*Simpler Solution - Use n'th root of unity* x^5 = 1 = e^(i 360° n) so, x = e^(i n 360° / 5) = e ^ (i n 72°) for n = 0, 1,2, 3,4 leading to 1 (the only real root) and 2 complex conjugate roots with angles ±72° and ±144° So, *the 5 roots are* 1, cos 72° ± i sin 72° and cos 144° ± i sin 144° (Use cos ±72° = sin 18° = (√5-1)/4) and (cos ±144° = -cos 36° = -(√5+1)/4) and similarly the sine counter-parts to express the solution without any un-computed trigonometric functions)
Exactly! And phi^2 could be replaced by phi+1, to make the roots look a bit simpler... And you don't need to put b=-1/a and replace and multiply by a to get the quadratic equation, because you have the sum S and the product P of a and b, and by theorem of Vieta, a and b ar the solutions to y^2 - Sy + P = 0
9:30 You can not only figure out sin(72) but also the sin and cos value of 72, 144, 216, 288. Those angles which you can raise to the fifth power to rotate back to 1 on the complex plane.
Why not use Eulers Identity? e^iπ = -1? Therefore: e^i2kπ = 1, x^5 = e^i2kπ, etc. In the end, it is a circle in the complex plane with roots at: (0°), (72°), (144°), (216°), and (288°).
At the very beginning he said he didn't want to use that method and wanted to show it in the more difficult algebraic way which is juuuust barely possible.
Since phi^2=phi + 1, it follows that phi^2 - 4=phi - 3, so the first two solutions can be presented as [phi+i*(3-phi)^1/2]/2 and [phi-(phi-3)^1/2]/2. For the other two, consider that -1 - 4*phi^2=-1 - 4*phi - 4=-5 - 4*phi. So we can present these solutions as [1+i*(5+4*phi)^1/2]/(2*phi) and [1-i*(5+4*phi)^1/2]/(2*phi).
An easier way to factor it is to use geometric series. (x^5 - 1) / (x -1) is a geometric series with first term 1, 5 terms, and ratio x. So 1 + x + x² + x³ + x⁴
And that was the only real root in the given solution set. What exactly is the problem here? Edit: Hey all, I will also take the opportunity to edit my comment. The original comment used to say that the video was straight-up wrong, but after my reply, this statement has been removed.
@@isavenewspapers8890 I really thought you commented long after I edited it. But if not, then youre right👍 because I did first comment implying/claiming there was a problem.
What an amazing video bprp! I am only an 8th grader but ever since I saw your calculus videos, I have been engrossed with your channel. Is it ok if you can make a video with a question featuring an improper integral, a derivative, a series and a limit all in one question where it is aimed at Calculus 1 and 2 students. Thank you!
Having 5 solutions, we can take 1, than we take the non arytmetic solution (with the module of a complex and its angle), select a numbrr with 1 as the module and with either 1/5 of 2π, or 2/5 , 3/5 and 4/5 of 2π as the angle Than just take cos and sin and you have your 5 solutions (And sorry for my bad english(
x is 1 the other 4 solutions (since its a power of 5 equation is has 5 solutions) are all equally spread around the complex plane (72 degrees apart) with a size 1, use trigonometry to find out the real and imaginary componants of each
thank you for the video and for correctly saying the name of the greek letter (not fai, but fee). Same should be for pi (which is pee, rather than pie).
The solution to x^n = 1 actually isnt difficult I did stuff with a unit circle and found: x = sin(Z(360/n)+90) + cos(Z(360/n)+90)i Where Z is an integer < 0.5n and greater than or equal to 0
No. The quartic is the end-of-the-line, when it comes to a closed-form formula in elementary functions as the master key to solve any polynomial of that degree. There are special case quintics where there exists a formula, but Galois proved there can be no such general formula for quintics, or anything beyond.
@@carultch OK, I don't know anything about Galois, but I asked about quartics, not quintics. When I write out the general equation x^4+ax^3+bx^2+cx+f I can use 4 variables, p, q, d and e in the form (x^2+px+d)*(x^2+qx+e) and that gives me de = f p + q = a pq+d+e = b pe+dq = c These reduce to the values in this video where d = 1 and e = 1. That should be solvable, right? It might not be trivial to solve, but there are 4 equations/4 unknowns.
