How to Solve Palindrome Equations

In this video I solved a palindrome equation using a technique that typically works for them

[@laman8914]

This dude is one of the best math teachers on YT. His trademark is to explain complex math problems in a step-by-step way that is simple and easy to follow. We never get tired or bored of watching his clips.

[@justine.3416]

you just earned a subscriber!! please never stop making these amazing, quality videos sir. your way of teaching complex equations in a digestible manner isn't just something you find easily on math tutorials these days haha

[@jwchem1984]

I really enjoy your KZbin program. Since I retired I have been watching a number of STEM KZbin programs. It’s amazes me how much math I have forgotten in 50 years. Anyway, the reason for this message is to suggest a change in how you use the chalkboard. I have been watching Michel van Biezen’s KZbin program. He manages his “chalkboard” by moving from right to left rather than than left to right when writing on the board. This allows the viewer to always be able to see the earlier writing because the writer to moving to blank board on left. Watch one of his programs and you see what I mean. Anyway, you have a great program and I really like your teaching style.

[@user-bt5xy4dj6e]

in a palindrome equation, if r is a root, then 1/r is also a root, this helps a lot, specially when using vieta's formula

[@thisrandomdude_]

this is so fascinating!! your smile is infectious and you're fantastic at explaining concepts too. what more could you possibly need from a teacher????

[@PrimeNewtons]

Thank you

[@ak-47520]

Sir, how is your handwriting that neat. I can not get over that

[@PrimeNewtons]

So nice of you

[@jadenredd]

i love your quotes at the end. can't wait to see your channel get even bigger!

[@Gremriel]

They're bible verses, and I don't understand why he is trying to inject religion into math videos.

[@jadenredd]

it’s his channel, he can do whatever he wants w it at the end of the day lmao

[@BartBuzz]

Another interesting and intriguing video.

[@temporarytemporary-fh2df]

dude i like the way your mind works...

[@Bedoroski]

Your videos are really memorisable. Plz keep it up

[@NadiehFan]

Of course I'm familiar with the method you demonstrate for solving palindromic equations, but for quartic palindromics I prefer completing the square twice, which is closely related to Ferrari's method for solving quartics in general. This does not require a substitution and has none of the drawbacks of other methods you mention at the beginning of the video. Here is how I would solve your equation: 2x⁴ − 13x³ + 24x² − 13x + 2 = 0 4x⁴ − 26x³ + 48x² − 26x + 4 = 0 (2x² + 2)² − 13x(2x² + 2) + 40x² = 0 (2x² + 2 − ¹³⁄₂x)² − ¹⁶⁹⁄₄x² + ¹⁶⁰⁄₄x² = 0 (2x² − ¹³⁄₂x + 2)² − ⁹⁄₄x² = 0 (2x² − ¹³⁄₂x + 2)² − (³⁄₂x)² = 0 (2x² − ¹³⁄₂x + 2 + ³⁄₂x)(2x² − ¹³⁄₂x + 2 − ³⁄₂x) = 0 (2x² − 5x + 2)(2x² − 8x + 2) = 0 2x² − 5x + 2 = 0 ⋁ 2x² − 8x + 2 = 0 2x² − 4x − x + 2 = 0 ⋁ x² − 4x + 1 = 0 (2x − 1)(x − 2) = 0 ⋁ (x − 2)² = 3 2x − 1 = 0 ⋁ x − 2 = 0 ⋁ x − 2 = √3 ⋁ x − 2 = −√3 x = ¹⁄₂ ⋁ x = 2 ⋁ x = 2 + √3 ⋁ x = 2 − √3

[@PrimeNewtons]

This is a gem. I love this!

[@joaomane4831]

Wow, I'm just amazed. I'm definitely trying out this technique. Thank you, sir.

[@holyshit922]

It is worth doing this for factoring polynomials of degree 6 or degree 8

[@joaomane4831]

@@holyshit922 But this method is only applicable if the coefficients of the equation are palindrome right

[@holyshit922]

@@joaomane4831 Method showed by Newton, yes only for palindromic coefficiients but for odd degree polynomial equation it is easy to guess one solution and reduce to one degree lower equation which is even With this method you reduce degree of eqation from 2n do n so for degreee 10 and greater you will not be able to solve it without some non-elementary functions like hypergeometric ones For degree 8 you can reduce it to degree 4 equation which can be solved with radicals For degree 6 you can reduce it to degree 3 equation which can be solved with radicals (There is one case of third degree equation which needs complex radicals which can be expressed in terms of trigonometric functions) For degree 4 it is possible to use method shown by Newton but as I showed in the comment there is no benefits over the method working for general quartic because as you can see resolvent cubic is partially factored and it is easy to find its roots

[@sleep1ngM0nsteR]

Amazing sir

[@nicolascamargo8339]

Hay que destacar que el método con una modificación pequeña solo es aplicable para los que tengan un número impar de términos con las potencias consecutivas de la variable

