Find all x

[@mohammadalvee8411]

"Those who stop learning stop living"🔥

[@Mohak-gq5sw]

"Those who stop living stop learning" 💀

[@mythyt3947]

Umm,why taking it so seriously 😅

[@ChOwToo]

@@Mohak-gq5sw The first affirmation is the general case, yours is a specific one.

[@jakehobrath7721]

With my first impression of your videos, I didn’t quite latch on. But as I kept with the channel, I discovered that the exercises you choose for your videos might be the best on KZbin. They all satisfyingly fit that “what’s the creative trick” pattern you look for in an exercise. I’m kinda hooked now

[@lornacy]

When I am looking for my mistakes, the first thing I check is whether I have messed up a positive or negative somewhere! Good to know I'm not alone ❤ Thank you for mentioning that 😊😊

[@jensraab2902]

It's often the little things that are tripping us up. And apparently it happens to the best of us. Just like you, I'm a bit relieved to see that I'm not alone in my stupid mistakes! 😁

[@Grecks75]

I solved the problem using the equivalent relation (x - 1)^4 = 12x^2. One thing I didn't like about the problem statement is that it doesn't specify the domain of x. Shall we solve it over the reals or the complex numbers? When solving in the complex plane, I found two positive real solutions and two non-real complex-conjugate ones in the second and third quadrant with a modulus of sqrt(1 + 4sqrt(3)) ~= 2.816 both. Their argument has an absolute value of approximately 105°. No nice numbers here. 😢

[@emanuellandeholm5657]

I tried out this problem the first time you posted it, lol! I got a quartic with almost Pascal's triangle coeffs, but the signs were off. It will be instructive to watch you solve this one for real. :) Edit: Yeah, the substitution t = x + 1/x cracks the case wide open. Nice!

[@westquote]

Just wanted to chime in that this was a truly great video. Sharing with my math nerd friends at work!!

[@jensraab2902]

I was looking for the link in the description of your video on palindromic polynomials and it seems you forgot to link it. (This video seems to be jinxed! 😅) Looked it up in your videos and wanted to post the link for the convenience of all those who wanted to check that video out first but YT apparently doesn't like links my comment and deleted it after just a few seconds. It's called "How to Solve Palindrome Equations". I guess plugging this into the search bar should bring up your video as one of the hits. Love your stuff, keep it coming!

[@jensraab2902]

On a rewatch I saw that the title of the video that I so proudly reproduced above because I was looking for it in the channel's video history is shown right there at the 3:24 in its full glory! 🤦🏽‍♂😅😅

[@arcangyal2269]

We have a lot of crazy math exercises, but I did not expect to see one in your channel

[@daniorugbani5914]

You’re really good in your presentations, excellent job.

[@petefritz5679]

Before dividing through by X^2 shouldn't we first confirm that X does not equal 0? It seems a minor point but maybe shouldn't be overlooked.

[@duckyoutube6318]

Pascal's triangle is amazing. I use it on anything beyond (a+-b)^4 Saves time and if you do it by hand its really easy to lose sight of signs and like terms. You end up with a page full of a's and b's.

[@ناصريناصر-س4ب]

After noting that 1 is not a solution to the equation, we can put (x+1)/(x-1)=y, so we get the equation 3y⁴-6y²-1=0. By solving this multiplying equation by square, we get the solutions to the original equation.

[@LearnmoreMoyo-q1o]

A beautiful approach....

[@vishalmishra3046]

*Another approach* I simply expanded (1+x)^4 using Pascal's triangle, to get 2 (1+x^4) = (1 + 4x + 6x^2 + 4x^3 + x^4) which simplifies to x^4 - 4x^3 + 6x^2 - 4x + 1 = 6x^2 + 6x^2 which is (x-1)^4 = 12 x^2 so (x-1)^2 = +/- √12 x On expanding LHS you get x^2 - (2 +/- √12) x + 1 = 0. The 2 quadratic equations give you all 4 solutions (2 real and 2 complex conjugate). x = (1 +/- √3) +/- √ [ (1+/- √3)^2 - 1 = 3 +/- 2√3 ], so 2 real solutions are (1+√3) +/- √(3 + 2√3) and 2 complex conjugate solutions are (1-√3) +/- i √ (2√3 - 3). *Simple. right* ?

