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A Nice Algebra Problem | How to simplify | world | Geendle
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zhlédnutí: 18

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Math Olympiad Question | Harvard Entrance Adapted Problem | How to Solve | Geendle
zhlédnutí 32Před 22 hodinami
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
zhlédnutí 19Před dnem
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
zhlédnutí 94Před 21 dnem
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
zhlédnutí 663Před 21 dnem
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
zhlédnutí 156Před měsícem
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
zhlédnutí 105Před měsícem
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
zhlédnutí 174Před měsícem
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
zhlédnutí 132Před měsícem
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
zhlédnutí 2,1KPřed měsícem
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
zhlédnutí 1,4KPřed měsícem
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
zhlédnutí 679Před měsícem
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
zhlédnutí 263Před měsícem
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
zhlédnutí 81Před měsícem
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
zhlédnutí 598Před měsícem
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
zhlédnutí 86Před měsícem
Math Olympiad Challenge | How to solve for (x, y) integers? | Geendle
Algebra | Fractional Equation | How to Solve | Maths Olympiad | Geendle
zhlédnutí 191Před měsícem
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Obvious mathematical solution | How to solve “x” in this tempting equation?
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Math Olympiad Inequality | How to solve? | Geendle
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Math Olympiad Challenge: A Cubic Equation Solved Without a Calculator | Geendle
A Nice Algebra Problem | How to Solve for x in this problem| world | Geendle
zhlédnutí 77Před měsícem
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Math Olympiad Challenge: Exponential Problem Solved Without a Calculator | Geendle
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Math Olympiad | A Nice Algebra Problem | How to Solve for (x,y) in This Problem | world | Geendle
zhlédnutí 149Před měsícem
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Unlocking Genius: Mind-Blowing Math Olympiad Moments! - Check Out The Ultimate solution | Geendle
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Unlocking the Secrets of Advanced Exponential Equations: A Step-by-Step Guide | Math Olympiad
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Math Olympiad | Why Did They Hide This Secret From Students? A Nice Approach| Geendle
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Math Olympiad | Why Did They Hide This Secret From Students? A Nice Approach| Geendle
Math Olympiad | A Nice Algebra Problem | How to Solve for (a,b) Positive Integers ?| world | Geendle
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Math Olympiad | A Nice Algebra Problem | How to Solve for (a,b) Positive Integers ?| world | Geendle
Math Olympiad Challenge | A Nice Algebra Problem | How to Solve for x and y in this Problem | USA
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Komentáře

  • @user-dq3uh6ee5w
    @user-dq3uh6ee5w Před 12 dny

    3.

  • @slavinojunepri7648
    @slavinojunepri7648 Před 18 dny

    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.

  • @MrKA1961
    @MrKA1961 Před 24 dny

    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.

  • @qnicks23434
    @qnicks23434 Před 24 dny

    What happens when x equal to 4, or 12, or 4/3? When a equals to 1?

    • @ArtemKreimer
      @ArtemKreimer Před 24 dny

      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.

    • @ArtemKreimer
      @ArtemKreimer Před 24 dny

      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.

  • @archangecamilien1879
    @archangecamilien1879 Před 26 dny

    x=2 works...

  • @DonRedmond-jk6hj
    @DonRedmond-jk6hj Před měsícem

    How do you know that 7^(m/3) + 6^(m/3) is an integer? If it isn't, then your solution fails.

  • @jacopomaccione7791
    @jacopomaccione7791 Před měsícem

    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

  • @theupson
    @theupson Před měsícem

    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.

  • @walterwen2975
    @walterwen2975 Před měsícem

    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

    • @Geendle
      @Geendle Před měsícem

      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!

    • @walterwen2975
      @walterwen2975 Před měsícem

      @@Geendle Thanks, 🙏

    • @HoSza1
      @HoSza1 Před měsícem

      @@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.

    • @walterwen2975
      @walterwen2975 Před měsícem

      @@HoSza1 Be glad to know if there is any more solution.

  • @DOTvCROSS
    @DOTvCROSS Před měsícem

    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.

    • @Geendle
      @Geendle Před měsícem

      Thank you!!! This would be a nice solution too!