I love how he says "now, this is the part where we have to think a little bit" So,what in the math have we been doing the whole time???? We did all this without thinking??? 😮😂
I don't understand the reason this method can't be used to solve something like x⁷=1? It feels like you can use the same method like factoring x-1 then factoring two cubics out of the sextic and solving for all? Which would give you a non transcendental solution to sin(π/7)
I was hoping for five values for x. Thanks for continuing these videos, it's been ages and I just came back to playing with math. You aged a little, but that makes it all the more beautiful ❤ Much love from here! ❤️ ❤
To solve the equation \( X^5 = 1 \) where \( X eq 1 \), we are looking for the fifth roots of unity. The solutions to this equation can be expressed in the form: \[ X = e^{2\pi i k / 5} \] where \( k \) is an integer. The fifth roots of unity are: 1. \( k = 0 \): \( X = e^{2\pi i \cdot 0 / 5} = 1 \) 2. \( k = 1 \): \( X = e^{2\pi i / 5} \) 3. \( k = 2 \): \( X = e^{4\pi i / 5} \) 4. \( k = 3 \): \( X = e^{6\pi i / 5} \) 5. \( k = 4 \): \( X = e^{8\pi i / 5} \) Since we want the solutions where \( X eq 1 \), the valid solutions are: 1. \( X = e^{2\pi i / 5} \) 2. \( X = e^{4\pi i / 5} \) 3. \( X = e^{6\pi i / 5} \) 4. \( X = e^{8\pi i / 5} \) In summary, the solutions to \( X^5 = 1 \) with \( X eq 1 \) are: - \( e^{2\pi i / 5} \) - \( e^{4\pi i / 5} \) - \( e^{6\pi i / 5} \) - \( e^{8\pi i / 5} \)
if u just want the answer its: \frac{1}{26}\left(5x-\ln\left(5\sin x+\cos x ight) ight)+C to get it start with u=tanx, should get a partial fraction situation and then abit of algebraic manipulation to get the simplified answer
That's the Gaussian integral. You can find no end of resources solving it online. Here's the basic process: 1. Square the integral. 2. Convert to double integral. 3. Switch to polar coordinates. 4. Solve. 5. Square root result.
2:33 “any polynomial can be factored in terms of linear and quadratic” Wouldn’t that mean any polynomial would have solutions expressible in terms of linear terms and square roots (possible complex roots)? That’s clearly not possible. x^3-2=0 for starters.
@@isavenewspapers8890 yeah, sorry about that. After I posted, I re-looked at the answers and saw the plus-minus signs and realized there were 5 solutions.
Suggestion: make paid courses on udemy for people Who want to learn calculus/Trigonometry/Advanced algebra and complex numbers. It will be Nice, at least for me.
Yeah, you got the point It's actually possible if you consider it as a pentagon in a 2d axis ,and it's rotated 90 degrees You can solve the equation with the geometric
@@文率 I played with similar concepts for encoding I/Q signals (basically quantized complex numbers) for efficient streaming. A few videos my personal channel.
You made a mistake. b is not simply the negative of [ 1 ± √(5) ] / 2, it is the RECIPROCAL of the negative. It didn’t affect your answer in the end though because you substituted for φ in the end, which then you went back to the a & b system of equations.
b wasn't written as the negative of (1 ± √5) / 2 in the first place. That would be (-1 ∓ √5) / 2, not (1 ∓ √5) / 2. This supposed mistake doesn't exist.
@@CaroSuon Actually, they said it's the *negative* reciprocal. Also, what makes you think I don't already understand the existence of this relationship?
dude no need for all these calculations. you can instantly tell it is 1 :D Maybe also consider -1 but -1^5 = -1 ... What is all that fancy stuff around that haha
Try my "extreme quintic equation" next: x^5-5x+3=0
ua-cam.com/video/GoGsVLnf8Rk/v-deo.htmlsi=WumfxJqZYwNfBWe_
Use demoiver's theorem 😊
Co-efficient (ab+2) should be with X^2.
Doesn’t ANY number above one increased when ^5? And ANY number under one decrease?
Hi, would you like doing number theory problems and proofs of its theorems?
@@Jordan-gt6gd yes
Everyone says that x⁵ = 0 is easier than this. But i think that x⁵ = x⁵ is much easier
How about x⁵=x⁵+1? Instead of everything, it's just nothing. Just as simple.
That's a violation.😢
It's not a quintic though
it's not quintic.
Quintic is ax⁵+bx⁴+cx³+dx²+ex+f=0.
