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Geendle
Приєднався 13 лис 2023
Geenddle is a mathematics channel with the aim of bringing knowledge to you, covering all possible subjects and helping people from all over the world to have access to the study of this essential tool for the development of science and for our daily lives. It is a channel that aims to serve students, teachers, enthusiasts and lovers of mathematics, and at the same time find fun while discussing our challenges.
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Geendle
We hope you feel very comfortable on our channel, expressing your comments and leaving your suggestions as well.
We invite you to subscribe to our channel and enjoy this incredible journey through never-ending science and truly "Go beyond math"
We appreciate you!
Geendle
A Nice Algebra Problem | How to simplify | world | Geendle
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#squareroot ,#crossmultiplicationmethod ,#division ,#fraction ,#maths ,#algebra ,#subtraction ,#calculus , #Math, #mathematics , #Algebra, #Geometry, #Calculus, #Trigonometry, #statistics , #probability , #linearalgebra , #numbertheory , #Topology, #mathproblems , #mathhelp , #mathtutorial , #MathLessons, #MathEducation, #MathTeacher, #mathstudent , #MathFun, #MathPuzzle, #mathhistory , #mathexplained , #MathForKids, #MathChallenge, #mathisfun
This comprehensive list covers a wide range of topics that can be addressed in math-related content.
Algebra:
Equations
Inequalities
Polynomials
Functions
Systems of Equations
Exponential and Logarithmic Functions
Geometry:
Euclidean Geometry
Coordinate Geometry
Trigonometry
Geometric Transformations
Solid Geometry
Geometric Proofs
Calculus:
Limits
Derivatives
Integrals
Differential Equations
Multivariable Calculus
Sequences and Series
Statistics and Probability:
Descriptive Statistics
Inferential Statistics
Probability Theory
Random Variables
Probability Distributions
Hypothesis Testing
Linear Algebra:
Vectors
Matrices
Determinants
Eigenvalues and Eigenvectors
Vector Spaces
Linear Transformations
Number Theory:
Divisibility
Prime Numbers
Modular Arithmetic
Diophantine Equations
Cryptography
Fermat’s and Euler’s Theorems
Discrete Mathematics:
Combinatorics
Graph Theory
Logic
Set Theory
Algorithms
Boolean Algebra
Topology:
Topological Spaces
Continuity
Compactness
Homotopy
Fundamental Group
Manifolds
Functional Analysis:
Banach Spaces
Hilbert Spaces
Linear Operators
Normed Spaces
Spectral Theory
Fourier Analysis
Mathematical Logic:
Propositional Logic
Predicate Logic
Model Theory
Proof Theory
Set Theory
Computability Theory
Abstract Algebra:
Groups
Rings
Fields
Modules
Galois Theory
Homomorphisms
Applied Mathematics:
Numerical Analysis
Optimization
Differential Equations
Mathematical Modeling
Control Theory
Game Theory
Recreational Mathematics:
Puzzles
Magic Squares
Mathematical Games
Mathematical Paradoxes
Mathematical Art
Fractals
History of Mathematics:
Ancient Mathematics
Medieval Mathematics
Renaissance Mathematics
Modern Mathematics
Biographies of Mathematicians
Development of Mathematical Theories
Mathematics in Computing:
Algorithms
Data Structures
Computational Complexity
Cryptography
Information Theory
Machine Learning
Mathematics in Natural Sciences:
Mathematical Physics
Mathematical Biology
Mathematical Chemistry
Environmental Mathematics
Financial Mathematics
Actuarial Science
Exam Preparation:
SAT Math
ACT Math
GRE Math
GMAT Math
Math Olympiad
College Entrance Exams
math olympiad
olympiad math
Math olympiad question
Math olympiad questions
olympiad math problem
algebra olympiad problems
a nice algebra problem
math tricks
imo
how to improve maths
how to solve any math problem
Thanks for visiting!
Please kindly ask for your signature. You help me a lot.
