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
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
@@TheDavidlloydjones, 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?
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)
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.
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)
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.
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
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.
(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)
√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
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.
@@ГаннаМаслакEl problema propuesto, como puedes observar, solo tiene sentido en C. De todas formas, volviendo a tu pregunta, en R, al elevar ambos miembros de una ecuación al cuadrado, independientemente de que uno de ellos sea negativo, en general, se obtiene una ecuación que NO es equivalente (esto se debe a que la función f(x)=x² no es inyectiva). Esta última contendrá las soluciones de la primera, pero puede tener más. Por eso, luego, hay que comprobarlas.
√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
Умножим обе части равенства на √а, получим а-а=32√а, или 32√а=0, а=0
Парадокс.
😂 Ну да. Надо было сразу сказать, что а - мнимое число вида xi
Если ты √a+√-a домножишь на √a у тебя получится а - a * i. А не а - а.
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
Конечно, именно такой вариант решения напрашивается в первую очередь.
Better if we square as it is, both sides at the first step:
a - a + 2ai = 1024
a = - 512i
√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
Squaring the equation is not an equivalent conversion. Thus the answer must be tested.
Yes, the function ! f(x) = x2 is not a injective function !
√-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
Great!
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
This result is correct but can be obtained in much fewer lines!🙂
Sometimes the steps are needed for better understanding.
Pięknie 😊z zastosowaniem wzorów skróconego mnożenia i przekształcaniem,skracaniem i redukcją 🧐podoba mi się 😊
Thanks you very much
A short route can be
√a(1 + i) = 32
√a = 32/(1 + i)
a =[32/(1+i) ]^2
= 1024/2i
= 512/i
= - 512 i
Trap equation. Notes...Wrong solution
Congs 👏👏👏The secret is proceeding step by step, needing no hurry👍
Square both sides. The a and (-a) cancel and after factoring the remaining term you get +/- 2ai = 32^2.
a = +/- 32*16i = +/- 512i
No. Let a be > 0. Then -a is
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
You will not record a 6 minutes video doing math in effective way.😂
per me' e' errato perché se a>0 allora -a
No existe en reales, pero sí en complejos@@mauriziocernigliaro3177
С точки зрения советской математической школы это бред!
The Soviet Union has gone where the bindweed twines, so maybe you'd better come up with a piece of actual mathematical reasoning...
@@TheDavidlloydjones, 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?
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)
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.
I did it in less lines. However this is the correct solution - tested by substitution.
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)
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)
(1+і)vā=32
Vā=16(1-i)=16 sqrt(2) exp(-pi/4)
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.
Super
Résoudre dans C!!!!
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
that was my approach as well. much more intuitive and faster.
Вот что я сразу увидел, так это каллиграфический почерк!!! Браво!
Thanks
High-level math peole are lucky af, can earn what they wish.
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
√a(1-1-1) =32 √a=32/-1=-32
a=(-32) ²=1024
How did ever make it through algebra Seeing problems and solutions like this.?
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.
√a(1+i)=32 => √a=32/(1+i) With this as the first step it takes very little to finish the calculation.
Squaring the lefter member we can obtain de answer faster.
Definitely
√a+√-a=32 √a(1+i) =32
a(1+-1+-1) =32 a(-1) =32
a=(-32i)²= 1024
2 pangkat 4
(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)
A correction : two imaginary solutions (not complex)
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.
Насколько нас хорошо учили в СССР! Хватило половины действий для ответа... Зачем так всё разжёвывать?
a-a+2ai=1024; i,e 2ai=1024 ; i,e ai=512; ; a=512/ii ; i,e a=-512i
The equation has no solution
If a > 0 them -a
√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
И что в итоге мы получили? Множество неявных корней и 9 минут потраченных в пустую. 😂😂😂 Вот что происходит, когда занимаешься херней😂😂
Correct result, long way. It is simple with complex numbers from the beginning.
Range of acceptable values of the equation
a>0 and -a>0,
a>0 and a
Made a simple equation a super complicated one!🤦🏻♀️
A²+2AB+B²=1024 -> a+-a+2√a√-a=1024 -> 2√-a²=1024 -> √-a²=512 -> -a²=262144 -> a²=-262144 -> a=√-262144 -> a=±512i.
Too complicated you must use isqare = - 1
So 32 square / (1-i) square= a
Then develop ( 1-i )square = -2i
So à = 32x32×i/-2
Thats all
Abs (a) = root sqr (a)
With luck.
