Find all real x

[@fryguy2009]

You have a very clear presentation style. Also, your chalkboard writing is very clean and legible. That’s rare in math instruction. I will continue to enjoy your presentations.

[@Christian_Martel]

Very nice! To find one solution might be easy, but proving this is the only solution is another game. Great job Coach!

[@nanamacapagal8342]

ATTEMPT: At about 5:40 is where my solution diverges from yours. I recognized the form of the equation a^2 - ab + b^2 - a - b + 1 = 0 as the equation of some conic section. I remember how to rotate conic sections, replace a with tcos(theta) - usin(theta) and b with tsin(theta) + ucos(theta). I selected theta = 45 degrees and scaled everything by sqrt(2) so the substitution would become a = t + u, b = t - u Substituting and simplifying leads to: t^2 + 3u^2 - 2t + 1 = 0 (t-1)^2 + 3u^2 = 0 The only way this works is if t = 1 and u = 0. Otherwise the left hand side will be bigger than 0. t = 1 and u = 0 means a = 1 and b = 1. thus, 1 = 2^x and 1 = 3^x. Clearly x = 0 is the only possible solution.

[@bobajaj4224]

yup, ellipse since the discriminant is

[@invisiblelemur]

Love that argument that three positive terms summing to zero means they're all zero.

[@Heater-v1.0.0]

Wonderful. You must be the coolest maths presenter on KZbin!

[@user-dq6jf9ru9e]

I solved this in a "brutal" way using complex numbers)) Starting with the equation at (5:14) a² - (b+1)a + (b²-b+1) = 0 and solving it we come to a=½(b+1 ±i√(3)(b-1)) Since a in left side is a real number then the imaginary part in the right side must be equal to zero. So b=1 and a=1.

[@daniorugbani5914]

I really enjoyed the way you solve math problems. You’re simply one of the best math teachers I have ever seen. Keep up the good work.

[@PrimeNewtons]

Thank you

[@jflamingo26]

This is very nice :) I used a similar strategy, but instead used the quadratic formula to write a in terms of b. This produces a negative determinant for all cases except b = 1, meaning that the only way we get a real solution is if b = 1 and hence x = 0.

[@rakeshrampal1587]

Very nice way to complete square terms and making a simple conclusion.

[@eduardoyamakawa1754]

I loved how you presented the solution, especially the part in which you multiplied by 2. Just something I noticed that would make it slightly shorter is that you could have broken down the +2 as 1+1 and you would have those perfect squares.

[@krisbrandenberger544]

I noticed that same thing.👍

[@FIST_3807]

YEP , I ALSO NOTICED THAT

[@alhazen8736]

Excellent approach to solve this fun problem!

[@manuelgonzales2570]

Excellent video. Thank you!

[@N00MAK_CA_Finalist]

Amazing😮 the way you explained sir was really really a practical approach and loved your way of teaching

[@ناصريناصر-س4ب]

The equation can also be written in the form (2a-b-1)²+3(b-1)²=0

[@PrimeNewtons]

I didn't see that

[@binaprasad6477]

Pls explain I don't get it

[@TEACHINGMATHEMATICSbyMrNKundu]

​@@binaprasad6477 It is very easy & clear.... just multiply 4 instead of 2 both sides, you get the result

[@venkatkrishnankoluman870]

I love your passion for maths. I am 67 years old and keep doing math sums to keep my brain ticking and pass my time. Your channel is a great motivator.

[@iankr]

Many thanks. I love your channel. Not just for the maths, but because I find your voice and delivery soothing and relaxing.

[@barence321]

I'm really enjoying your videos, and I loved your simple algebraic proof. Very nice!

[@Jeremy-i1d]

Beautiful use of CTS to produce the sum of three squares = 0 to narrow possible solutions down to just x=0 at a stroke - love it ❤

[@Manas_12]

I’m genuinely amazed by the way you solved this problem algebraically! It’s fascinating to see how different approaches can lead to the same solution. I tackled this problem by graphing it, which gave me a clear visual representation, but your algebraic method is both elegant and insightful. Great work!

[@satyapalsingh4429]

Your method of solving the question is amazing .I like your method of expression .You are so cool .Keep it up !!!

[@kaenemorerinyane9392]

My thought process was that imagining the graph of that function. It tends to negative infinity as x gets bigger and tends to 0 as x goes to negative infinity. After finding the solution of x=0, an extremely close number less than 0 gives you a value less than 1 as with a value slightly greater than 0 I conclude that the graph has a maximum at x=0 i.e the graph never meets the line y=1 ever again so x=0 is the only real solution

[@ronaldjensen2948]

Taking the derivative of the function gives ln(2)2^x+ln(3)3^x-2ln(2)4^x+(ln(2)+ln(3))6^x-2ln(3)9^x which is positive for x < 0 and negative for x > 0. This shows that x = 0 is the absolute maximum and therefore the only real solution.