#squareroot ,#crossmultiplicationmethod ,#division ,#fraction ,#maths ,#algebra ,#subtraction ,#calculus , #Math, #mathematics , #Algebra, #Geometry, #Calculus, #Trigonometry, #statistics , #probability , #linearalgebra , #numbertheory , #Topology, #mathproblems , #mathhelp , #mathtutorial , #MathLessons, #MathEducation, #MathTeacher, #mathstudent , #MathFun, #MathPuzzle, #mathhistory , #mathexplained , #MathForKids, #MathChallenge, #mathisfun
This comprehensive list covers a wide range of topics that can be addressed in math-related content.
Algebra:
Equations
Inequalities
Polynomials
Functions
Systems of Equations
Exponential and Logarithmic Functions
Geometry:
Euclidean Geometry
Coordinate Geometry
Trigonometry
Geometric Transformations
Solid Geometry
Geometric Proofs
Calculus:
Limits
Derivatives
Integrals
Differential Equations
Multivariable Calculus
Sequences and Series
Statistics and Probability:
Descriptive Statistics
Inferential Statistics
Probability Theory
Random Variables
Probability Distributions
Hypothesis Testing
Linear Algebra:
Vectors
Matrices
Determinants
Eigenvalues and Eigenvectors
Vector Spaces
Linear Transformations
Number Theory:
Divisibility
Prime Numbers
Modular Arithmetic
Diophantine Equations
Cryptography
Fermat’s and Euler’s Theorems
Discrete Mathematics:
Combinatorics
Graph Theory
Logic
Set Theory
Algorithms
Boolean Algebra
Topology:
Topological Spaces
Continuity
Compactness
Homotopy
Fundamental Group
Manifolds
Functional Analysis:
Banach Spaces
Hilbert Spaces
Linear Operators
Normed Spaces
Spectral Theory
Fourier Analysis
Mathematical Logic:
Propositional Logic
Predicate Logic
Model Theory
Proof Theory
Set Theory
Computability Theory
Abstract Algebra:
Groups
Rings
Fields
Modules
Galois Theory
Homomorphisms
Applied Mathematics:
Numerical Analysis
Optimization
Differential Equations
Mathematical Modeling
Control Theory
Game Theory
Recreational Mathematics:
Puzzles
Magic Squares
Mathematical Games
Mathematical Paradoxes
Mathematical Art
Fractals
History of Mathematics:
Ancient Mathematics
Medieval Mathematics
Renaissance Mathematics
Modern Mathematics
Biographies of Mathematicians
Development of Mathematical Theories
Mathematics in Computing:
Algorithms
Data Structures
Computational Complexity
Cryptography
Information Theory
Machine Learning
Mathematics in Natural Sciences:
Mathematical Physics
Mathematical Biology
Mathematical Chemistry
Environmental Mathematics
Financial Mathematics
Actuarial Science
Exam Preparation:
SAT Math
ACT Math
GRE Math
GMAT Math
Math Olympiad
College Entrance Exams
math olympiad
olympiad math
Math olympiad question
Math olympiad questions
olympiad math problem
algebra olympiad problems
a nice algebra problem
math tricks
imo
how to improve maths
how to solve any math problem
Thanks for visiting!
Переглядів: 18
Відео
Math Olympiad Question | Harvard Entrance Adapted Problem | How to Solve | Geendle
Переглядів 329 годин тому
Math Olympiad Question | Harvard Entrance Adapted Problem | How to Solve PLEASE SHOW YOUR SUPPORT BY SUBSCRIBING TO THE CHANNEL. THANK YOU VERY MUCH. #algebra, #mathematics , #olympiad , #maths ,#simplification
Solving a Challenging Irrational Equation Using Substitution | Geendle
Переглядів 19День тому
Hello!!! Please, Subscribe! #maths #education #mathematics #matholympiad #algebra In this video, "Solving a Challenging Irrational Equation Using Substitution | Geendle," we dive into the process of solving a complex irrational equation through the technique of substitution. Substitution is a key method for simplifying and solving equations that might initially seem daunting.