[@tcmxiyw]

Nice video and great solution. Note that all four of the solutions are complex numbers, two of them are real numbers and two of them are imaginary numbers.

[@XTRA14LIVE]

The video is a correction of a previous video where the author made a mistake in solving a problem. The author uses the binomial expansion and Pascal's triangle to solve the problem, which is their preferred method. The polynomial equation is palindromic, which allows the author to use a specific solving strategy. The author uses substitution and the quadratic formula to find the real solutions to the equation, and also identifies the complex solutions. The author emphasizes the importance of continuous learning and never stopping to learn.

[@MONDLIMTHOMBENI-tt2vl]

That nodd after -6 =0 (last line ) hits. I was like oh yes😂

[@chepkwuruihoseaben1336]

You are a genius

[@afifamyouni673]

i solved it in a different way. i didn't use the substitution method. i developed it 1st, then i used the difference of 2 squares formula, then the quadratic formula, and i got the same answers.

[@misterj.a91]

Great. My confusion is now gone

[@samdean1966]

A slightly different route would be to transform it into (x-1)⁴=12x² and go (x-1)²=±4√3x, and this can be phrased as a quadratic in x-1

[@octobermathematics]

Thank you from this video I learnt about Palindromic equation today 🙏

[@SALogics]

Nice problem, Nice solution ❤❤

[@hobojoe6453]

You’re the man

[@dikkedorus]

Where does the "rad" nomenclature come from with the square roots, I never seen that before? I would love to see the solutions plotted on a complex plane, it looks like it would create a nice geometric interpretation somewhere in there.

[@AvoidsPikes-]

I had a Canadian professor in my university who regularly said "rad." Since then, I have believed it to be a non-American, english speaking thing.

[@PrimeNewtons]

'Radical' for irrational roots. Also, it is shorter than 'square-root of'.

[@shlokthosar8094]

2:03 very good

[@SorosKornel]

Wow, I didn't think I would meet a Hungarian excercise here. (I'm from Hungary by the way)

[@nousernameleft999]

my approach would be to rewrite it as 0 = (x-1)^4 - 12x^2 = ((x-1)^2+ sqrt(12)x)*((x-1)^2 -sqrt(12)x) from difference of squares, so then its just two quadratics

[@johnlv12]

awesome approach

[@okaro6595]

This really cries for the use of the pq-formula. When a=1 and b is even it simplifies things.

[@m.h.6470]