[@baidonchandipo2804]

I love the level you are in terms of dealing with numbers

[@Grecks75]

Before watching: (1) Let a = 2^x and b = 3^x. The equation can then be written as: (a + b) + ab - (a^2 + b^2) = 1. (2) a and b can both be considered exponential functions of x with bases greater than 1. Therefore they both are either > 1, = 1, or < 1, both at the same time (same x). (3) Now let's consider the product (a - 1)(b - 1). We have: (a - 1)(b - 1) = ab - (a + b) + 1 = 2ab - (a^2 + b^2) = -(a - b)^2, where the second equality is according to the equation (1). (4) According to (2) we know that (a - 1)(b - 1) >= 0. We also have -(a - b)^2

[@Grecks75]

What a beautiful problem! ❤

[@eastonrocket兀]

Bro is truing to take away his views from his videos by cheating in the comments 🗣️🔥

[@kylecow1930]

a+b+ab-a2-b2=1 a+b-ab-1=a2+b2-2ab (a-1)(1-b)=(a-b)2 (a-1)(b-1)

[@orni8519]

Thanks for very interesting math problem, and I really love your style of speaking. Love from Russia

[@AxillaryPower2]

Your handwriting on the chalk board is the cleanest I've ever seen!

[@TheMariusMan]

Wonderful equation. It took me 10 minutes and I found another way, but yours seems more adequate. Great video!

[@ManojkantSamal]

Respected Sir, Good evening....You're the master key of mathematics......

[@MichaelGrantPhD]

I love this content, but at first, I found it to go too slow. And then I realized that if I play it back at 1.5x it's absolutely perfect. If you feel the same tension, try it!

[@ishantagarwal1082]

Here's what I did: Let 2^x=t Let 3^x=k Treating the given equation as a quadratic ak²+bk+c=0, the discriminant, D = √{-3(t-1)²} This implies the only real solution is when t=1. The rest is elementary.

[@jo555444]

The beauty of Algebra. Wonderful solution.

[@shlokthosar8094]

Beautiful problem sir I love your channel..❤❤❤❤

[@fplancke3336]

That was a beautiful problem with an unexpected solution

[@하태욱-b2x]

Many thanks for your passion on math. From a+b-a^2+ab-b^2=0, where a,b are real number, apply quadratic formula about a, then the expression will show complex solution, then b must be 1 to be real a, which also confirms a=1, then x=0

[@marielcanela7228]

This was so crazyyy!!! Loved it

[@nikolayguzman331]

Excellent as usually!

[@math9333]

Hi man I just saw your channel, and I subscribed immediately I was wondering if you could do some videos on evaluating the partial sum for some combined geometric and arithmetic series like for example 1/1 + 2/2 + 3/4 + 4/8 + 5 /16 ..... and possibly shows a cool trick as you always do

[@Symyouner]

Love from Kashmir ❤

[@LITHICKROSHANMS-gw2lx]

I enjoy mathematics very much and solution is too great. Sir can you solve this lim(infinite tower of 'x') (x→0)

[@amiraliazimi6355]

What a beautiful equation! ❤

[@kugathasasarmasivakumar5083]

First, I found trivial case x=0, then I proved x

[@subhashpatil2833]

Sir your videos are excellent.❤

[@matematikgokseldir]

please don't lose this energy that you have, teacher!

[@peterjansen4826]

A simple answer for a problem with some tricky rewriting. It shows how much mathematics education fails, you don't practice this rewriting enough at a school, they keep it limited to just solving the same standard 2nd-degree equations all the time.

[@alexandermorozov2248]

The solution is great! But additional questions arise :) Does this equation have complex solutions, how many of them and how to find them?

[@Simpson17866]

Only if there are complex values of "x" that give 2^x = 3^x = 1 EDIT: No, wait, you're right - if a-1, b-1, and a-b are complex, then they don't each need to be 0 for the squares to add up to 0. if a = A+Bi and b = C+Di, then we get (A+Bi-1)^2 + (C+Di-1)^2 + (A+Bi-C-Di)^2 = 0

[@m.h.6470]