Math olympiad problem solving a crazy exponential without calculator | Geendle
Переглядів 9421 день тому
Hello friends! Welcome to Geendle! Please kindly ask for your signature. You help me a lot. #squareroot ,#crossmultiplicationmethod ,#division ,#fraction ,#maths ,#algebra ,#subtraction ,#calculusandmathematicslearning , #subscribetomychannel ,#crossmultiplicationmethod ,#subtractiontricks , #cuberoot
Functional Equation | Geendle
Переглядів 66321 день тому
Hello friends! Welcome to Geendle! Please kindly ask for your signature. You help me a lot. In this video I solve a functional equation. #squareroot ,#crossmultiplicationmethod ,#division ,#fraction ,#maths ,#algebra ,#subtraction ,#calculus , #Math, #mathematics , #Algebra, #Geometry, #Calculus, #Trigonometry, #statistics , #probability , #linearalgebra , #numbertheory , #Topology, #mathproble...
Math Olympiad Question | Algebra Math Problem| Geendle
Переглядів 156Місяць тому
Math Olympiad Question | Algebra Math Problem| Geendle #maths #algebra #mathematics #olympiad #olympiadmathematics Please show your support by liking and subscribing to the channel. Let's make this channel a great place to enjoy math! #matholympiad #algebra #mathematics #maths #math #olympiad #matholympiad #algebra#math#simplification #Exponents#vedicmath#viralmathproblem #howto#mathematics...
Solving A Nice Math Algebra Problem | How to solve for variables "a "and "b"? | Geendle
Переглядів 105Місяць тому
Hello Friends! Thanks for Watching! PLEASE, SUBSCRIBE! In this Vídeo I solve this symultaneous equations a b=4; a^5 b^5 =464 #maths, #mathematics ,#education ,#matholympiad, #usa ,#algebra , #algebratricks
Math Olympiad Challenge | Symplify | Geendle
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Hello friends! Welcome to Geendle! Please kindly ask for your signature. You help me a lot. #squareroot ,#crossmultiplicationmethod ,#division ,#fraction ,#maths ,#algebra ,#subtraction ,#calculus , #Math, #mathematics , #Algebra, #Geometry, #Calculus, #Trigonometry, #statistics , #probability , #linearalgebra , #numbertheory , #Topology, #mathproblems , #mathhelp , #mathtutorial , #MathLessons...
Math Olympiad | A Nice Algebra Problem | How to Solve for (x)| world | Geendle
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Hello friends! Welcome to Geendle! Please kindly ask for your signature. You help me a lot. #squareroot ,#crossmultiplicationmethod ,#division ,#fraction ,#maths ,#algebra ,#subtraction ,#calculus , #Math, #mathematics , #Algebra, #Geometry, #Calculus, #Trigonometry, #statistics , #probability , #linearalgebra , #numbertheory , #Topology, #mathproblems , #mathhelp , #mathtutorial , #MathLessons...
Math olympiad problem solving an exponential without calculator | Geendle
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Hello friends! Welcome to Geendle! Please kindly ask for your signature. You help me a lot. #squareroot ,#crossmultiplicationmethod ,#division ,#fraction ,#maths ,#algebra ,#subtraction ,#calculusandmathematicslearning , #subscribetomychannel ,#crossmultiplicationmethod ,#subtractiontricks , #cuberoot
Math olympiad problem a square root without calculator | Geendle
Переглядів 1,4 тис.Місяць тому
Hello friends! Welcome to Geendle! Please kindly ask for your signature. You help me a lot. In this video I solve the square root of a huge number without using a calculator. #squareroot ,#crossmultiplicationmethod ,#division ,#fraction ,#maths ,#algebra ,#subtraction ,#calculus
Math Olympiad Challenge: Exponential Problem Solved Without a Calculator | Geendle
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Hello friends! Welcome to Geendle! Please kindly ask for your signature. You help me a lot. In this video I solve an exponential without calculator a calculator. #squareroot ,#crossmultiplicationmethod ,#division ,#fraction ,#maths ,#algebra ,#subtraction ,#calculus , #Math, #mathematics , #Algebra, #Geometry, #Calculus, #Trigonometry, #statistics , #probability , #linearalgebra , #numbertheory...