Solution: (1 + x⁴) / (1 + x)⁴ = 1/2 First of all, x ≠ -1 Then (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴ so (1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴ if we artificially add and remove (4x + 6x² + 4x³)/(1 + 4x + 6x² + 4x³ + x⁴), we end up with 1 - (4x + 6x² + 4x³)/(1 + 4x + 6x² + 4x³ + x⁴) = 1/2 |-1 -(4x + 6x² + 4x³)/(1 + 4x + 6x² + 4x³ + x⁴) = -1/2 |*-1 (4x + 6x² + 4x³)/(1 + 4x + 6x² + 4x³ + x⁴) = 1/2 But that literally means, that 1 + x⁴ = 4x + 6x² + 4x³ as both the original left term and the new left term are equal to 1/2, so you can multiply both terms with (1 + x)⁴ and end up with the above equation. Putting it all on one side and you get x⁴ - 4x³ - 6x² - 4x + 1 = 0 To solve this tricky polynomial, first we divide everything by x² x² - 4x - 6 - 4/x + 1/x² = 0 Then we rearrange the terms a bit (x² + 2 + 1/x²) + (-4x - 4/x) - 8 = 0 (x + 1/x)² - 4(x + 1/x) - 8 = 0 Now we substitute x + 1/x = u u² - 4u - 8 = 0 u = 2 ± √(4 + 8) u = 2 ± √12 u = 2 ± 2√3 Now we resubstitute x + 1/x = 2 ± 2√3 (x² + 1)/x = 2 ± 2√3 |*x x² + 1 = (2 ± 2√3)x |-(2 ± 2√3)x x² - (2 ± 2√3)x + 1 = 0 x₁,₂ = 1 + √3 ± √((1 + √3)² - 1) x₁,₂ = 1 + √3 ± √(1 + 2√3 + 3 - 1) x₁,₂ = 1 + √3 ± √(3 + 2√3) x₃,₄ = 1 - √3 ± √((1 - √3)² - 1) x₃,₄ = 1 - √3 ± √((1 - 2√3 + 3 - 1) x₃,₄ = 1 - √3 ± √(3 - 2√3) And since 2√3 > 3, we can make it -1 * (2√3 - 3), leading to two complex solutions x₃,₄ = 1 - √3 ± i√(2√3 - 3)

[@surendrakverma555]

Very good. Thanks 🙏🙏🙏🙏

[@octobermathematics]

Thank you

[@mathmachine4266]

2(1+x⁴)=(1+x)⁴ 2+2x⁴=1+4x+6x²+4x³+x⁴ -1+4x+6x²+4x³-x⁴=0 I dont feel like solving the quartic equation. But luckily, since this is symmetric, I can apply the following transformation to turn it into a quadratic (well, technically 2 quadratics): -1/x²+4/x+6+4x-x²=0 (Assuming x≠0) -(x+1/x)²+4(x+1/x)+8=0 x+1/x = 2±2√(3) x²-2x(1±√(3))+1=0 x=1±√(3)+√(3±2√(3)) or x=1±√(3)-√(3±2√(3)) √(3+2√3)=⁴√(3)√(2+√3) √(a+b) = √((a+√(a²-b²))/2) + sgn(b)√((a-√(a²-b²))/2) √(2+√3) = (√(3/2)+√(1/2)) = (√(3)+1)/√(2) √(3-2√3)=⁴√(3)i√(2-√3) = ⁴√(3)i*(√(3)-1)/√(2) All 4 solutions are x=(1+√(3))(1±⁴√(3/4)) x=(1-√(3))(1±⁴√(3/4)i)

[@tam31433]

This was cute❤

[@jacobgoldman5780]

is there a nice simplification for the sqrt(3+2sqrt(3)) as that doesnt look as nice. Otherwise solid solution!

[@AvoidsPikes-]

That was driving me nuts! 😅

[@tcoren1]

If there is it's not of the form a+b*sqrt(3) with a and b real numbers. Not sure if a more complicated form exists

[@rmandra]

Thanks!

[@PrimeNewtons]

Thank you!

[@allanmarder456]

After you get the equation x^4 - 4x^3 -6x^2 -4x +1=0 make the substitution x=y+1 to eliminate the cubic term: (y+1)^4 - 4(y+1)^3 -6(y+1)^2 -4y +1 simplifies to y^4 - 12y^2 -24y -12=0 Rewrite this as y^4 = 12y^2 +24y +12 =(sqrt(12)*y^2 +sqrt(12))^2. Thus y^4 - (sqrt(12)*y^2 +sqrt(12))^2 =0 The difference of 2 squares gives you 2 quadratics in y. Solve for y values then back substitute x=y+1 for the final answers.

[@HBGsantra]

NICE PROBLEM 👍👍

[@afuyeas9914]

Alternatively, the equation is equal to (x-1)^4-12x^2 which immediately leads to the two quadratics you end up finding

[@tambuwalmathsclass]

Great work 👍 At 10:18 Next time subtract that 8 from 4ac separately because of that equality 🟰 sign.

