yeah but thats not the question or the answer just the process is kinda same so its more of a different problem than being the same problem@@normalbattlecat8088
@@akarshkumar6300 I know very well but if you solve this question for finding roots so this guy correct but when you find the answer for an other questions you get directly get the solution 2
Positives: -1+i√3 theta = arctan(√3/-1) (In forth quadrant as it's consider to be having negative coefficient for the imaginary part) theta = -π/3 (not in second quadrant) (add π) theta = 2π/3 r = √[(√3)²+(-1)²] r = √4 = 2 => r[cos(theta)+isin(theta)] = 2[cos(2π/3)+isin(2π/3)] => [2[cos(2π/3)+isin(2π/3)]^3 = 2^3[cos(3(2π/3))+isin(3(2π)3)] = 8[cos(2π)+isin(2π)] 2π is a full circle, which is also could be 0 in the plane (or write is as sums; cos(2kπ±t)=cos(t) = 8[cos(0)+isin(0)] = 8[1+0] = 8 The negatives answer will play a similar thing out Edit: you can also use expansion rule (x+y)^3 = x^3+3x^2y+3xy^2+y^3 (-1+i√3) = -1+3(1)(i√3)+3(-1)(-1(3))-3i√3 = -1+3i√3+9-3i√3 = 8
he's completing the square, a common practice that works as an alternative to the quadratic formula and is where said formula was derived from he's adding (b/2)^2 which in this case would be (2/2)^2 and then as you can infer, it would become 1^2 which is just 1
@@jamesharmon4994 I’m aware, it’s just that many variables are also letters, so the OP might want to subconsciously (or something similar, I’m not a psychologist) put the i behind out of habit
limitless numbers, limitless solutions! limited intelligence, limited discoveries. 1000 years ago we discovered how to solve second degree equations, we are now defining sq root of negative numbers. if humanity could survive, who knows what we would discover in the next few decades.
I know for the second answer you can solve that by plugging it in the quadratic formula. I don't understand the method used here, like where that +1 come from.
The polar form would be easier: x = r * e^jt x^3 = r^3 * e^3jt = 2^3 = 2^3 * e^0 => r = 2 and 3t = 0 + 2pi*k => t = 2pi/3 * k => t = 0, 2pi/3, -2pi/3 => x = 2, 2*e^j2pi/3, 2*e^-j2pi
Hola, el símbolo "i" significa la unidad imaginaria, que es un número que al elevarlo al cuadrado da como resultado -1. Los números imaginarios se combinan con los números reales para formar los números complejos, que tienen muchas aplicaciones en la ciencia y la tecnología. Saludos desde España.
It depends to the Set which X belongs. If X belongs to R, the only solution is 2. And if X belongs to C as it's implicitly assumed here then their are 3 solutions as shown in this video
its the imaginary unit you can search about imaginary and complex numbers basically i is sqrt(-1) and square root of negative numbers i.e. multiples of i are called imaginary numbers ex - 3i, 4i, pi*i, sqrt(2)i etc and real numbers plus imaginary numbers r called complex numbers like 2+4i
@@Rando2101, that's how everyone does it. I don't know what you were taught, but I know that's very understandable and detailed. How is that not enough? With x²=4, you put a squareroot sign on x² and 4 to get the answer. Like, what process do you use?
Pov: you have 100k+ hours in a game and you decide to play it again
Cube roots were invented in 499CE
People before 499CE :
other 2 sol wouldve been missed
Damn. I didn't think of imaginary solutions...