  • @foerza649
    @foerza649 Před měsícem

    just brute force it 😂. 559 is not really a big number

    • @Geendle
      @Geendle Před měsícem

      🤣🤣🤣🤣🤣 you're funny! I Love your comment!

  • @krwada
    @krwada Před měsícem

    prime factorization gets answer very fast 1156 = 2x2x17x17 =34^2

  • @agentm10
    @agentm10 Před měsícem

    This is probably the easiest olymp8ad problem ever.

  • @manudude02
    @manudude02 Před měsícem

    They should do one of these where it isn't conveniently an integer.

    • @Geendle
      @Geendle Před měsícem

      I'll solve one of this soon! thanks!

  • @HoSza1
    @HoSza1 Před měsícem

    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. 😂

    • @Geendle
      @Geendle Před měsícem

      😍 Thank you!

    • @keescanalfp5143
      @keescanalfp5143 Před 12 dny

      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 .

  • @tortinwall
    @tortinwall Před měsícem

    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.

    • @keescanalfp5143
      @keescanalfp5143 Před 12 dny

      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 .

  • @lucho2868
    @lucho2868 Před měsícem

    32^2=1024 and 33^2 is odd so (asuming the answer is an integer number) 34^2=1156

    • @kicorse
      @kicorse Před měsícem

      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.

    • @lucho2868
      @lucho2868 Před měsícem

      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.

    • @kicorse
      @kicorse Před měsícem

      @@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!

    • @lucho2868
      @lucho2868 Před měsícem

      @@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.

    • @lucho2868
      @lucho2868 Před měsícem

      @@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.

  • @lightning77125
    @lightning77125 Před měsícem

    34

  • @beyondoftrue
    @beyondoftrue Před měsícem

    Nice.

  • @ben_adel3437
    @ben_adel3437 Před měsícem

    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

    • @Geendle
      @Geendle Před měsícem

      Thank you! That's a nice way too!

  • @willvu3581
    @willvu3581 Před měsícem

    You have to state that a, b, and c are natural numbers as well

    • @Geendle
      @Geendle Před měsícem

      Thanh you bro! I'll consider that in the next videos.

  • @jacopomaccione7791
    @jacopomaccione7791 Před měsícem

    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

    • @Geendle
      @Geendle Před měsícem

      Well done! Thank you! Your method your method is awsome.

    • @jacopomaccione7791
      @jacopomaccione7791 Před měsícem

      @@Geendle Thank youuuuuuu

  • @atanasiumirela7187
    @atanasiumirela7187 Před měsícem

    i did it in my head in like 20 seconds when I saw the thumbnail. It's really fun!

    • @Geendle
      @Geendle Před měsícem

      That's nice Thank you!

  • @jacopomaccione7791
    @jacopomaccione7791 Před měsícem

    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

    • @Geendle
      @Geendle Před měsícem

      Very good! I'l try to explain one similar video using your idea! Thank you!

    • @jacopomaccione7791
      @jacopomaccione7791 Před měsícem

      @@Geendle i would be honored

  • @gerryiles3925
    @gerryiles3925 Před měsícem

    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...

    • @Geendle
      @Geendle Před měsícem

      Thanks! the other hashtags are intended for the video to reach more people.

  • @TheEulerID
    @TheEulerID Před měsícem

    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.

    • @Geendle
      @Geendle Před měsícem

      Nice! Thank you. I was just trying to show another way of solving it.

  • @Cynicalgeek743
    @Cynicalgeek743 Před měsícem

    Took me a whole 5 seconds to solve. 9mins and 7 secs of my life saved not following the solution

  • @edmx
    @edmx Před měsícem

    This is mental. So slow! Anyone doing a math olympiad would just look at it and know the answer, surely?

  • @mouradbelkas598
    @mouradbelkas598 Před měsícem

    You had 7^(m/3) + 6^(m/3) = 13, hence 13 = 6+7 then m/3 = 1 and m=3

    • @Geendle
      @Geendle Před měsícem

      Yes, that's a beautiful observation! Thank you!

  • @ericfielding668
    @ericfielding668 Před měsícem

    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.

    • @Geendle
      @Geendle Před měsícem

      Thats great! Thank you!

  • @Krmpfpks
    @Krmpfpks Před měsícem

    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

    • @beyondoftrue
      @beyondoftrue Před měsícem

      I also thought the same way. When I realized that the base is two, I intuitively remembered converting decimal numbers to binary.