@@YarinGD you win!
Does anyone else miss the pokeball mic? Maybe a special return of the pokeball episode before the end of the year?
2:35
His face when he double thinks about what he said is gold.
“Tf did I just say”
Wouldn't x^5 = 0 be the easiest? 😂😂
We want the simplest, not the freeest
@@Musterkartoffel😂😂
That would be the most trivial
Unless you're willing to extrapolate and include x⁵=x⁵ I guess.
!No
I love the connection between phi and this quintic/pentagons
Is it just me that saw this as a tiktok search thing? UA-cam has done this with other shorts
@@Chessbutmostlyrandomstuff Nice profile picture :)
@@Chessbutmostlyrandomstuff
I think it's like some random words just get that search thing, it's a bit odd
4:26 THAT X² IN THAT STEP STAYED AS "X" FOREVER RAAAAAAAA
Unsatisfactory 😭 once you see it, you can't unsee it
I saw it immediately because he actually mentioned it but didn't fix it
Yes, I spotted that, too, and wanted it to be corrected as BPRP did correcting a previously missed one at 4:24.
9:24 You mean sine of 72 DEGREES.
Granted 72 radians is a ridiculous statement
isn't x⁵=0 simpler ?
@@charlievane 😂
Thats not simpler,thats free
x⁵=1 is still the simplest one that still gives a challenge, the definition of 'simple' and 'challenge' is left as an exercise to the reader.
@@Musterkartoffel How is it free if x can't be any value?
@@anglaismoyen0?
we can also divide by x^2 and reduce the equation into quadratic by substituitng x+1/x as t
He has done that before in previous videos on the exact same quartic equation.
Only by assuming x0.
@@alanclarke4646 it is not assumed because we can very easily verify from the original equation that x cannot be 0
*Simpler Solution - Use n'th root of unity*
x^5 = 1 = e^(i 360° n) so, x = e^(i n 360° / 5) = e ^ (i n 72°) for n = 0, 1,2, 3,4 leading to 1 (the only real root) and 2 complex conjugate roots with angles ±72° and ±144°
So, *the 5 roots are* 1, cos 72° ± i sin 72° and cos 144° ± i sin 144°
(Use cos ±72° = sin 18° = (√5-1)/4) and (cos ±144° = -cos 36° = -(√5+1)/4) and similarly the sine counter-parts to express the solution without any un-computed trigonometric functions)
1:10 I am surprised this is my first time seeing this type of method
Hence why I like mutlivalued functions. x=⁵√1, write out the 5 values by rotating by 2π/5 each time from 1, and we're done.
Como se usa esse texto azul?
@@oKrybianiAs far as I noticed, those are added automatically by UA-cam
Ok but he said he wouldn’t use polar forms
@@oKrybiani I don't. I think UA-cam adds it automatically. Which text was blue?
@@xinpingdonohoe3978 the smol 5
So satisfying! I was never good at this kind of math, so it's really interesting to see how you go about solving these equations.
"-1/phi" could just be called "1-phi".
Exactly! And phi^2 could be replaced by phi+1, to make the roots look a bit simpler...
And you don't need to put b=-1/a and replace and multiply by a to get the quadratic equation, because you have the sum S and the product P of a and b, and by theorem of Vieta, a and b ar the solutions to y^2 - Sy + P = 0
Phi low so phi.
1-sqrt(5)/2 is usually the conjugate of phi, or phi bar.
9:30 You can not only figure out sin(72) but also the sin and cos value of 72, 144, 216, 288. Those angles which you can raise to the fifth power to rotate back to 1 on the complex plane.
Why not use Eulers Identity? e^iπ = -1? Therefore: e^i2kπ = 1, x^5 = e^i2kπ, etc.
In the end, it is a circle in the complex plane with roots at: (0°), (72°), (144°), (216°), and (288°).
At the very beginning he said he didn't want to use that method and wanted to show it in the more difficult algebraic way which is juuuust barely possible.
Wow! That was really enlightening.
Since phi^2=phi + 1, it follows that phi^2 - 4=phi - 3, so the first two solutions can be presented as [phi+i*(3-phi)^1/2]/2 and [phi-(phi-3)^1/2]/2.
For the other two, consider that -1 - 4*phi^2=-1 - 4*phi - 4=-5 - 4*phi. So we can present these solutions as [1+i*(5+4*phi)^1/2]/(2*phi) and [1-i*(5+4*phi)^1/2]/(2*phi).
