrect,

[@rainerzufall42]

Two obvious solutions: a = ± 512 i. Are there more? Probably not.

[@rainerzufall42]

My quick check was to get a 45° complex number z = √a with Re(z) = 32/2 = 16. That is 16 + 16 i (negative 16 - 16 i). But a = z² = (16 + 16 i)² = 256 (1 + i)² = 256 * 2 i = 512 i (negative - 512 i).

[@mmsit99]

✓a+✓ai=32 ✓a(1+i)=32 a(1-1+2i)=32×32 2ai=32×32 ai=16×32 -a^2=512 a=✓-512 a=✓512i a=16✓2i

[@lvr7188]

I don't know why you made it so complicated: (√a+√-a)²=±32² (√a)²+(√-a)²+2√a*√-a=±32² a-a+2√-a²=±32² 2a√-1=±32² a=±512i

[@tonyennis1787]

What did I do wrong? sqrt(a)+sqrt(-a) = 32 sqrt(a)+i*sqrt(a)=32 sqrt(a)(1+i)=32 ! factor the sqrt of a sqrt(a)=32/(1+i) +-a = 32^2/(1+i)^2 ! square both sides +-a=1024/(1+i+i+-1) ! FOIL the denominator +-a = 1024/2i +-a=512/i a=+-512/i womp

[@Ivan__Vladimirovich]

И что в итоге мы получили? Множество неявных корней и 9 минут потраченных в пустую. 😂😂😂 Вот что происходит, когда занимаешься херней😂😂

[@sahasukanta]

Do u get any extra money by using an unnecessary lengthy process?

[@befekadudebebe6833]

“It is easy, when you know it”. How about absolute beginners or who did math 30 years ago?

[@PrajwalM-t1u]

Sir can you answer for this question ie if 1÷a^2+1÷b^2 =1÷13 then a^2+b^2=

[@haiderlughmani]

Explain in next video.

[@PrajwalM-t1u]

@haiderlughmani thank you sir

[@PavloDurov]

Can be simplified if we square both sides at the beginning, no? √a + √(-a) = 32 SBS a - a + 2√a√(-a) = 1024 2√a√(-a) = 1024 √a√(-a) = 512 SBS a(-a) = 512² -a=512² a=-(512²) a=√(-512²) a=±512i

[@p.narayananmulleria5416]

4x4=16,4x - 4 = 16 16 + 16 = 32

[@bigneiltoo]

Why is he writing the same opening line again?

[@prof.samirghosh3811]

3 line ka math 1 page bana diya?

[@carfer]

There is only one solution, not two, because the given solution is squaring the expression, and this method adds an additional solution that must be discarded. I think it's simplier to resolve doing the next: Sqr(a)+Sqr(-a) = 32 --> Sqr(a)+Sqr(a)*i = 32 --> Sqr(a)(1+i) = 32 --> Sqr(a)(1+i)*(1-i) = 32*(1-i) --> Sqr(a)*2 = 32*(1-i) --> Sqr(a)= 16*(1-i) --> Now squaring, for get "a": a = 16*(1-i)*16*(1-i) = 16^2*(1-i)^2 --> a = 2^8*(1-2*i-1) = -2^9*i = -512i (unique solution)

[@dimkad744]

Range of acceptable values ​​of the equation a>0 and -a>0, a>0 and a

[@IvoNovacek-jq2lz]

What the SQURING means?

[@anis786]

Let sqrt(a)=x. Then sqrt(-a)=xi. x+xi=32 x(1+i)=32 x=32/(1+i) ×(1-i)/(1-i) x=16-16i a=+/-512i

[@tvrashid]

√a+√-a=32 (√a+√-a)²=32² a+2ai-a=1024 2ai=1024 ai=512 a=512i

[@chandrababu3070]

Super

[@munirdisi9206]

Can be solved in a much shorter way

[@gerardgalissie2546]

La otation "racine carrée" est inappropriée pour les nombres complees

[@WernerDurandi]

Genau!