Math Olympiad | How to solve? | Geendle
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Hello friends! Welcome to Geendle! Please kindly ask for your signature. You help me a lot. #squareroot ,#crossmultiplicationmethod ,#division ,#fraction ,#maths ,#algebra ,#subtraction ,#calculus , #Math, #mathematics , #Algebra, #Geometry, #Calculus, #Trigonometry, #statistics , #probability , #linearalgebra , #numbertheory , #Topology, #mathproblems , #mathhelp , #mathtutorial , #MathLessons...
Math olympiad equation | Geendle
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Hello friends! Welcome to Geendle! Please kindly ask for your signature. You help me a lot. #squareroot ,#crossmultiplicationmethod ,#division ,#fraction ,#maths ,#algebra ,#subtraction ,#calculus, #nigeria ,#nigerianmovies, #nollywoodmovies , #angola ,#caboverde ,#exponents, #southafrica ,#india ,#squareroot ,#matholympiad, #equation , #mathstricks
Math Olympiad Great Challenge | Geendle
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Hello friends! Welcome to Geendle! PLEASE SUBSCRIBE!!! #squareroot ,#crossmultiplicationmethod ,#division ,#fraction ,#maths ,#algebra ,#subtraction ,#calculus , #Math, #mathematics , #Algebra, #Geometry, #Calculus, #Trigonometry, #statistics , #probability , #linearalgebra , #numbertheory , #Topology, #mathproblems , #mathhelp , #mathtutorial , #MathLessons, #MathEducation, #MathTeacher, #math...
Math Olympiad Challenge | How to solve for (x, y) integers? | Geendle
Переглядів 86Місяць тому
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Algebra | Fractional Equation | How to Solve | Maths Olympiad | Geendle
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Переглядів 577Місяць тому
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Obvious mathematical solution | How to solve “x” in this tempting equation?
Переглядів 378Місяць тому
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Math Olympiad | A Nice Algebra Problem | How to Solve for (a,b) Positive Integers ?| world | Geendle
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Math Olympiad Challenge | A Nice Algebra Problem | How to Solve for x and y in this Problem | USA
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3.
Excellent approach. I like the way you factor the left hand side of the equation. You could have considered all factorizations of 1241 to show that 3 is the only solution. You used the factorization 17x73 to show that 3 is a solution of the equation.
You sort of "guessed" the solution. Why not a direct guess: for m=2 the LHS is far less than 559, then try m=3 and you're done.
What happens when x equal to 4, or 12, or 4/3? When a equals to 1?
The solution as presented needs to exclude the points at x=4, x=12, and x=4/3. But it also needs to separately include the cases for x=-2 and x=2, which are missing as is.
I guess a=-2 and a=2 happen at exactly points where x needs to be excluded anyway, so my second clause doesn't really matter.
x=2 works...
How do you know that 7^(m/3) + 6^(m/3) is an integer? If it isn't, then your solution fails.
7^1 + 6^1 = 13, and 559 = 13 x 43
Hi, I also solved it with the same method but how do you prove that 823543=7^7? In my solution I use this idea: the last term is 3 so the repeated number could be 3 or 7 because they are the only numbers that if repeated generate a 3: for 3 the pattern is 3^(1+4k) and for 7 it is 7^(3+4k). So 3^3 is not a solution because 1+4k cannot be 3 (if k is natural), but if you do it for 7 you find the solution because 3+4k is equal to 7 when k is 1, so the result is 7 ^7. The idea could also be extended to numbers that contain 3 or 7 as a final term such as 13, 27 or 10023...I thought that with 10^10 we exceed the digits of 823543 and this allows us to stay under 10
because f(a,b) = f(b,a), i started by using a = 2+ x, b=2 -x ... solutions for x will be in equal and opposite pairs, and this substitution into the second given will result in a quartic. that quartic must have factors of the form (c - x^2) only because of the first observation, meaning it is really quadratic in x^2. distributing and combining (2+x)^5 + (2-x)^5 is demeaning and time-consuming, but is most of the work in the problem.
Math Olympiad Problem: 7^m + 6^m = 559; m = ? 7^m + 6^m = 559 = (13)(43) = (7 + 6)(49 - 6) = (7 + 6)(7² - 6) = 7³ + 6(7²) - 6(7 + 6) = 7³ + 6(7² - 13) = 7³ + 6(36) = 7³ + 6(6²)= 7³ + 6³; m = 3 Answer check: 7^m + 6^m = 559; Confirmed as shown Final answer: m = 3
www.youtube.com/@walterwen2975, Wow! That's a nice solution!!! I'm saving it and I will post a video solving a similar problem in two different ways soon. Thank you!