Solution: 2^x + 3^x - 2^x * 2^x + 2^x * 3^x - 3^x * 3^x = 1 2^x = a 3^x = b a + b - a² + ab - b² = 1 a = b = 1 is an obvious solution, leading to 2^x = 1 = 2^0, so therefore x = 0 a + b - a² + ab - b² = 1 a - a² + ab + b - b² = 1 a - a² + b(a + 1) - b² = 1 |*-1 -a + a² + b(-a - 1) + b² = -1 |+a -a² b(-a - 1) + b² = a - a² - 1 |+((-a - 1)/2)² (completing the square on the left side) ((-a - 1)/2)² + 2b(-a - 1)/2 + b² = ((-a - 1)/2)² + a - a² - 1 ((-a - 1)/2 + b)² = (-a - 1)²/4 + a - a² - 1 ((-a - 1)/2 + b)² = ((-a - 1)² + 4a - 4a² - 4)/4 ((-a - 1)/2 + b)² = (a² + 2a + 1 + 4a - 4a² - 4)/4 ((-a - 1)/2 + b)² = (-3a² + 6a - 3)/4 ((-a - 1)/2 + b)² = 3(-a² + 2a - 1)/4 |√ (-a - 1)/2 + b = √3√(-a² + 2a - 1)/2 |-(-a - 1)/2 b = (√3√(-a² + 2a - 1) - (-a - 1))/2 b = (√3√(-a² + 2a - 1) + a + 1)/2 So for all values a, such as -a² + 2a - 1 > 0, there are real values for b -a² + 2a - 1 > 0 |*-1 a² - 2a + 1 < 0 (a - 1)² < 0 This is a contradiction, as a square can't be negative So in conclusion, the only real solution - in fact the ONLY solution - is x = 0

[@DavidSantiagoPoloSilveraBio]

Math problem-solving is exciting. Your channel is addictive. And your are damn HOT 🔥, professor (sorry for this harassment, but I just had to let it flow). Have a nice day, professor!

[@PrimeNewtons]

Thank you.

[@محمدالنجفي-ظ1ه]

My god that's so impressive performance you just made it simple

[@dirklutz2818]

Beautiful!

[@raimis0u]

Keep up the enthusiasm ❤

[@MathTutor1]

Very nice!

[@internallyinteral]

Thank you for keeping math cool 😎

[@bwcbiz]

weLL-played, sir.

[@AshenWijethilaka-j4i]

Sir, could you please make a video on Clairaut's equation and Ricatti equation ?

[@andrasferencz7948]

This one is brilliant!

[@dondisco6399]

Very nice trick!❤️

[@alohamark3025]

Perfecto.

[@PriyanshuShil_]

Great solution

[@moonwatcher2001]

Awesome ❤

[@nothingbutmathproofs7150]

Amazing!

[@tmrapper6378]

Beautiful solution

[@Chestno58]

Thank you Sir...

[@GreenMeansGOF]

Can you share the other way you tried to factor? I promise I can handle it.

[@nyyakko]

holy hell this one is so cool!

[@sakshamsingh7325]

We can easily do by odd or even…for any other power of x except zero left hand side is even but right hand side is odd…only when x is zero is the equation equal

[@Maths__phyics]

Hello, İ discovered phyics formula, And I wanna to shoow this formula to America, I wanna help from you, please help me.❤❤❤

[@nicolasb11]

Wow Prime Newtons bravo 😂🎉

[@jimmeade2976]

It's been a long time since I did math, but I would have approached this by using natural logarithms. Since a^x=xln(a), you get x(ln(2)+ln(3)-ln(4)+ln(6)-ln(9)=1 or x(.693+1.098-1.368+1.791-2.197)=1 or x(0.017)=1 or x=1/0.017=58.82. That's a completely different answer. What did I do wrong?

[@futureiskey6049]

Remember that (a+b+c..)^x ≠ a^x + b^x ….

[@jimmeade2976]

@@futureiskey6049 I know that, and it's not what I am doing. I'm converting each item in the list to its natural logarithm, then factoring out the x, leaving a sum of natural logarithms.

[@futureiskey6049]

@@jimmeade2976 what you did is ln(2^x+3^x…). You can’t pull the x out of that logarithm. To pull the logarithm out, the inside argument must be (a +b + c)^x not (a^x) +b^x etc

[@jimmeade2976]

@@futureiskey6049 If a^x=x*ln(a), why can't I do what I did? I'm taking the natural log of each term, not the sum of the terms

[@futureiskey6049]

@@jimmeade2976 you can’t take the natural log of each term. When you apply a function to both sides of an equation, the function applies to everything. So if I have a+b+c = 1 then, ln(a+b+c) = ln(1) Not ln(a) + ln(b) + ln(c) = ln(1) You can’t distribute a function over each term.

[@nebulousquartz2712]

Thank you ❤

[@rubinkatz9850]

you are amazing

[@KaivalyaChess]

sir you are the best

[@WillamGorsuch]

1:25 The answer, when x = 1, is -2, not 1.