[@MarcosRegalado-h7m]

a½+(-a)½=32 a½+(-1)½a½=32 a½(1+(-1)½)=32 a½=32/(1+(-1)½) (a½)²=32²/(1+(-1)½)² a=1024/1+2(-1)½-1 a=1024/2(i²)½ a=1024/2i a=±512i

[@stpat7614]

I got the same results (a different way), but I don't know how to verify that sqrt(2^9*i) + sqrt(-2^9*i) = 32, or that sqrt(-2^9*i) + sqrt(2^9*i) = 32.

[@maison5.0]

The equation has no solution If a > 0 them -a

[@michael.a.covington]

Your background music is a well-known American Christian hymn, "Leaning on the Everlasting Arms." No offense taken, but I had never heard it in quite this context before :)

[@AP-qt9cp]

i did it in 3 lines by squaring the original equation.🙃

[@emha7100]

Das Quadrat jeder reellen Zahl kann nur positiv sein. Es gilt also nur der Hauptwert einer Quadratwurzel. Wenn diese Bedingungen beachtet werden, erhalte ich lediglich die Lösung 2 * ia = 1024 und somit a = -512i

[@rainerzufall42]

Das ist aber falsch! Wir ziehen eine Wurzel in IC, da muß man höllisch aufpassen! Die wesentliche Gleichung ist nicht 2 a i = 1024, sie ist ± 2 a i = 1024. Daraus ergibt sich a = ± 512 i. Überprüf's! Der Knackpunkt ist mMn 2 √a √-a = 2 |a i|, da die Wurzeln konjugiert sind. Wir suchen den positiven Realteil! Dieser wird aus Symmetriegründen sowohl mit a als auch mit -a erreicht! Der Imaginärteil der Summe ist dagegen 0.

[@rainerzufall42]

Das ist übrigens der Grund, warum fast jeder lieber mit Polarkoordinaten rechnet! Beim ersten Überfliegen hatte ich unbesehen mit 2 √a √-a = 2 a i weitergerechnet und den Fehler erst bemerkt, als dies eine positive Zahl sein sollte. Obacht!

[@emha7100]

@rainerzufall42 Danke, dass du mich auf meine Fehler aufmerksam gemacht und meine Antwort korrigiert hast. Leider lernt man nur durch Fehler. Darum möchte ich mich für meine Narrheit entschuldigen!

[@rainerzufall42]

@@emha7100 Es gibt keine Narrheit, wenn man weiter nach der Wahrheit sucht. Freut mich, daß meine Ausführungen weitergeholfen haben. Es ist ein schmaler Grat zwischen hilfreich sein oder besserwisserisch. Hier gab es eine Stolperfalle, die wirklich böse ist! Wie Du bereits vorab korrekt erwähnt hattest, geht es darum, den Hauptwert der Wurzeln(n) zu beachten. Kurz gesagt: Die komplexe Wurzel erhält das Vorzeichen des Imaginärteils, egal was der Realteil sagt (außer für Re z = 0). D.h. in der komplexen Ebene bleiben Werte oberhalb der x(Re)-Achse dort und unterhalb bleibt dort. Werte im 1. und 2. Quadranten landen im 1. Quadranten und Werte im 3. und 4. Quadranten landen im 4. Quadranten. (rein negative realle Zahlen haben ihre Wurzel auf der positiven y(Im)-Achse) Was schon bereits eine Zahl kompliziert ist, wird zum Höllentrip bei Produkten. Bereits √-4 √-9 = ? ist tricky! Nix mit √a √b = √ab (die Antwort ist -6, also -√ab).

[@Superbettor]

Abs (a) = root sqr (a)

[@krishg6047]

Does olympiad want to waste time of students on one problem instead of using shortcut methods?

[@Profe_Lauro]

Está bien papa. Parece un examen de secundaria.

[@БелАлекс]

Ну вроде как обычная задача. На олимпиадную точно не тянет.

[@andrekapamba3081]

Résoudre dans C!!!!

[@rachid8475]

Très très long : dès le début l'équation peut s'écrire : racine carré de(a)+ i (racine carré de a) = 32 Ça sera plus simple

[@ashar139]

Thanks

[@MsIryLondon]

Made a simple equation a super complicated one!🤦🏻‍♀️

[@ЮрийСеменов-м8и]

a>=0, a