OMG I LOVE THAT PHI POPPING OUT AS RELATES TO THE COMPLEX CIRCLE
An easier way to factor it is to use geometric series. (x^5 - 1) / (x -1) is a geometric series with first term 1, 5 terms, and ratio x. So 1 + x + x² + x³ + x⁴
Last four roots are complex. ANY number >1 will increase when raised to 5th power. And ANY number
And that was the only real root in the given solution set. What exactly is the problem here?
Edit: Hey all, I will also take the opportunity to edit my comment. The original comment used to say that the video was straight-up wrong, but after my reply, this statement has been removed.
@@isavenewspapers8890i didnt say there was a problem, just that 1 is the only real root and the others are complex.
@@ibperson7765 You know UA-cam shows when your comments are edited, right?
@@isavenewspapers8890 I really thought you commented long after I edited it. But if not, then youre right👍 because I did first comment implying/claiming there was a problem.
@@ibperson7765 I see.
So you have derived a closed form of cos 2n pi/5, sin 2n pi/5 n 0..4 . Thats nice
IS THE FISRT TIME I NOTICED YOU TYPED ON YOU T SHIRT AL GEBRA ARABIC "I AM VERY PROUD"
Can you make a video explaining how to solve this?
√(2x²) - √(2x) = 3
1:03 I know how to divide polynomials the long way but I don't recall ever being taught this way (which you call "synthetic division").
What an amazing video bprp! I am only an 8th grader but ever since I saw your calculus videos, I have been engrossed with your channel. Is it ok if you can make a video with a question featuring an improper integral, a derivative, a series and a limit all in one question where it is aimed at Calculus 1 and 2 students. Thank you!
there's a mistake! b=-1/a so b=2/(1-+sqrt(5)) and not b=(1-+sqrt(5))/2
Same thing.
Wait a second...
Oh, when you wrote "b=2/(1-+sqrt(5))", I think you forgot a minus sign.
They're the same
Having 5 solutions, we can take 1, than we take the non arytmetic solution (with the module of a complex and its angle), select a numbrr with 1 as the module and with either 1/5 of 2π, or 2/5 , 3/5 and 4/5 of 2π as the angle
Than just take cos and sin and you have your 5 solutions
(And sorry for my bad english(
Oh wow I didnt even notice I was early becuase I was so invested! Amazing video
Cu stilul tău de a prezenta problemele vei reuși să îndepărtezi mulți vizitatori de matematica adevărată.
So good!
When I saw this I immediately thought "roots of unity". But yeah this is a creative way. (Also why not just use the quintic formula? oh wait... :)
Also, why not just use the geometric series formula for x^4+x^3+x^2+x+1? 🤔
@@sudoer-Ht Because this is not a geometric series? A series goes on forever, up to infinity.
@@bjornfeuerbacher5514 series can be finite. Search for geometric series sum formula
@@bjornfeuerbacher5514 No, the term "series" does not mean it goes on forever. The two terms "infinite series" do.
@@tfg601 A series is, by definition, an infinite sum.
Me: Mr bprp can I please go to the washroom:
Mr bprp: 2:38
Isnt it b = -2/(1+√5) ?
Same thing.
The video was (1-√5)/2
@@nurrohmadi7852 Same thing.
@@isavenewspapers8890 you right
x is 1
the other 4 solutions (since its a power of 5 equation is has 5 solutions) are all equally spread around the complex plane (72 degrees apart) with a size 1, use trigonometry to find out the real and imaginary componants of each
@@md-sl1io he explicitly said he wasn't doing that because it was too easy
plz do septic equation
thank you for the video and for correctly saying the name of the greek letter (not fai, but fee). Same should be for pi (which is pee, rather than pie).
That can unfortunately cause ambiguity, given that the letter P is pronounced the exact same way.
Why not draw a circle and divide into 2pi/5 sectors?
The solution to x^n = 1 actually isnt difficult
I did stuff with a unit circle and found:
x = sin(Z(360/n)+90) + cos(Z(360/n)+90)i
Where Z is an integer < 0.5n and greater than or equal to 0
You can also rewrite it as an exponential
Yup just roots of unity
Isn't the thumbnail saying الجبر in Arabic?
Or i am seeing things ?
Yeah he has a shirt that says algebra in arabic (which is (الجبر))
So it says Al-Jabr? As in, Al-Khwarizmi's famous book?