@@Geendle Thanks, 🙏
@@walterwen2975 ok you found a solution after quite a bit of mental juggling and how do you know you got all the solutions? That part is still missing tho.
@@HoSza1 Be glad to know if there is any more solution.
If a person learns sequences of quartics, just like most remember sequences of squares, {1,4,9,...}. By mental recall of memorized arithmetic: Whatever the sum be, the sum of 6 and 7's as squares will be too small, as a sum of quartic's, HA, a single 6 OR 7 is too much. Edit start: If that has been said already: I paused at @0:03 - took longer to type then solve.
Thank you!!! This would be a nice solution too!
just brute force it 😂. 559 is not really a big number
🤣🤣🤣🤣🤣 you're funny! I Love your comment!
prime factorization gets answer very fast 1156 = 2x2x17x17 =34^2
Yes!!!!
This is probably the easiest olymp8ad problem ever.
They should do one of these where it isn't conveniently an integer.
I'll solve one of this soon! thanks!
6, 36, 216. 7, 49, 243. m=3. Is there anorher solution? No, because f(x)=6^x+7^x is a function that increases monotonically. Yes calculus. 😂
😍 Thank you!
yeah who was, at secondary school, advised to memorize • all squares from 11 up to 20 . • all powers under 1024 of 2 up to 10 . so incl. 2¹⁰ , 3⁶ , 4⁵ , 5⁴ , 6³ ,7³ , 8³ , 9³ . that is , in the question is immediately visible 343+216. then indeed considering that 6^m and 7^m are monotonously increasing functions of m, meaning that the found m = 3 should be the only possible answer .
Easy. 559 is odd so m must also be odd as 6^m is even. 7^5 is far too big so m must be 3. I’ll save time and not watch the video.
well that's funny . and wouldn't 7^4 be odd . could you please show whether it will be even . . btw it is equal to 2401, like 49² .
32^2=1024 and 33^2 is odd so (asuming the answer is an integer number) 34^2=1156
I had the same reasoning but verified it by noting that 1024 + 4*33= 1156. In general, x^2+4(x+1) = (x+2)^2.
To verify you can just multiply with the standard algorythm: 34^2 = 30*34+4*34 = 1020+136 = 1156 . There's no need of looking for algebraic identities since it's an arithmetic problem that doesn't require further generalisation.
@@lucho2868 of course you can, but if you know rules for nearby perfect squares than my way is quicker. You might as well say that you didn't need to use your knowledge that 32^2 = 1024 in the first place!
@@kicorse You also used 1024^.5=32. In fact you wrote a proof around the fact that that 32^2=1024. And your answer is formally incorrect since you didn't even proove that 32^2=1024. Your solution is useless because it requires differences, elementary algebra and (over all that) a long mutiplication whilst mine only require the long multiplication to check at the end. Making it shorter, more rigurous, and more generally applycable than yours.
@@kicorse for example, if you wanted approximate the value of any square root by hand the long multiplication's result of a nearby approximation could be used as an input in a Bakhshali-kind formula to get a good approximation result whilst the method you suggest would work less generally.
34
That's rigth!
Nice.
Thanks, bro!
1156 is between 900 and 1600 and squared numbers that end in 6 are either X4 or X6 so the answer is either 34 or 36 and after that you just need to check these
Thank you! That's a nice way too!
You have to state that a, b, and c are natural numbers as well
Thanh you bro! I'll consider that in the next videos.
20=(8343-7^a)^(1/a), now i think this, if 20^k (k is a natural number) finishes with a 0 for all k, it means that 8343-7^m has to finished with a 0. So 7^m has to finished with a 3 because the last 3 of 8343 minus the last term of 7^m has to be equal 0. Now I study the pattern of 7^k and it is: 7^1=7, 7^2=...9, 7^3=...3, 7^4=...1, 7^5=...7. Now i know that every 4 times the pattern restart and i know that for 7^3 i have a number that finished with a 3. So 7^a=7^(3+4n) where n is natural. But for n>1 8343-7^a is negative so my only option is n=0 that means a=3 and if you try to put a=3 in 20^a+7^a it becomes 8343
Well done! Thank you! Your method your method is awsome.