Do u get any extra money by using an unnecessary lengthy process?
Решение надо дать во множестве комплексных чисел,с введением числа i.
too much fuss to arrive at a simple solution.
(Va + V-a)^2 = 32^2
(Va +-JVa)^2 = 1024
a^2 - a^2 +-2Ja = 1024
a=1024/+-2J = -+J512
Il existe une démonstration bien plus simple, pour arriver au résultat.😮
Correct but too long a solution. Just divide the origibal equation by squareroot of a and bang, you have the solution in just two lines after that.
(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
(32/(1+i))^2 =>1024/(1+2i-1) => 1024/2i = -512i?
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.
What a tediously tasking work at problem one
Условии надо поставить:решить во множестве комплексных чисел.,т.к.во множестве действительных чисел решением является пустое множество.
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
Procure o professor Denis Rodrigues da helpeengenharia que vc. Vai aprender a ensinar.
La otation "racine carrée" est inappropriée pour les nombres complees
a=2^6
Easy.
Проверим? √512i +√-512i=32 мне одному кажется что они не равны? Слева комплексное число справа натуральное да ещё и целое и тоже будет с -512i
а сразу возвести в квадрат правую и левую часть? Бастрее получится.
What the SQURING means?
А где проверка этого решения, подстановкой ответов в уравнение?
a appartient à R car après calcul a²=-513²
Ну вроде как обычная задача. На олимпиадную точно не тянет.
Complex number
If you take 10 minutes to do each math olympiad problém, you're not going to pkace very high
Imaginary
Muy complicado
Pourquoi ne pas écrire directement
2a/2 = (512-32✓-a)
sans diviser les 2 membres par 2. On gagne une ligne
16+16 = 32. a = 16^2
🙋🇧🇷
-512
and now solve √a+√(-a)+√ia+√(-ia)=b 😮
😂😂😂😂
झीला दिया भाई
यदि किसी ने 11th में maths पढ़ा हो तो बिना म्यूजिक के इसे 30 सेकंड में सॉल्व कर देगा😂
А где область определения,которая равна числу о?
(-512*i)^1/2 = ?
Can you solve this?
Нет решений в действительных числах, листайте дальше
Goof
2/4
Why is he writing the same opening line again?
Se a>0 -a
Comon!!! +/- 512i ??? Are you serious??? There's no right answer. Oh comon.
Полный идиотизм,особанно здорово глядеть как он в столбик умножает32 на 32
Никого не смущает, что корень из 512 не равен 16 ?
1/2
(-a). not € lR
-512^2 =-512 × -512 = 1024 = (512)^2 = a^2 , Hence = a = 512
Vetem me te pare: a= + - 4
Хіба можна підносити до квадрату ліву і праву частину рівняння, якщо одна з них від'ємна?
En el cuerpo de los números complejos no tiene sentido decir que un número sea "positivo" o "negativo".
@@ГаннаМаслакEl problema propuesto, como puedes observar, solo tiene sentido en C. De todas formas, volviendo a tu pregunta, en R, al elevar ambos miembros de una ecuación al cuadrado, independientemente de que uno de ellos sea negativo, en general, se obtiene una ecuación que NO es equivalente (esto se debe a que la función f(x)=x² no es inyectiva). Esta última contendrá las soluciones de la primera, pero puede tener más. Por eso, luego, hay que comprobarlas.
a=32kwadrat karena akar angka minus=0
For heaven's sake cut out that ridiculous music. It does not behove a maths problem to have hurdy gurdy music playing in the foreground.
√a+√(-a)=32 → a = ??
√a + √(-a) = 32 √a + i √a = 32 √a (1+i) = 32 → √a = 32/(1+i) = z a = z^2
z = ρ(cos θ+i sin θ ) z^n = ρ^n (cos nθ+i sin nθ ) de Moivre
z = 32/(1+i) = 32/(1+i) ∙ (1-i)/(1-i) = 32/2 (1-i) = 16√2∙√2/2 (1-i)
z = 16√2 (√2/2 - i √2/2) = 16√2 [cos(-π/4) + i sin(-π/4) ] ρ =2^4 √2 θ = -π/4
a = z^2 = ρ^2 (cos 2θ + i sin2θ) = 2^9 (cos(-π/2) +i sin(-π/2) )
a = -2^9 i= -512 i
Зачем во такое бессмысленное выставлять в ютубе?
No
Безграмотность
Stupid long method
Time wasting!
a>=0, a