Everything is possible in math
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- Опубликовано: 7 фев 2025
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#maths #math
Why do you overcomplicate?
x⁴=-4 ⇒ x²=±2i ⇒ x=±(1+i) for x²=2i and x=±(-1+i) for x²=-2i. And so x=±1±i. Done, good place to stop.
How did u solved x^2=2i
@@khaledf3977 A method for calculating square roots for complex numbers: halve the argument and take the square root of the lenght. for 2i, the arg equals 𝜋/2 and the length is 2, so its square root is a complex number r whose arg is 𝜋/4 and its length is √2. The polar form of r is r = √2(cos(𝜋/4)+isin(𝜋/4)) = √2(√2/2+i√2/2) = 2/2+2i/2 = 1+i. And of course -1-i is also a root of 2i. Checking the validity: (1+i)² = 1+2i-1 = 2i and (-1-i)² = 1+2i-i = 2i.
@@khaledf3977 x^4=-4, so x^2=sqrt(-4)=sqrt(4)*sqrt(-1)=+/-2i.
After seeing this equation, one could easily say : It will surely have complex solutions. Great work! 👍👍😄😄
YES : The equation must be expressed in polar form. So X^4 = 4eiπ(2k+1). 4^(1/4) = 2^1/2; les racines sont alors : x(0,1,2,3) = Sqrt(2).e[iπ(2k+1)[/4...What gives in algebraic form for k=0,1,2,3: x0 = 1+i, x1=-1+i, x2= -1-i, x3=1-i. No needs 11 mn 07, in 2 minutes it's done. By the way this demonstrates the computational power of algebra in C
Since it has complex solution(answer is iota) so it is indeed complex😅
that's great, but he did this question in such a great way that a class 9 student can also understand ( as me ). but it was a easy question ,easily solved as well.
Thank you for watching. Much love and respect. Have a great day! What do you think about this question?❤❤❤
Another good video 👍
How about a Real solution? Is It even possible?
If you don’t know about complex numbers, this equation has no real solutions.
If you know about complex numbers, then you are familiar with Moivre formula, Euler formula (exp iπ = - 1), and the roots of unity…
So, letting Y = X / √2, the equation becomes Y^4 = -1.
The Y solutions are the fourth roots of - 1, i.e. exp (iπ / 4 + ikπ / 2) for k = 0, 1, 2 and 3.
Coming back to X, the solutions are : {1 + i, -1 + i, -1 - i, 1 - i} .
Thanks for your videos !
@@jpl569 Right. So, since you know complex numbers and don't care about real world, go to your employer and ask him to pay you in imaginary units. Dork.
@@pelasgeuspelasgeus4634bro really got salty because someone mentioned complex numbers
@@Pestana02 You can't do the same if you like. Or explain complex numbers by using marvel characters...
That equation has no solution in the real world.
lol, I see what you did there
*That's why the complex world exists*
@@mr.d8747 You mean science fiction world where non countable numbers produce countable results. Think about it a bit...
Just imagine the answer
@@atharvaSafew😂 Precisely. An imaginary solution that has no tangible, countable, actual depiction in real world. Usable only for youtube videos and Academia papers...
x^4 = -4
add 4 to each side
x^4 + 4 = 0
Notice how it’s not a sum of fourths? We’ll use it anyways.