[@doowadiwadi]

I could be wrong but seems to me that when entering x=0 in the original equation one gets 2+3-4+6-9 = 1=> -2 = 1. Which is not true.

[@joeschmo622]

✨Magic!✨

[@Modou_Desu]

Morning digest

[@hervesergegbeto3352]

Merci professeur

[@surendrakverma555]

Thanks Sir

[@swiftswallow3879]

Mathematics for breakfast 🥣

[@alainrogez8485]

Mazette !

[@arian-i5c]

But this can be easily solvable by logarithms When use log in both sides of equation we can easily get x=0

[@RyanLewis-Johnson-wq6xs]

x=0 final answer

[@AmanPrajapati-o2r]

Any one from India ❤ ❤ ❤ AttEnDaNcE HeRe 😊😊😊😊😊❤. ❤ .❤

[@SidneiMV]

2ˣ = a 3ˣ = b a + b - a² + ab - b² = 1 (a + b) - (a² - ab + b²) = 1 (a + b)² - (a³ + b³) = (a + b) a³ + b³ = (a + b)² - (a + b) a³ + b³ = (a + b)³ - 3ab(a + b) a³ + b³ = (a + b)² - (a + b) (a + b)² - 3ab = (a + b) - 1 (a + b)² - (a + b) + 1 - 3ab = 0 (a + b) = [1 ± √(12ab - 3)]/2 ..... I don't know how to continue .....

[@RyanLewis-Johnson-wq6xs]

2^x+3^x-4^x+6^x-9^x=1 x=0

[@kianushmaleki]

❤️

[@justekiara1953]

Math is easy with you as a teacher.

[@ToanPham-wr7xe]

😮

[@danielbranscombe6662]

I manipulated the equation into 2^x+3^x+6^x=1+4^x+9^x both sides are strictly increasing. Thus there can be at most 1 solution. Since x=0 is a solution it is the only solution.

[@jay_sensz]

That's not a valid argument. Two strictly increasing functions can intersect an arbitrary number of times, e.g. sin(x)+2x = cos(x)+2x.

[@niloneto1608]

​@@jay_senszExcept if all terms are increasing individually, then it's valid. However, op's original argument fails when x

[@ruthlee9819]

2^x+3^x+6^x=1+4^x+9^x 2^x+3^x+6^x=2^2x+3^2x+1^2 6^X=1 ?

[@ruthlee9819]

X=0

[@RAWLEDGE]

There should be (-1) instead of (+1)

[@pyprem]

He brings everything to the RHS of the line above (as he says at around 4:38) but then also 'sneakily' switches the sides when writing the second last line down, so +1 is correct (since it stays on the RHS which he just writes down first). Or alternatively, you could multiply the line above with -1 and then bring the -1 from the RHS to the LHS.

[@RAWLEDGE]

​@@pypremThanks Doubt cleared

[@rasikajayathilaka3516]

X = 0

[@aidensharp6515]

Uh, if you take the equation modulo two, you get only the 3^x + 9^x left which becomes 1+1 = 0 modulo 2. But this isn't 1 so x = 0.

[@absolutezero9874]

Sadly no response from you

[@emyrronain6983]

There should be a second logarithmic solution that includes imaginary number Right?

[@xinpingdonohoe3978]

There should be so, yes. The conclusion that each term must be equal to zero is because he wanted them to be real quantities, so their squares would be nonnegative. Complex numbers have no such issues. Remember that 2^x and 3^x for complex x will be multivalued, except when x is an integer. And if x is not rational, this will mean countably infinite values. There will definitely be some, possibly a finite amount or possibly an infinite amount, where a specific branch (determined by the logarithm's branch) makes the equation equal to 1.

[@alipourzand6499]

Actually the solution is exp(2*n*pi*i) ☺

[@WillamGorsuch]

x = 0. It is not hard.

[@baddogproductions9722]

My equation: (a-1)(1-b)=(a-b)² => a-1=a-b and 1-b=a-b, from which a=b=1

[@richardassal7788]

Just looking at the problem it was obvious that x=0. Nothing else would make sense.

[@raymondseligman7003]

You have 4^X = (2^x)^2. Wouldn’t that be? 4x^2?

[@xinpingdonohoe3978]

@@raymondseligman7003 (a^b)^c=a^(bc) (a^b)^c≠a^c b^c The x is in the exponent of the 2, it's not multiplied by it. So we can't have the x suddenly multiplying the 2.

[@-aqua-marine-]

Bruh, just take log on both sides and then we can easily find x=0

[@williamspostoronnim9845]

Гениально изящно!