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.

[@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.

[@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.

[@Heater-v1.0.0]

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

[@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

[@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.

[@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 🗣️🔥

[@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

[@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

[@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 ❤

[@iankr]

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

[@satyapalsingh4429]

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

[@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!

[@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!

[@barence321]

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

[@rakeshrampal1587]

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

[@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.

[@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

[@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

[@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)

[@baidonchandipo2804]

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

[@TheMariusMan]

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

[@nikolayguzman331]

Excellent as usually!

[@rodro3429]

Here’s a beautiful method I found which I think is fascinating. 2^x+3^x-4^x+6^x-9^x=1 Let 2^x=a and 3^x=b then: a+b-a^2-b^2+ab-1=0 Negating both sides: a^2+b^2-a-b-ab+1=0 Factoring: (a-b)^2-(a-1)(b-1)=0 Let the left hand side be Δ, so that: Δ=0 Now consider the following polinomilal: P(x)=1/2(a-1)x^2+(a-b)x+1/2(b-1) Notice Δ is the discriminant of P(x). And since Δ=0, it follows that P(x) has exactly 1 real solution ∀a∈R For this to happen, the quadratic coefficient must be 0. So (a-1)/2=0=>a=1. Analogously it can be found that b=1. a=b=1 so x=0

[@orni8519]

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

[@marielcanela7228]

This was so crazyyy!!! Loved it

[@ManojkantSamal]

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

[@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

[@AshenWijethilaka-j4i]

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

[@amiraliazimi6355]

What a beautiful equation! ❤

[@GreenMeansGOF]

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

[@jo555444]

The beauty of Algebra. Wonderful solution.

[@emilie375]

Well done ! I made the a, b substitution but I didn't found the good factorization

[@fplancke3336]

That was a beautiful problem with an unexpected solution

[@AxillaryPower2]

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

[@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.

[@shlokthosar8094]

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

[@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.

[@Aatif45]

Love from Kashmir ❤

[@matematikgokseldir]

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

[@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.

[@subhashpatil2833]

Sir your videos are excellent.❤

[@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.

[@dirklutz2818]

Beautiful!

[@kugathasasarmasivakumar5083]

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

[@LITHICKROSHANMS-gw2lx]

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

[@mazenzidieh]

Very very nice. Thanks

[@raimis0u]

Keep up the enthusiasm ❤

[@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

[@WillamGorsuch]

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

[@bwcbiz]

weLL-played, sir.

[@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

[@Maths__phyics]

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

[@moonwatcher2001]

Awesome ❤

[@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

[@MathTutor1]

Very nice!

[@silver-ep8wn]

beautiful solution

[@PriyanshuShil_]

Great solution

[@Chestno58]

Thank you Sir...

[@dondisco6399]

Very nice trick!❤️

[@alohamark3025]

Perfecto.

[@andrasferencz7948]

This one is brilliant!

[@tmrapper6378]

Beautiful solution

[@internallyinteral]

Thank you for keeping math cool 😎

[@nebulousquartz2712]

Thank you ❤

[@محمدالنجفي-ظ1ه]

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

[@arian-i5c]

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

[@joeschmo622]

✨Magic!✨

[@nyyakko]

holy hell this one is so cool!

[@nothingbutmathproofs7150]

Amazing!

[@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 .....

[@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.

[@nicolasb11]

Wow Prime Newtons bravo 😂🎉

[@hervesergegbeto3352]

Merci professeur

[@KaivalyaChess]

sir you are the best

[@surendrakverma555]

Thanks Sir

[@Modou_Desu]

Morning digest

[@rubinkatz9850]

you are amazing

[@alainrogez8485]

Mazette !

[@RyanLewis-Johnson-wq6xs]

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

[@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_13875]

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_13875Except if all terms are increasing individually, then it's valid. However, op's original argument fails when x

[@RyanLewis-Johnson-wq6xs]

x=0 final answer

[@swiftswallow3879]

Mathematics for breakfast 🥣

[@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

[@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.

[@justekiara1953]

Math is easy with you as a teacher.

[@kianushmaleki]

❤️

[@AmanPrajapati-o2r]

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

[@ToanPham-wr7xe]

😮

[@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

[@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) ☺

[@rasikajayathilaka3516]

X = 0

[@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.

[@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