[@OctalMicrobe71s]

This is quite similar to a MAT question from this year X^7+2x^6-5x^5-7x^4+7x^3+5x^2-2x-1=0

[@PrimeNewtons]

I'll give it a try

[@quzpolkas]

Nice problem! Uses the same concepts, but takes them a step further. Pair up monomials with opposite sign coefficients (X^7 with -1, 2X^6 with -2x, e.t.c). Factor out the highest power of X in each pair. Use the fact that X^N-1=(X-1)(1+X+X^2+...+X^(N-1)) for positive integers N (geometric progression sum formula). Divide everything by (X-1)*X^3 (X=1 is a root, X=0 is not). And finally, show that, using the same substitution as in the video, X^3 + 1/X^3 = T^3 - 3T. After that the equation can be transformed into a cubic in terms of T = X + 1/X, which can be easily factored by grouping. That's how I solved it. Answers: 1, (-3±sqrt(5))/2, (1±sqrt(5))/2, (-1±sqrt(5))/2 [sqrt(5) means square root of 5]

[@lawrencejelsma8118]

Is that an advanced mathematics course in Numerical Analysis problem using Newton's Iterations to solve polynomial sized order 6 reduced to a cubic polynomial solution but still iterative with a t= x + 1/x substitution. I don't have a graphing calculator so I don't know if the roots other than the obvious x=1 root have to numerical analysis studies in Newton's Iterations algorithm to find the other 6 roots ... Therefore a required computational solution using a graphing calculator problem? You can't solve this problem other than for x = 1 and leaving a 6 power polynomial for solutions that don't have integer further solutions.

[@lawrencejelsma8118]

@@quzpolkas ... I see the Golden Ratio solutions and Golden Ration squared solution by your answers. I was left with up to your X^3 + 1/X^3 = T^3 - 3T the T^3 - 2T -15 = 0 cubic polynomial to solve. Unless we know Golden Ratio and Golden Ratio squared are solutions which they apparently are then Newton's Iterative Root finder algorithm may be skipped in Numerical Analysis mathematics for graphing calculations.

[@quzpolkas]

@@lawrencejelsma8118 After all the factoring and substitutions I got this equation for T: T^3 + 3 * T^2 - 5*T - 15 = 0 which is factorable by grouping: (T + 3)*(T^2 - 5) = 0.

[@nicolascamargo8339]

Muy interesante el video

[@beapaul4453]

Can you post a video on pentation please, that would be great to see.

[@ganesanthenappan5366]

Your Videos are Quite Interesting. Please upload some videos on Topology

[@lukaskamin755]

I never heard palindrome applied to an equation, while I know the term "symmetrical equation". Is it accepted name used in schools ( of which country?)? Very curious

[@lagomoof]

If I remember correctly (and I might not) "symmetrical" usually refers to a function on multiple variables doing the same thing with each variable. e.g. f(x,y) = √(x² + y² + 4) is symmetrical in x and y because everything done to x is also done to y, or otherwise, swapping the input values doesn't change the result. Compare with a g(x,y) = √(x²/2 + y²/3 + 4) where x is not treated the same as y, thus g is not symmetrical. Or consider a h(x,y,z) = √(x²/2 + y²/3 + z²/2) which is symmetrical with respect to x and z, but not with the other pairings (or all three). Anyway, if that definition is right, it might explain why the name symmetrical might not be used for polynomials with palindromic coefficients. On the other hand, it wouldn't be the first time a word is used in multiple ways in mathematics!

[@allozovsky]

"A self-reciprocal polynomial is also called _palindromic_ because its coefficients, when the polynomial is written in the order of ascending or descending powers, form a palindrome" (from wiki).

[@lukaskamin755]

@@lagomoof symmetrical is a very popular word in mathematics, I believe it might be used in various branches of the science with different meanings. I studyied math in Russian/UKrainian and we used such term

[@lukaskamin755]

@@allozovsky never heard of reciprocal polinomials, not to mention self-reciprocal , nice to know, we never used them

[@lukaskamin755]

@@allozovsky Funny if you ttranslate the same article in Russian, you'll find the name symmetrical there in the meaning of palindromic polynomial, but the word coreesponding to palindromic was also there, it's a surprise for me😄

[@hafizusamabhutta]

In our country they're called reciprocal equations in which when x is replaced by 1/x they remain same.

[@kiracaasi]

Hello,. Thank you for this video but i just wanna ask if how can i answer this problem about integrals of (2sin^-1 (x/2)) / sqrt of 4-x^2 about given yielding inverse trigonometric function po siya.

[@MegaNerd117]

Beautiful palindrome equation. I love the Bible verse at the end. Proverbs 13:20 Amen.

[@giorgiobarchiesi5003]

In Italy license plates are composed of two letters, three digits, two letters. So, it is possible that a license plate is palindrome. What is the probability?