[@tmrapper6378]

This is a nice problem

[@MrCool-fm5ty]

Youre a nice problem

[@AvoidsPikes-]

This problem is very naughty. He solved it nicely though. I would have been dumbfounded. 😅

[@aljawad]

A very nice problem! Have you tackled problems involving “dual numbers”?

[@mathe-schrittfurschritt3717]

There is another way: Original equation is equivalent to 2+2x^4 = (x+1)^4 2 + 2x^4 = x^4 + 4x^3 + 6x^2 + 4x + 1 x^4 - 4x^3 - 6x^2 - 4x + 1 = 0 x^4 - 4x^3 + 6x^2 - 4x + 1 = 12x² (x-1)^4 = 12x² (x-1)^2 = sqrt(12)|x| ...and from there on, it should be pretty obvious (cases x0). Keep up the good work! Regards :-))

[@antonionavarro1000]

Good work. Thanks.

[@Grecks75]

Thanks for showing the t = (x + 1/x) substitution for the palindromic quartic equation. That was interesting, I didn't know it. I solved the problem using the equivalent relation (x - 1)^4 = 12x^2. That's another simple and direct way to tackle it.

[@alexs3243]

What is the intro song?

[@davidzanetti1269]

This is an off-topic challenge from a non- mathematician: can you prove the divisibility rules for 9 and 3 and use the proof to show that 12461(base 36) is not prime.

[@study_math]

x^4-4x^3-6x^2-4x+1=(x-1)^4-12x^2=((x-1)^2+2√3x)((x-1)^2-2√3x)

[@Nikioko]

(1 + x⁴) / (1 + x)⁴ = 1/2 2 (1 + x⁴) = (1 + x)⁴ 2 + x⁴ = 1 + 4x + 6x² + 4x³ + x⁴ 4x³ + 6x² + 4x − 1 = 0

[@ME-po1dr]

Wrong you forgot to factorise the 2 so it is 2x⁴ instead of x⁴

[@Nikioko]

@@ME-po1dr You are right, thanks. 2 + 2x⁴ = 1 + 4x + 6x² + 4x³ + x⁴ x⁴ − 4x³ − 6x² − 4x + 1 = 0 But doesn't help, either.

[@BRUBRUETNONO]

Hi thanks for your insteresting problem that I solved as here below. Of course, I didn't look at your solution. Tell me if you like mine. (1+x^4)/(1+x)^4=1/2 2(1+x^4)=(1+x)^4 x^4+4x^3+6x^2+4x+1=2x^4+2 -x^4+4x^3+6x^2+4x-1=0 x^4-4x^3-6x^2-4x+1=0 (x^4-4x^3+6x^2-4x+1)-12x^2=0 (x-1)^4-12x^2=0 [(x-1)^2]^2-[2sqrt(3)x]^2=0 [(x-1)^2-2sqrt(3)x][(x-1)^2+2sqrt(3)x]=0 [x^2-2(sqrt(3)+1)x+1][x^2+2(sqrt(3)-1)x+1]=0 Then [x^2-2(sqrt(3)+1)x+1]=0 or [x^2+2(sqrt(3)-1)x+1]=0 Delta=(sqrt(3)+1)^2-1.1 or Delta=(sqrt(3)-1)^2-1.1 Delta=3+2sqrt(3)>0 or Delta=3-2sqrt(3)

[@benhassineiem]

I USED A DIFFERENT APPROACH USING DEVELOPPEMENT OF (x-1)^4 SINCE IT HAS THE SAME COEFF. AS THOSE OF (x+1)^4. ADD OR SUBTRACT AND YOU WILL HAVE SOMETHING LIKE THIS: ( )^4=( )^2

[@guruone]

AsymptoticSum[(-Log[1 - t x]/t)/z /. t -> n/z, {n, 1, z}, z -> Infinity]

[@mmfpv4411]

Is that your answer? Confused