same my mind was screaming 2 2 2 the whole time
this is normal or bare minimum on differential equation
Always expect the possibility of having n solutions on an x^n. It's not always the case tho
It's impossible to solve in the real numbers section you can check it on delta
@@KermitSF yo, its actually exactly the case. Fundamental theorem of algebra
He did quadratic formula without actually doing it 😯
Completing the square
Actually, completing the square is where the quadratic formula comes from
simpler version
x³ = 8
2³ = 8
x = 2
It is better to write it this way:
x³ = 8
x³ = 2³
x = 2
bruh dont just leave the other 2 solutions
More like incomplete version
Stupid version you mean
Even simpler version
x^3 = 8
Cube root both sides
x = 2
Wow.. it is so complex 😁
Oh yes, the three roots of unity
Bruh its the 3 roots of 8 not 1
@@dazai826same idea, just different radius of circle
yeah but thats not the question or the answer just the process is kinda same so its more of a different problem than being the same problem@@normalbattlecat8088
Directly x=2
Then you will not get other 2 solutions
@@akarshkumar6300 I know very well but if you solve this question for finding roots so this guy correct but when you find the answer for an other questions you get directly get the solution 2
It's actually + or -2
@@3dsaulgoodmanactually it's just +2 a.nd not-2
@@3dsaulgoodman only +2 because when you take cube of +2 you get 8 and when you take cube of -2 you get -8 which not statisfy
Do the hardest one you can possibly find. Make a whole video of you tryina solve it
Wouldn't the second qnser not give you 8 if plugged in?
No real number would be an answer to the second equation,but there exists complex solutions
Positives:
-1+i√3
theta = arctan(√3/-1) (In forth quadrant as it's consider to be having negative coefficient for the imaginary part)
theta = -π/3 (not in second quadrant) (add π)
theta = 2π/3
r = √[(√3)²+(-1)²]
r = √4 = 2
=> r[cos(theta)+isin(theta)]
= 2[cos(2π/3)+isin(2π/3)]
=> [2[cos(2π/3)+isin(2π/3)]^3
= 2^3[cos(3(2π/3))+isin(3(2π)3)]
= 8[cos(2π)+isin(2π)]
2π is a full circle, which is also could be 0 in the plane (or write is as sums; cos(2kπ±t)=cos(t)
= 8[cos(0)+isin(0)]
= 8[1+0]
= 8
The negatives answer will play a similar thing out
Edit: you can also use expansion rule
(x+y)^3 = x^3+3x^2y+3xy^2+y^3
(-1+i√3) = -1+3(1)(i√3)+3(-1)(-1(3))-3i√3
= -1+3i√3+9-3i√3
= 8
De Moivre's theorem💯
Where are you getting the +1 from when you add it to -4
he's completing the square, a common practice that works as an alternative to the quadratic formula and is where said formula was derived from
he's adding (b/2)^2
which in this case would be (2/2)^2
and then as you can infer, it would become 1^2 which is just 1
@@lemon.linguist I see ty
I ALWAYS want to put the I *behind* (for some reason?!!)
Likely bc it’s where most variables end up going
It' probably because the "standard form" is "a+bi".
@@Whatheman28Except I isn't a variable, it's a constant.
@@jamesharmon4994 I’m aware, it’s just that many variables are also letters, so the OP might want to subconsciously (or something similar, I’m not a psychologist) put the i behind out of habit
@Whatheman28 I want to put the I behind because the standard form is a + bi, where in this case a = -1 and b= sqrt(3).
Perfectly you.
难点ai处处永机出 筒口自己都纳闷了（因不等价啊）❤❤❤❤❤
limitless numbers, limitless solutions! limited intelligence, limited discoveries. 1000 years ago we discovered how to solve second degree equations, we are now defining sq root of negative numbers. if humanity could survive, who knows what we would discover in the next few decades.
X^3 = 8
X^3 = 2^3
X = 2
Predict the unpredictable moment😅
I know for the second answer you can solve that by plugging it in the quadratic formula. I don't understand the method used here, like where that +1 come from.
It called completing the square. He added 1 to the both sides, so he could make (x+1)^2 from x^2+2x+1.
Thanks. It's been so long, when you mentioned "completing the square" it finally clicked. @user-rn2cp8qh6b
You should smile more often
The polar form would be easier:
x = r * e^jt
x^3 = r^3 * e^3jt = 2^3 = 2^3 * e^0
=> r = 2 and 3t = 0 + 2pi*k
=> t = 2pi/3 * k
=> t = 0, 2pi/3, -2pi/3
=> x = 2, 2*e^j2pi/3, 2*e^-j2pi
X³ = 8
X = 3√8
X = 2
Roots of unity are faster
I got purple
Where does the i come from?