[@Simpson17866]

Let's start with "AB 121 BA" as the general form. What are the odds that the second A is the same as the first? What are the odds that the second B is the same as the first? What are the odds that the second 1 is the same as the first?

[@YakobWakjera]

Sir ..... I have better understanding of math ... Since I start following you a week ago.....

[@blodinbramo1938]

i don't understand why at the end 4+ -2√3 all over 2 is 2+ - √3, because like that, to me, u devide the denominator with both 4 and 2 on the numerator, tho i don't think u can do it, or can u?, anyway love from italy, u are really good on teaching this things

[@adw1z]

u can because 4 + 2sqrt(3) = 2[2+sqrt(3)], and the 2's cancel

[@blodinbramo1938]

@@adw1z got it, tysm

[@ashutoshparihar7086]

You can indeed divide both terms by two. As you can take 2 as common factor from numerator and then cancel it out with the denominator. However it doesn't work the other way around in case the numerator and the denominator are switched.

[@HaliPuppeh]

Does this method work with palindromic equations of higher degree or would you divide by something different? Also, Taco Cat spelled backwards is Taco Cat

[@lawrencejelsma8118]

At 11:09 mark you erroneously criss crossed your variables x and t but continued on solving for x in the remaining solution and your brain readjusted the criss crossing error of x being 2 or x being 1/2 not t as you did! 😲😂

[@JourneyThroughMath]

I just did one of these on my own channel. Yours is better though!

[@allozovsky]

0:22 When Adam first met Eve, he said to her: - _Madam, I'm Adam._ And she answered: - _Eve._ This was the first palindromic dialogue on Earth.

[@temporarytemporary-fh2df]

she sould be single at that time but watever.

[@Maths_3.1415]

Lol, just believe in fictional stories. Are you seriously trying to claim that some guy named Adam discovered palindromic numbers? Nice one.

[@sciphyskyguy4337]

Never odd or even

[@holyshit922]

Quartic polynomial in this form is easy to factor using general approach ax^4+bx^3+cx^2+bx+a = 0 x^4+b/a*x^3+c/a*x^2+b/a*x+1 = 0 (x^4+b/a*x^3)-(-c/a*x^2-b/a*x-1) = 0 (x^4+2*b/(2a)*x^3+b^2/(4a^2)x^2) - (-c/a*x^2-b/a*x-1) = 0 (x^2+b/(2a)x)^2 - ((b^2/(4a^2)-c/a)*x^2-b/a*x-1) = 0 (x^2+b/(2a)x+y/2)^2 - ((y+b^2/(4a^2)-c/a)*x^2+(b/(2a)*y-b/a)*x+y^2/4-1) = 0 4*(y^2/4-1)(y+b^2/(4a^2)-c/a) - (b/(2a)*y-b/a)^2 = 0 (y^2 - 4) (y+b^2/(4a^2)-c/a) - b^2/(4a^2)(y-2)^2 = 0 (y - 2)(y+2)(y +b^2/(4a^2) - c/a)-b^2/(4a^2)(y-2)(y-2) = 0 (y-2)((y+2)(y +b^2/(4a^2) - c/a)-b^2/(4a^2)(y-2)) = 0 y = 2 will give us difference of two squares but it may sometimes leads us to the complex coefficient factorization Lets assume that y=2 is a good choice (x^2+b/(2a)x+1)^2 - (2+b^2/(4a^2)-c/a)x^2 = 0 (x^2+b/(2a)x+1)^2 - (2+(b^2-4ac)/(4a^2))x^2 = 0 (x^2+b/(2a)x+1)^2 - (8a^2+b^2-4ac)/(4a^2)*x^2 = 0 (x^2+b/(2a)x+1)^2 - (sqrt(8a^2+b^2-4ac)/(2a)*x)^2 = 0 a(x^2+(b-sqrt(8a^2+b^2-4ac))/(2a)x+1)(x^2+(b+sqrt(8a^2+b^2-4ac))/(2a)x+1) = 0

[@robertveith6383]

That is not "easy to factor!" That is messy as hell, and you used hundreds of characters to write it out.

[@lukaskamin755]

Let me correct you, not fir any equation, but rather for any power equation( I'm not sure of the correct name of the equations of "polynomial=0"-type). Anti-example cos x --1 =0 or exp(x)-1=0

[@allozovsky]

What exactly was the statement?

[@allozovsky]

Did you mean this one: "x cannot be equal to zero for any equation that has a constant term"?

[@allozovsky]

Yes, this statement concerns _algebraic_ equations (aka _polynomial_ equations), where the LHS can be expressed as a *finite* sum of terms, each of which is a constant multiplied by a variable raised to a non-negative integer power. The cosine function and the exponential function may be expressed only as an *infinite* sum of monimial terms.

[@allozovsky]

And they both have a "+1" constant term, so when you subtract 1, the constant term turns to zero.

[@lukaskamin755]

@@allozovskyyes

[@MichaelAdjei-up2ce]

First ❤🎉