Hello everyone! The book that I told you about in the video is a good resource but that's not where this problem comes from. This problem comes from "Solving Problems in Algebra and Trigonometry" by V. Litvinenko. (correct book name in description as well!) I think this is a very interesting problem (Why would I pick it otherwise, right? :) I hope you enjoy the video. Thoughts, ideas? 😁😉😁
@resilientcerebrum
3 жыл бұрын
Can I send pdf of the book from which you can take questions from? The questions are really interesting!!!
@SyberMath
3 жыл бұрын
@@resilientcerebrum Sure. Email me at SyberMathBiz@gmail.com
@MINEXKILLER
3 жыл бұрын
another way would be to say if a quadratic has one sol. This means it's a perfect square and 9 is 3 squared you multiply by 2 and this should be the coefficient of x TL;DR : a-6=2×3 because of perfect squares
@SyberMath
3 жыл бұрын
👍
@MINEXKILLER
3 жыл бұрын
@@SyberMath TL;DR I guess. anyway still one of my favorite KZitemrs
@DrQuatsch
3 жыл бұрын
What I found most interesting about this, is that indeed for a = 12 you get one solution (x = 3). But that when you can take any negative value for a, you'll get an x in the range (-3, 0), because obviously x has to be negative as well to get an answer from log(ax), but log(x+3) also has to give an answer, which means for negative x it has to be between -3 and 0.
@tmacchant
2 жыл бұрын
I consider intersection, f(x)=(x+3)^2 (x>-3) and g(x)=ax (x>0 when a>0, x
@mohamedihab4599
3 жыл бұрын
I really appreciate how you introduce new types of questions and new ideas gradually, thank you, SyberMath. (Also, great problem, I find logarithmic equations very enjoyable).
@SyberMath
3 жыл бұрын
You're very welcome! Thank you! 💖
@platformofscience9790
3 жыл бұрын
It's amazing that in youtube we have this guy that helps us in these equations. Infinite thank you
@SyberMath
3 жыл бұрын
My pleasure! 💖
@Vladimir_Pavlov
3 жыл бұрын
Since, there must be a*x >0, then in the first system of inequalities for the roots x1 and x2, we should write 0>x1>-3, and since it turns out that a
@elias69420
2 жыл бұрын
(x + 3)^2 = ax x^2 + 6x + 9 = ax x^2 + (6 - a)x + 9 = 0 For this quadratic equation to have one root, its discriminant must be equal to 0 (in other words, there is a double root) The discriminant is (6 - a)^2 - 4*1*9 = 36 - 12a + a^2 - 36 = a^2 - 12a = 0 So a = 0 or a = 12, but if a = 0, we have 2*log(x + 3) = log(0) which is undefined. So a = 12, and x is 3
@yoavmor9002
2 жыл бұрын
Many things overlooked in this video. Namely, checking that the value of x when the discrimination is 0 is actually in the equation's domain, and accounting for negative values for a in the other cases
@SSGranor
3 жыл бұрын
The second case that you write at 7:38 is unnecessary. The way that you've defined x_1 and x_2 they will always satisfy x_1>x_2 whenever they are real, so it's not possible to simultaneously have x_1=y for any number y.
@SyberMath
3 жыл бұрын
I know but it's kind of fun to look at! 😁
@ramaprasadghosh717
3 жыл бұрын
(x+3)^2 = ax or x*x +2 (3 -a/2 ) x + (3-a/2)^2 = (3-a/2)^2 -9 In case there exists only one root, the RHS would be zero i.e. a = 12 and x = a/2 -3 = 3
@user-mx8sj1nc6v
3 жыл бұрын
It is nice to "feel" what is going on. I drew y=2*log(x+3). x=-3 asimptotus. Now we see clearly that if a12 we have 2 solutions based on delta. If we take a=1000 just to "feel", then log(ax) will go up "fast" and eventually intersect again with y.
@SyberMath
3 жыл бұрын
Absolutely!
@ananintesarbinfaiz863
3 жыл бұрын
Great problem! After a hectic physics exam this was quite a refresher 😅
@SyberMath
3 жыл бұрын
Glad it helped!
@242math
3 жыл бұрын
good to see you tackling these interesting problems, great job Syber
@SyberMath
3 жыл бұрын
Thanks!
@TheAlcolade
3 жыл бұрын
If a quadratic and a line have only one point of intersection, they will be tangent at that point. The "log" of these curves will also be tangent at that point. Taking the derivative on both sides: 2 / (x+3) = 1 / x has one solution x = 3. Using x = 3 in original equation simplifies to log 12 = log a. ...does this need more justification anywhere?
@SyberMath
3 жыл бұрын
Wow! That's interesting. I do not think so
@davidseed2939
3 жыл бұрын
at 2:13 you have x^2 -(a-6)x +9. If there is one solution then the equation will the form of a perfect square. and by vieta's formula the root product is 9 so each root is +-3, -3 would give log(0) so thats not a solution. So the double root is x=3 or (x-3)^2 =0 or x^2 -6x +9=0 so -6x =-(a-6)x so a=12
@SyberMath
3 жыл бұрын
Good!
@BiscuitZombies
3 жыл бұрын
I got the a
@tarunmnair
3 жыл бұрын
Nice video, to find that single feasible solution, when a=12, x=3 and when a=0, x=-3 right ?
@SyberMath
3 жыл бұрын
a=0?
@advaykumar9726
3 жыл бұрын
Nice video! Can you do a video on IMo problems?