@@Geendle Thank youuuuuuu
i did it in my head in like 20 seconds when I saw the thumbnail. It's really fun!
That's nice Thank you!
Niiiiiiceeeee solution, i find another process: 6=(559-7^m)^(1/m), now i think this, if 6^k (k is natural number) 6^k has to finished with a 6, for example 6*6=36, 36*6=216,..., it means that 559-7^m has to finished with a 6. So 7^m has to finished with a 3 because the 9 of 559 minus the last term of 7^m has to be equal 6. Now I study the pattern of 7^k and it is: 7^1=7, 7^2=...9, 7^3=...3, 7^4=...1, 7^5=...7. Now i know that every 4 times the pattern restart and i know that for 7^3 i have a number that finished with a 3. So 7^m=7^(3+4n) where n is natural. But for n>0 559-7^m is negative so my only option is n=0 that means m=3 and if you try to put m=3 in 6^m+7^m it becomes 559
Very good! I'l try to explain one similar video using your idea! Thank you!
@@Geendle i would be honored
1156 is a very long way from being a "huge number". As others have pointed out, this answer takes only a few seconds to work out in your head (divide by 2, divide by 2, left with 289 which is 17^2, answer is 2*17). Also, you didn't need to multipy by 4/4, what you had was (10^2 + 2) ^ 2, giving the result of 102/3 after the square root which is 34. Also, also, what's with the unrelated #-tags? The only one relevant to this video is #squareroot...
Thanks! the other hashtags are intended for the video to reach more people.
My approach is simply to find the prime factors. It's obviously divisible twice by 2, leaving us only to find the square root of 289. It's also obviously not divisible by 3. 5 can be rules out as 5^4 is 625 which means we must hop that there's an exact square root of 289, or it gets very messy. It's clearly greater than 13 squared (169) which most people know, and the next prime is then 17, the square of which is 289. So, we have sqrt(2*2*17*17) = 34. nb. I don't think merits being an olympiad question.
Nice! Thank you. I was just trying to show another way of solving it.
Took me a whole 5 seconds to solve. 9mins and 7 secs of my life saved not following the solution
Nice!😉
This is mental. So slow! Anyone doing a math olympiad would just look at it and know the answer, surely?
Probably yes!
You had 7^(m/3) + 6^(m/3) = 13, hence 13 = 6+7 then m/3 = 1 and m=3
Yes, that's a beautiful observation! Thank you!
I used the "by hand" method, which only took two steps = less than a minute. Grade 4 students used to be taught this method, which is similar to long division.
Thats great! Thank you!
It is waay simpler to solve by converting it to binary. Find the largest power of 2 that fits in 400 -> 2^8 = 256 Same for the remaining 144 -> 2^7 = 128 Same for the remaining 16 -> 2^4 The approach shown in the video serves no better purpose, it’s neither faster nor more general
I also thought the same way. When I realized that the base is two, I intuitively remembered converting decimal numbers to binary.
I'll post another video using this method! Thank you!
a,b,c \in {4,7,8} would be better
Thanks! I will improve!
I tried beginning to solve it before opening the video, my method was to try to find the prime factors of 1156 to if possible remove most of the work from the square root, found that you could divide it by 2 two times wich gives 2√289, then I had to think for a bit longer until I reached 17 wich fits perfectly, giving √1156 = 34, ngl I was not expecting the answer to be an integer
Nice! Thank you!
I used the method I learned in freshman year of high school. I had the answer in less than a minute.
Wow! that's nice!
I was way faster by testing numbers
It is not.
@@francoismusic_ But its literally just taking the max. allowed exponent 3 times
What?!?! I figured out that it is 34 in about 2 seconds, in my head. 30 x 30 = 900.....40 x 40 = 1600. So the number must fall between 30 and 40. Only 4 squared and 6 squared end in 6, so it has to be one of those two numbers in the singles position. Since 1156 is much closer to 900 than 1600, the singles number must be 4. Therefore 34 is the answer. Multiplied 34 x 34 to verify--also in my head. I am not a mathematician or an engineer, and I am 74 years old. Your method is far too complicated. I stopped watching after about a minute.