@@xinpingdonohoe3978 Yep. Al-Jabr is the namesake of algebra. Our word alcohol is also an Arabic-inspired word.
@@xinpingdonohoe3978
Exactly
As in that book
since b = -1/a shouldn't b be equal to -2/(1 + sqrt(5))?
Does that approach for solving the quartic work for all quartics? That was sweet.
No. The quartic is the end-of-the-line, when it comes to a closed-form formula in elementary functions as the master key to solve any polynomial of that degree. There are special case quintics where there exists a formula, but Galois proved there can be no such general formula for quintics, or anything beyond.
@@carultch OK, I don't know anything about Galois, but I asked about quartics, not quintics. When I write out the general equation
x^4+ax^3+bx^2+cx+f
I can use 4 variables, p, q, d and e in the form
(x^2+px+d)*(x^2+qx+e)
and that gives me
de = f
p + q = a
pq+d+e = b
pe+dq = c
These reduce to the values in this video where d = 1 and e = 1.
That should be solvable, right? It might not be trivial to solve, but there are 4 equations/4 unknowns.
At 7:06 it shouldve been b=2/(1-sqrt(5))
You forget (ab+2)x²
Yes, I spotted that, too, and wanted it to be corrected as BPRP did correcting a previously missed one at 4:24.
The square term would be ab+2, which he already has.
Co-efficient (ab+2) should be with X^2.
Can you do one video in polar form?
De moivres theorem and symmetry around the unit circle for an elegant approach?
I love how he says "now, this is the part where we have to think a little bit"
So,what in the math have we been doing the whole time???? We did all this without thinking??? 😮😂
That statement probably means critical thinking. You don't need critical thinking to apply a formula.
I wasn't expecting the Golden Ratio to pop out of this at some point.
I don't understand the reason this method can't be used to solve something like x⁷=1? It feels like you can use the same method like factoring x-1 then factoring two cubics out of the sextic and solving for all? Which would give you a non transcendental solution to sin(π/7)
@@iaroslav3249 Who said sin(π/7) is transcendental? It's not.
Oh wait it can be written with cubic and square roots, for some reason I thought it transcendental.
I was hoping for five values for x.
Thanks for continuing these videos, it's been ages and I just came back to playing with math.
You aged a little, but that makes it all the more beautiful ❤
Much love from here! ❤️ ❤
+-
You got five values, the second and third boxes are pairs of complex conjugates
Isn't Using De Movier theorem more easier?
I felt weird when he forgot to close the bracket and I said "close it!", he stopped for a sec then closed it.
Great, but i wish you had kept going to show that the other 4 answers were imaginary, and maybe what the curve looks like in complex space
To solve the equation \( X^5 = 1 \) where \( X
eq 1 \), we are looking for the fifth roots of unity. The solutions to this equation can be expressed in the form:
\[
X = e^{2\pi i k / 5}
\]
where \( k \) is an integer. The fifth roots of unity are:
1. \( k = 0 \): \( X = e^{2\pi i \cdot 0 / 5} = 1 \)
2. \( k = 1 \): \( X = e^{2\pi i / 5} \)
3. \( k = 2 \): \( X = e^{4\pi i / 5} \)
4. \( k = 3 \): \( X = e^{6\pi i / 5} \)
5. \( k = 4 \): \( X = e^{8\pi i / 5} \)
Since we want the solutions where \( X
eq 1 \), the valid solutions are:
1. \( X = e^{2\pi i / 5} \)
2. \( X = e^{4\pi i / 5} \)
3. \( X = e^{6\pi i / 5} \)
4. \( X = e^{8\pi i / 5} \)
In summary, the solutions to \( X^5 = 1 \) with \( X
eq 1 \) are:
- \( e^{2\pi i / 5} \)
- \( e^{4\pi i / 5} \)
- \( e^{6\pi i / 5} \)
- \( e^{8\pi i / 5} \)
Sir
Can you please solve this problem
Integration of 1/5+cotx
Please
Is that (1/5)+cotx or 1/(5+cotx)?
@@Speak22wastaken 2nd one
@@thekingishere2007 Have you tried a Weierstrass substitution?
if u just want the answer its:
\frac{1}{26}\left(5x-\ln\left(5\sin x+\cos x
ight)
ight)+C
to get it start with u=tanx, should get a partial fraction situation and then abit of algebraic manipulation to get the simplified answer
7:18 something is going wrong here...b=-2/(1+- sqrt 5)...