@SyberMath
3 жыл бұрын
I probably can! 😁
@kivicode
3 жыл бұрын
The funny thing for me is that my school final exam contains similar parametric expressions/systems
@pardeepgarg2640
3 жыл бұрын
Can you do a problem on Girard - Newton identities combined with Vietes Formula for polynomial roots
@kpt123456
3 жыл бұрын
It is obvious that x1 is greater than x2. As in 1st one we are adding a positive value and in 2nd one we are subtracting
@SyberMath
3 жыл бұрын
Sure!
@allmight801
3 жыл бұрын
Can i get a help with log question If log base 5 do 8 = p and log base 5 od 9 = q then calculate log base 5 of 6
@SyberMath
3 жыл бұрын
Sure. Here is my solution: twitter.com/SyberMath/status/1392942966361116672
@allmight801
3 жыл бұрын
@@SyberMath thank you very much this helps a lot.
@rafael7696
3 жыл бұрын
Nice. Anyway I think that in case 2, should be proven (but it is obvious) that for a
I agree. Handpicked for you from among thousands! 😁
@holyshit922
3 жыл бұрын
(x+3)^2=ax x+3 > 0 ax > 0 ?
@vivekbhutada3049
3 жыл бұрын
Nice one
@SyberMath
3 жыл бұрын
Thanks 🔥
@pardeepgarg2640
3 жыл бұрын
Are you ready to solve Reimann hypothesis
@SyberMath
3 жыл бұрын
Good question! 😁
@aahaanchawla5393
3 жыл бұрын
you mean attempt and fail to solve the Reimann hypothesis?
@pardeepgarg2640
3 жыл бұрын
@@aahaanchawla5393 Nope , really trying to solve ang get success , even if he failed , nothing bad in coz many great mathematician Even fail to prove and so this is a very famous problem since 200yrs I even tried but failed ;/
@pardeepgarg2640
3 жыл бұрын
THE METHOD I USE IS SO COMPLEX AND THEN I FOUND THAT I STARTED FROM WRONG STATEMENT THO Xd
@manojsurya1005
3 жыл бұрын
Good one
@SyberMath
3 жыл бұрын
Thank you!
@daviscuevaquinto8882
Жыл бұрын
a mi me salió a =16 y x=1
@necro5379
2 жыл бұрын
Easy one
@chessdev5320
3 жыл бұрын
find the integral of x²/(x+1) and then confirm your answer with Wolfram Alpha, you will find something Weird in the answer they have given.
@roy2615
3 жыл бұрын
I got x²/2 - x + ln( |x + 1| ) + C is the weird think you're talking about the 2/3?
@chessdev5320
3 жыл бұрын
Exactly. how does that -3/2 comes?
@PS-mh8ts
3 жыл бұрын
@@chessdev5320 It's solved by a computer. We do not know how a computer arrives at an integral because it's a symbolic computation (as opposed to numerical). A computer might not follow the exact method that a human would. Even when we humans solve an integral, sometimes it so happens that we get two or more constants, which we then combine into one constant as we write our final result. The answer given by Wolfram may not be the same answer that you get, but it's still correct in this case despite the introduction of a constant -3/2.
@chessdev5320
3 жыл бұрын
@@PS-mh8ts it has happened to me for the first time that my answer is different from 'Computer'. also, I don't know about symbolic or numerical Computation. can you plz elucidate?
@PS-mh8ts
3 жыл бұрын
@@chessdev5320 Algebra is mainly symbol manipulation. Look at your ordinary calculator. It can readily give you (3+1)² whereas it won't be able to tell you that (x+1)² = x²+2x+1, even though it's common knowledge. Deriving the value of (3+1)² is an example of numerical computation (where computers excel) whereas the latter (deriving x²+2x+1 from (x+1)²) is an example of symbolic computation. Special algorithms are required for symbolic computation, and it's far more complex than numerical computation. Another example would be that of definite integral. If you're asked to find the definite integral of cos(θ) between bounds 1 and 2, for example, you would readily write it as (sin 2) - (sin 1). But a calculator might give you the result as 0.0678. It might use a method such as Simpson's rule to arrive at this answer, without ever having to find the integral of cos(θ) at all. This is the difference between symbolic computation and numerical computation. If you're familiar with computer programming, you might know that languages such as Lisp, Scheme, Prolog, etc., had to be invented to facilitate symbolic computation whereas traditional languages (such as Fortran, Pascal, C, C++, etc.,) are mainly geared toward numerical computation. Hope I've answered your question.
@nadanaobrow9672
3 жыл бұрын
Interessante , legal que alem de matemática também posso praticar um pouco de inglês no seu canal xd . Enfim acho que você não vai entender nada, mas ótimo video bro hehehe
@SyberMath
3 жыл бұрын
Obrigado pelas amáveis palavras. Fico feliz em ouvir isso!
@anshumanagrawal346
3 жыл бұрын
a=12 is my answer without solving or watching the video
@anshumanagrawal346
3 жыл бұрын
I forgot about the extraneous roots🙃
@SyberMath
3 жыл бұрын
😁
@fitrihakim1205
3 жыл бұрын
Do u support Palestine? Btw, great videos!
@SyberMath
3 жыл бұрын
Thank you!
@sergeigrigorev2180
3 жыл бұрын
First :)
@SyberMath
3 жыл бұрын
Good job! 💖
@mathevengers1131
3 жыл бұрын
For the most beautiful equation made by me search MathElite and see his latest video. Video's title is the most beautiful equation.... Due to some problem I am not able write link here.
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