    • @Geendle
      @Geendle Před měsícem

      I'll post another video using this method! Thank you!

  • @user-zr7ie6so1q
    @user-zr7ie6so1q Před měsícem

    a,b,c \in {4,7,8} would be better

    • @Geendle
      @Geendle Před měsícem

      Thanks! I will improve!

  • @_.__-._-_.-..-...
    @_.__-._-_.-..-... Před měsícem

    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

    • @Geendle
      @Geendle Před měsícem

      Nice! Thank you!

  • @JohnMackenroth-mg6jc
    @JohnMackenroth-mg6jc Před měsícem

    I used the method I learned in freshman year of high school. I had the answer in less than a minute.

    • @Geendle
      @Geendle Před měsícem

      Wow! that's nice!

  • @triforce9856
    @triforce9856 Před měsícem

    I was way faster by testing numbers

    • @francoismusic_
      @francoismusic_ Před měsícem

      It is not.

    • @triforce9856
      @triforce9856 Před měsícem

      @@francoismusic_ But its literally just taking the max. allowed exponent 3 times

  • @tominmo8865
    @tominmo8865 Před měsícem

    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.

    • @krabkrabkrab
      @krabkrabkrab Před měsícem

      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.

    • @Geendle
      @Geendle Před měsícem

      That's nice! Thank you!!!

  • @ericsonmario5337
    @ericsonmario5337 Před měsícem

    🎉

    • @Geendle
      @Geendle Před měsícem

      Thanks pidimo!😉

  • @ronbannon
    @ronbannon Před měsícem

    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.

    • @Geendle
      @Geendle Před měsícem

      Thanks! I'll check that!

  • @MatsCooper
    @MatsCooper Před měsícem

    The terms multiply to 17 therefore one of the terms is 17 and the other must be one. Such nonsense!!

    • @Geendle
      @Geendle Před měsícem

      Yes! Thank you!

  • @ericsonmario5337
    @ericsonmario5337 Před měsícem

    🎉

  • @AA-st6of
    @AA-st6of Před měsícem

    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

    • @Geendle
      @Geendle Před měsícem

      www.youtube.com/@AA-st6of, Thank you! This is really nice! I'll try this method in a different video!!!

    • @AA-st6of
      @AA-st6of Před měsícem

      @@Geendle 👍

  • @Essentialsend
    @Essentialsend Před měsícem

    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.

    • @Geendle
      @Geendle Před měsícem

      www.youtube.com/@Essentialsend,Thank you!😍 I will bring some more similar problems to share ideas about them.

  • @uncannyroaches5933
    @uncannyroaches5933 Před měsícem

    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.

    • @Geendle
      @Geendle Před měsícem

      www.youtube.com/@AA-st6of, you are right in your analysis. Thank you!!!

  • @ronnietwelvetoes1876
    @ronnietwelvetoes1876 Před měsícem

    Borring

  • @ThoMas-dq7pg
    @ThoMas-dq7pg Před měsícem

    x=1 didnt work so x=2 was my next guess

    • @Geendle
      @Geendle Před měsícem

      That's right!!!

  • @pukulu
    @pukulu Před měsícem

    x =2 by inspection. The technical solution is preferable of course.

    • @Geendle
      @Geendle Před měsícem

      That's right. Thank you!!!

  • @tejpalsingh366
    @tejpalsingh366 Před měsícem

    On solving 4^2-4&(-3)^2-(-3) gives solns .. but power value cant be -ve hence first x=2 only works

    • @Geendle
      @Geendle Před měsícem

      Yes! That's the answer. Thanks!

  • @user-bx8tu3sj3w
    @user-bx8tu3sj3w Před měsícem

    Congratulations ! What does it means? The ''factors'' of 175 that lead us to a positive integer solution are just 7 and 25. Thanks!

    • @Geendle
      @Geendle Před měsícem

      www.youtube.com/@user-bx8tu3sj3w, that's it! Thank you!

  • @agranero6
    @agranero6 Před měsícem

    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.

  • @user-yt4fn9pj8l
    @user-yt4fn9pj8l Před měsícem

    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.

    • @Geendle
      @Geendle Před měsícem

      I like your analysis. Thank you for sharing your ideas!