And of course the reason you can use this to find the sin, cos, and tan of 72 is because of the polar form.
Who knows in thumbnail on his shirt algebra is written
Really 😲
I think the octic power gives way nicer roots
Yeah, but those are also more trivial and less likely to make for an interesting video. It does sound like a nice bprp fast video, though.
Not enough for you? I think solving it the long way like he did would be funny. Otherwise the 16th power is quite similar
All the solutions of this equation exist on my channel ❤️
Does logarithm not come under algebra?
How do we solve the integration which is on his shirt?
That's the Gaussian integral. You can find no end of resources solving it online.
Here's the basic process:
1. Square the integral.
2. Convert to double integral.
3. Switch to polar coordinates.
4. Solve.
5. Square root result.
If that’s the simplest one, I’d hate to see what a difficult one looks like
Now try doing it using the quartic formula (and use up the entire marker to finish) 😂.
2:33 “any polynomial can be factored in terms of linear and quadratic”
Wouldn’t that mean any polynomial would have solutions expressible in terms of linear terms and square roots (possible complex roots)? That’s clearly not possible. x^3-2=0 for starters.
Okay, I didn’t fully understand that, but it looked to me that he only showed 3 answers for x. Should there be 5 answers for x?
There are two plus-minus signs, and each choice of plus or minus corresponds to a different solution. So yes, there are 5 solutions altogether.
@@isavenewspapers8890 yeah, sorry about that. After I posted, I re-looked at the answers and saw the plus-minus signs and realized there were 5 solutions.
Why so much work, it's a starfish in polar form lol.
that's it
الجبر
Nice 🤟
nice algebra الجبر
Can you send to me one of the الجبر T-shirts ?
Suggestion: make paid courses on udemy for people Who want to learn calculus/Trigonometry/Advanced algebra and complex numbers. It will be Nice, at least for me.
Plus/minus too many roots
Can anyone explain me why φ is suddenly stealing the show? Like i get that it's there. But is there any visual representation?
Palindromic quartic equation, solve it with the palindromic style 😂
1
Now for X^5 = i
Yeah, you got the point
It's actually possible if you consider it as
a pentagon in a 2d axis ,and it's rotated 90 degrees
You can solve the equation with the geometric
@@文率 I played with similar concepts for encoding I/Q signals (basically quantized complex numbers) for efficient streaming. A few videos my personal channel.
😊😊
So... Students, what do we learn from this... 😮
You made a mistake. b is not simply the negative of [ 1 ± √(5) ] / 2, it is the RECIPROCAL of the negative. It didn’t affect your answer in the end though because you substituted for φ in the end, which then you went back to the a & b system of equations.
Two mistakes that canceled
b wasn't written as the negative of (1 ± √5) / 2 in the first place. That would be (-1 ∓ √5) / 2, not (1 ∓ √5) / 2. This supposed mistake doesn't exist.
@@isavenewspapers8890you didn’t understand the comment, he’s talking about the relation between a and b being reciprocals to each other
@@CaroSuon Actually, they said it's the *negative* reciprocal. Also, what makes you think I don't already understand the existence of this relationship?
@@CaroSuon If it's unclear, (1 ∓ √5) / 2 and -2 / (1 ± √5) are the same thing.
Simple? It's a trap!
How would you solve x^5 + x = 1 ?
WolframAlpha gives a truly horrifying solution.
@@isavenewspapers8890 i don't have a subscription so i only could see the answer...
tbh I do not get his but still noice
Bro pretends like φ² ≠ φ + 1 💀💀
I miss the imaginary Solution or is it?
You mean the nonreal solutions? Those are covered by the last two equations that were written, where each equation represents two solutions.
X squared.
Surprise phi.
His shirt in the thumbnail says algebra in Arabic
Not the simplest, it is beaten by x^5 = 0.
I like the shit in the thumbnail it says algebra in Arabic
Answer: x=
Sorry i Dorgot
This one doesn’t count. Try x^5-2=0.
dude no need for all these calculations. you can instantly tell it is 1 :D
Maybe also consider -1 but -1^5 = -1 ...
What is all that fancy stuff around that haha
Math feels just made up sometimes
Omg just use cis !
Why don't you use the computer
Because the point is to do it by hand.
@@isavenewspapers8890 Thanks for explanation