a = x, b = 4^1/4
(a^2 + sqrt2(ab) + b^2)(a^2 - sqrt2(ab) + b^2) = 0
(x^2 + sqrt2((4^1/4)x) + (4^1/4)^2)(x^2 - sqrt2((4^1/4)x) + (4^1/4)^2) = 0
Simplify the 3rd term in each bracket, and 4^1/4 to sqrt2
(x^2 + sqrt2(sqrt2(x)) + 2)(x^2 - sqrt2(sqrt2(x)) + 2) = 0
Multiply the 2nd term
(x^2 + 2x + 2)(x^2 - 2x + 2) = 0
Solve for each case
Case 1:
(x^2 + 2x + 2) = 0
Use the quadratic formula
a = 1, b = 2, c = 2
D = b^2 - 4ac
D = 2^2 - 4(1)(2)
D = 4 - 8
D = - 4
x = (-b +- sqrtD)/2a
Substitute in values
x = (-2 +- sqrt(-4))/2(1)
Expand the denominator and sqrt the -4
x = (-2 +- 2i)/2
Simplify
x = -1 +- i
Case 2:
(x^2 - 2x + 2) = 0
Use the quadratic formula
a = 1, b = -2, c = 2
D = b^2 - 4ac
D = (-2)^2 - 4(1)(2)
D = 4 - 8
D = - 4
x = (-b +- sqrtD)/2a
Substitute in values
x = (-(-2) +- sqrt(-4))/2(1)
Expand the denominator, sqrt the -4 and simplify the -(-2)
x = (2 +- 2i)/2
Simplify
x = 1 +- i
4 complex solutions:
x = - 1 - i, 1 - i, - 1 + i, 1 + i
The equation must be expressed in polar form. So X^4 = 4eiπ(2k+1). 4^(1/4) = 2^1/2; les racines sont alors : x(0,1,2,3) = Sqrt(2).e[iπ(2k+1)[/4...What gives in algebraic form for k=0,1,2,3: x0 = 1+i, x1=-1+i, x2= -1-i, x3=1-i. No needs 11 mn 07, in 2 minutes it's done. By the way this demonstrates the computational power of algebra in C
I would have used Euler exponentiation to get to those results (-4 = 4 * e^i*Pi).
it could be simply
YES : The equation must be expressed in polar form. So X^4 = 4eiπ(2k+1). 4^(1/4) = 2^1/2; les racines sont alors : x(0,1,2,3) = Sqrt(2).e[iπ(2k+1)[/4...What gives in algebraic form for k=0,1,2,3: x0 = 1+i, x1=-1+i, x2= -1-i, x3=1-i. No needs 11 mn 07, in 2 minutes it's done. By the way this demonstrates the computational power of algebra in C
Such simple equations are most easily understood with an Argand Diagram (on the complex plane). But then I always did like algebraic geometry :).
I could only imagine such a solution
It was honestly pretty trivial to just solve in my head.
x^4=-4
so x^2=+/-2i
so x=sqrt(+/-2i)
sqrt(2i)=1+i, -1-i
sqrt(-2i)=1-i,-1+i
You are amazing!
lot of work. easier to use compex arg and complex angle.
x4=-4
(x2)2=-4
(-1).(x2)=√4
-1.x2=2
-1=2/x2=>-1/2=1/x2=>sobs -√1/2=1/x
=>-1/√2=1/x =-√2=x ie
x=√2i
x = √2exp(π(2k+1)i/4) = √2(cos(π(2k+1)/4) + isin(π(2k+1)/4)), where k = 0,1,2,3. i.e. x = ±1 ± i
it could be simply, I've done exactly that, and solution in 2 minutes
x⁴=-4
Let x=a+bi {a,b∈R}
(a+bi)⁴=-4
a⁴+4a³bi-6a²b²-4ab³i+b⁴=-4+0i
Comparing the real and imaginary parts of both sides of the equation
a⁴-6a²b²+b⁴=-4 ①
and
4a³b-4ab³=0
4ab(a²-b²)=0
4ab(a-b)(a+b)=0 => 4 cases
case 1: a=0 (rejected: from ① b⁴=-4, but b∈R)
case 2: b=0 (rejected: from ① a⁴=-4, but a∈R)
case 3: a-b=0 => a=b =>
from ①: -4b⁴=-4 => b⁴=1 => b=±⁴√1=±1 (complex roots rejected, b/c b∈R)
if b=1 ==> a=1 => x₁=1+i
if b=-1 ==> a=-1 => x₂=-1-i
case 4: a+b=0 => a=-b =>
from ①: -4b⁴=-4 => b⁴=1 => b=±⁴√1=±1 (complex roots rejected, b/c b∈R)
if b=1 ==> a=-1 => x₃=-1+i
if b=-1 ==> a=1 => x₄=1-i
pree test , ((-2)^(1/2))^2=? , / 1/2 *2=1 / -> ( -2)^1= - 2 , ((x)^(1/4))^4 = - 4 , -> x= (-4)^(1/4) , partial result , and no complex ...
Assalam o Alaikum,
at this step when we have this expression
(x^2+3)-4x^2=0
we can put square root on B.S
so,
x^2+-2x+2 is our Q.E
and we can directly use this eq
Jazak Allah khair ❤❤
Sigma math 🦅
Cool equation
X belongs to phi and
X is equal to fourth root of -4
i love this
It's easier to apply the De Moivre's Theorem.