Smart
x^2n = m , m < 0
m = - k = k i^2 , k > 0
i^2 = - 1 .
👍👍👍
2, 2w, 2w²
Привет из России, мне всегда было интересно что означает символ "i" в математике поскольку мы такое не изучали
Hola, el símbolo "i" significa la unidad imaginaria, que es un número que al elevarlo al cuadrado da como resultado -1. Los números imaginarios se combinan con los números reales para formar los números complejos, que tienen muchas aplicaciones en la ciencia y la tecnología. Saludos desde España.
'i' is the imaginary unit which is defined solely by the property that its square is -1.
Ну комплексные числа это довольно интересная область)))
Эта i называется "мнимая единица", можешь загуглить
Pas de solutions dans R
Ey Ey Ey, solo pregunte la hora
He could have just done like cube root of 8 is 2
Sorry if anyone else had commented on this before me but like I felt he is overcomplicating it
then just leave the other 2 solutions???
there are 3 solutions to this because the x has a power of 3
It depends to the Set which X belongs. If X belongs to R, the only solution is 2. And if X belongs to C as it's implicitly assumed here then their are 3 solutions as shown in this video
Well on yt i think we all assume the situation with more/harder solutions is the case@@user-jl2kq7np9c
chill
x = 2 , i√3 - 1, -(i√3 + 1)
Thank you,sir
? strange i would say you had it up until 4th line..... i would say just div squ 8/8/8 soooo like 8/8 = 1 right then 1/8 = ????? 0.125... i dfk
8/8/8 is not the same thing as cubed root of 8
then you could have simply written (blablabla) instead of (x²+2x+4).😂
Why are you overcomplicating it? Take the cube root of both sides...x=2!!!
Okay, now where are the other two solutions?
Because, the number of solutions depends upon the exponent
@@isavenewspapers8890только 2 дает 8 при возведение в третью степень
@@Red_Fox_MK not really since there are imaginary solutions that give 8 as well, but it depends if they are saying that x is a real number Or not
@@RGGBMGO теперь понял. Сам не заметил как попал в эту логическую ловушку, что ответ не может быть комплексным. Признаю ошибку
Not me, saying “bro please just take the cube root”
用背的吗？不要误导，老师，你只是做题多了，
Bro x=2 …. 🤷🏽
and what abt other 2 solutions??
il risultato è x = 2
It’s just 2 dude
its 2 and 1+3i and 1-3i tho
Hlo mister..
Whats the i?
Square root of minus one
Tengkiu sir👍
its the imaginary unit you can search about imaginary and complex numbers basically i is sqrt(-1) and square root of negative numbers i.e. multiples of i are called imaginary numbers ex - 3i, 4i, pi*i, sqrt(2)i etc and real numbers plus imaginary numbers r called complex numbers like 2+4i
@@dazai826 tengkiu sir🫡
Why?
Bcz he can do it
To find all 3 answers
X^3=8
8=2^3=(-2)^3
x=2,-2
(-2)^3 does not equal 8
The left side is correct answer X = 2, the right side is formula again so no answer (my teacher said if no answer equal wrong answer)😂😂
X³=8
X³=2³
X=2
and what abt the other 2 solutions?? if you did this irl or in an exam you'd be wrong n get a 0(maybe 0.5 or 1 if ur teach is genorous)
N'importe quoi
ทำไมต้องทำให้ยุ่งยาก
Can't I just do:
x³= 8
∛(x³) = ∛8
Cancel ∛ and ³ on x so you get:
x = ∛8
Since 2•2•2 = 8,
So, x = 2
Not enough solutions
@@Rando2101, no. You don't get to say that. I'm not letting you. Like, how is that even incomplete? That's like so detailed.
@@thehell2173 It's like x²=4. Square root is not enough
@@Rando2101, that's how everyone does it. I don't know what you were taught, but I know that's very understandable and detailed. How is that not enough? With x²=4, you put a squareroot sign on x² and 4 to get the answer. Like, what process do you use?
@@thehell2173 the way you do it gives you a solution that is correct, which is fine.
You just need to do some more steps to get other solutions