I did too. I used 32^2=1024; well-known if you're familiar with powers of 2. So it's higher than 32, but ends in 6, so the rest is as you stated.
That's nice! Thank you!!!
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Thanks pidimo!😉
You msde a mistake, the real solutions are (5, 2) amd (-2,-5) You are also making a mistake with the complex solutions, x^2 -3x+19=0, x= 3/2 \pm i sqrt(67)/2.
Thanks! I'll check that!
The terms multiply to 17 therefore one of the terms is 17 and the other must be one. Such nonsense!!
Yes! Thank you!
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how about this: let 3^(2x-1/x+1) = t 3^(3x-x-1/x+1)=t 3^(3x/x+1)=3t thus, 3^(5x-1/x+1) = t・3t =3t^2 which means that 3t^2+t-30=0 t must be real so the only solution is t=3 hence 2x-1/x+1=1 i.e. x=2
www.youtube.com/@AA-st6of, Thank you! This is really nice! I'll try this method in a different video!!!
@@Geendle 👍
on the one hand I don't like problems where imagination has go soooo far. I am a teacher and would not have found every step. Would have solved the quadratic in a different way. However. This is an example I really like to teach my students. It's too complex to solve it alone. but its a perfect example to train all different kind of methods and keep an eye open to all of them at once. Learning this problem by heart is in itself a good lesson. So thank you. I copied it already in latex to make a lesson out of it.
www.youtube.com/@Essentialsend,Thank you!😍 I will bring some more similar problems to share ideas about them.
27 and 3 are the only possible values. so equate powers to (3,1) and (1,3) and see which value yields a coherent solution. thjnk about it powers of 3 are 1,3,9,27. any power higher than that cannot yield a sum of 30. and we cannot have irrational numbers as the sum of two irrational numbers canot be rational(integer). so fractional powrrs are out of the picture. and, negative powers are also out of picture. you cannot possiblt have 1/3 and 89/3, both are not power of 3.
www.youtube.com/@AA-st6of, you are right in your analysis. Thank you!!!
Borring
x=1 didnt work so x=2 was my next guess
That's right!!!
x =2 by inspection. The technical solution is preferable of course.
That's right. Thank you!!!
On solving 4^2-4&(-3)^2-(-3) gives solns .. but power value cant be -ve hence first x=2 only works
Yes! That's the answer. Thanks!
Congratulations ! What does it means? The ''factors'' of 175 that lead us to a positive integer solution are just 7 and 25. Thanks!
www.youtube.com/@user-bx8tu3sj3w, that's it! Thank you!
A Physicist(or any common sense person) will realize the positive part just crescent and so x should be small, just throwing integers will find x=2 and it is the only one. Even if x was not integer is just throw integers on it until the left size would be bigger than the left. Why spend 12 minutes at it? It takes about 7 to 10 seconds. You don't need to be a Ramanujan to solve this.
Sorry man, but when you are rewriting to (a-b)*(a+b)=12 and saying that (a-b) = 1 or 2 or 3 and (a+b) = 12 or 6 or 4 your are making mistake. (a-b) or (a+b) are integer values only for even values of 'x'. So you are solving the equation for even 'x'. If you take odd 'x' then (x^x)/2 will be non-integer, e.g. x=3 -> (a+b) = 3^13.5+3^1.5; or different value: x=5 -> (a+b) = 5^1562.5+5^2.5. This isn't integer. Yeah, you got lucky with this math problem that both your multipliers (a+b) and (a-b) are integers and that for growing x>1 you get ever growing function of f(x)=x^x^x-x^x. But for another similar problem like x^x^x-x^x=7625597484960 this your trick wouldn't work. (a+b)*(a-b)=7625597484960 doesn't imply, that (a+b) is integer and (a-b) is integer. For this case, where x=3, this will be 2761453.635828058 * 2761443.2435232126 = 7625597484960.
I like your analysis. Thank you for sharing your ideas!