This is 4 solutions Complexes
Just do
4rt(-4) or sqrt(2i)
Bam done
How about ±√(2i) ?
This are only 2 solutions where I would expect 4 ...
yeah , think that principally both
x = ±√(2i) and
x = ±√(-2i)
are possible roots .
however in a so called solution we are just supposed to work away the i and the minus sign from under the root sign .
suggest you'll get those four
±1 ± i ,
so with arguments
¼π, ¾π, 5π/4, 7π/4 .
oh , and with modulus
|¹²³⁴x| = √2 .
What about x= (2i)^1/2
X = ±√-4
That solution of yours is wrong.
Prove it wrong then because (√-4)² is -4
With all respect for your skills if you cannot simplify further than this, then in an exam situation you probably will run out of time for other questions.
@@Nguyễn-j9q
Let’s substitute your answer
(sqrt(-4))^4 = -4
rewrite -4 as a power which you rewrite as, a^(exponent/denominator)
((-4)^1/2)^4 = -4
Simplify the left side of the equation, using the (a^m)^n = a^mn rule
(-4)^((1/2)(4)) = -4
Simplify the exponent
(-4)^(4/2) = -4
Simplify it again
(-4)^2 = -4
16 =/= -4
Using the negative answer will yield the same result since sqrt(x^2) = |x|
Your error was that you saw the “x^4” as “x^2”.
Your second error was not simplifying your answer to ±2i
Ok
I found solution to be √2 please tell me if it's correct
X=-4^1/4
Multiplying RHS power by 2/2 gives.
16^1/8
Converting it to lowest from gives √2.
Will anyone tell me if it's correct please
@@sushmagovindraodarokar8727 If you do ^4 with any real number you will always get a positive result. You need a complex number to get the negative result "-4" when doing ^4 . You are right, that √2 is the magnitude of the solution, but it has a real and a compex part.
Why x is not √(2i) ???
How big is sqrt(2i)? Is it bigger or smaller than 1?
x⁴=-4
x⁴+4=0
(x²)²-(2i)²=0
(x²-2i)(x²+2i)=0
(x²-(√2√i)²)(x²-(√2i√i)²)=0 but √i=1/√2+i/√2 (principal root)
(x-√2(1/√2+i/√2))(x+√2(1/√2+i/√2))(x-√2i(1/√2+i/√2))(x+√2i(1/√2+i/√2))=0
(x-1-i)(x+1+i)(x+1-i)(x-1+i)=0
so
x₁=1+i
x₂=-1-i
x₃=-1+i
x₄=1-i
x^4 = -4
x^4 = 4∙(-1)
x^4 = 4∙e^((1+2k)πi) k∈Z
x^4 = 4∙e^((1+2k)πi) / ^(1/4)
x = √2∙e^((1+2k)πi/4)
for k=0: x₁ = √2∙e^(πi/4) = √2∙(cos(π/4)+i∙sin(π/4)) = √2∙(1/√2+i/√2) = 1+i
for k=1: x₂ = √2∙e^(3πi/4) = √2∙(cos(3π/4)+i∙sin(3π/4)) = √2∙(-1/√2+i/√2) = -1+i
for k=2: x₃ = √2∙e^(5πi/4) = √2∙(cos(5π/4)+i∙sin(5π/4)) = √2∙(-1/√2-i/√2) = -1-i
for k=3: x₄ = √2∙e^(7πi/4) = √2∙(cos(7π/4)+i∙sin(7π/4)) = √2∙(1/√2-i/√2) = 1-i
You spent 11’ and several page scrolls for something really elementary. Just express -1 in polar form, for k = 0…3, probably you’ve heard about Euler identity…
Is the rudeness necessary? This is a community for math enthusiasts, please don't contaminate the space with hubris and discourage sharing. Some of us are still learning other methods and your manner is distracting from your math skills. You could also just leave the channel.
Maths taught me everything has a solution.
Then you need to return to basics.
Trol me why you are not using comp and programming? You can learn how to spend your life tíme better.
Scam
I will go back to study identities. We do not divide by zero and your fast explanation was not clear. Bye!
This is simple and obvious. Think of rotating/ multiplying vectors. It is a matter of how you look at the problem. The obvious thingis that the magnitude must be the 4th root of 4. No problem The next thing is to divide 180 deg by 45 and get 45 degr
Really? So, having in mind we live in a 3d world with 3 real axes, can you describe where is the location of imaginary axis?
1,414213562373095i ???