Пікірлер

• 
  @GeorgeSickinger4 сағат бұрын

  X=1. This is an incredibly complex solution for what is the simplest of equations Not an elegant solution

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  @ChidexOnlineMathClass015 сағат бұрын

  ❤❤❤❤ You can join my membership club or send super thanks in order to support this channel. Thanks

• 
  @harnisha915 сағат бұрын

  Improve your handwriting first ! a looks like 9.

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  @ChidexOnlineMathClass015 сағат бұрын

  @@harnisha91 I will. Thanks

• 
  @ggmoyang5 сағат бұрын

  9+b=4, b=5. easy. But wait, 9 - b is 1? hmmm

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  @ChidexOnlineMathClass015 сағат бұрын

  @@ggmoyang Ohhh, it's actually a not 9. Please l am sorry for my writing

• 
  @ChidexOnlineMathClass015 сағат бұрын

  Please support my channel by joining my youtube membership club. ❤❤❤ Thanks You can also send a super thanks ❤❤❤❤❤❤❤❤❤❤

• 
  @RomiMsikaMichel20 сағат бұрын

  3.75

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  @ChidexOnlineMathClass0119 сағат бұрын

  @@RomiMsikaMichel Yes exactly

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  @FridayJacob-bq8tt23 сағат бұрын

  Congratulations

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  @ChidexOnlineMathClass0122 сағат бұрын

  @@FridayJacob-bq8tt Thanks

• 
  @cfu11erКүн бұрын

  Another faster but perhaps less general approach: observe this is a sum of powers of two. This suggests finding the binary representation by simply taking the greatest power less than 148 and repeating recursively: 148-128=20, 20-16=4, and the log base2 of these is your a,b,c. This really only works with base two, unless the RHS cooperates and only has digits 0,1 in its representation in that base.

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  @ChidexOnlineMathClass01Күн бұрын

  @@cfu11er Yes exactly, thank you

• 
  @markfournette2483Күн бұрын

  Terrific! Another approach: x^3 = exp(i π +2πn) for n=0, 1, 2

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  @ChidexOnlineMathClass01Күн бұрын

  @@markfournette2483 Yes, nice one

• 
  @davidusa47Күн бұрын

  Okay, but wouldn’t God say only solve for the positive solutions?

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  @ChidexOnlineMathClass01Күн бұрын

  @@davidusa47 Yes of course, we can solve for real solutions only

• 
  @Dfire4422 күн бұрын

  Boss

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  @ChidexOnlineMathClass012 күн бұрын

  @@Dfire442 My dear brother 😍😍

• 
  @ElzaAmbrus2 күн бұрын

  Creezi confuzie; la numitor trebuia sa scrii 2a!

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  @alone-h7o2 күн бұрын

  Good Explanation sir ❤🎉

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  @ChidexOnlineMathClass012 күн бұрын

  @@alone-h7o Thanks so much

• 
  @ChidexOnlineMathClass012 күн бұрын

  @@alone-h7o Please stay connected, there will be a live class here on my channel tomorrow or next

• 
  @alone-h7o2 күн бұрын

  @@ChidexOnlineMathClass01 Sir are you on telegram ? Can you send me ur telegram username 😊

• 
  @Alacrity236882 күн бұрын

  Nice, you can also solve by substituting sqrt(x/1-x)=u Why are you wearing gloves, is it so cold there?

• 
  @ChidexOnlineMathClass012 күн бұрын

  @@Alacrity23688 Yeah, thanks

• 
  @SidneiMV2 күн бұрын

  x³ + 9x - 10 = 0 x³ - 1 + 9(x - 1) = 0 (x - 1)(x² + x + 10) = 0 *x = 1* x² + x + 10 = 0 => complex solution

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  @ChidexOnlineMathClass012 күн бұрын

  @@SidneiMV Nice

• 
  @hansgruber53912 күн бұрын

  Why not simply square each side of the equation? (and repeat three times) sqrt(x * sqrt(x * sqrt(x))) = 3 x * sqrt(x * sqrt(x)) = 9 x^3 * sqrt(x) = 81 x^7 = 6561 x = 6561^(1/7) = 3.51

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  @ChidexOnlineMathClass012 күн бұрын

  @@hansgruber5391 But you later arrived at an answer that is not well simplified.

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  @ChidexOnlineMathClass012 күн бұрын

  @@hansgruber5391 Well, is also good way to approach this kind of problem

• 
  @renesperb3 күн бұрын

  one should notice that the solutions of x+1/x = √3 i.e. 1/2(√3 + or - i) are on the unit circle in the complex plane.Take e.g. + sign this can be written as exp[ i π/3]. Now any integer power n can easily be computed. For n= 21 one has 21 * π/3 =7 π = 2*2 π+1. Now exp[2 π i ] = 1 and therefore x^21 = exp[ i π ] = -1 . 1/x is just exp[- i π/3 ], hence the rest is easy .

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  @ChidexOnlineMathClass013 күн бұрын

  @@renesperb Yeah, exactly

• 
  @arvnagar49603 күн бұрын

  4 is also an answer

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  @ChidexOnlineMathClass013 күн бұрын

  @@arvnagar4960 No, If you substitute 4, we get 2² = 4 ≠ 2

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  @myronbuck24363 күн бұрын

  Your introduction does not prove x is not R, it assumes x is not an integer. Showing three cases (really just two, since 0 is an even number) is not proof, although it is accurate.

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  @ChidexOnlineMathClass013 күн бұрын

  @@myronbuck2436 I am sorry for that

• 
  @sujalGhorse-xh8bs3 күн бұрын

  Is this really an olympiad level quesiton, looks like a basic class 10th question.

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  @ChidexOnlineMathClass013 күн бұрын

  @@sujalGhorse-xh8bs Junior Olympiad

• 
  @DreamAuras3 күн бұрын

  You forgot abs

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  @ChidexOnlineMathClass013 күн бұрын

  @@DreamAuras From where please

• 
  @sarfaroz28005 күн бұрын

  \|3+\|2

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  @ChidexOnlineMathClass015 күн бұрын

  ❤❤Good

• 
  @kayrabolat31325 күн бұрын

  Clever solution 👏🏻 Would you try to integrate 1/(1+sin(x)^2) by only substitution?

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  @ChidexOnlineMathClass015 күн бұрын

  @@kayrabolat3132 I will, probably today. Just subscribe to my channel and turn on notification button

• 
  @davestillhere41695 күн бұрын

  Thank you for doing this problem.

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  @ChidexOnlineMathClass015 күн бұрын

  @@davestillhere4169 Thanks so much brother

• 
  @pairadeau5 күн бұрын

  beautiful work. thanks a lot.

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  @ChidexOnlineMathClass015 күн бұрын

  @@pairadeau Thanks so much

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  @restablex5 күн бұрын

  x³ + 1/x³ = 0 x⁶ = -1 x³ = ± i also 1/x³ = -x³ Then x²¹ = (x⁶)³ x³ = (-1)³ x³ = -x³ Finally x²¹ - 1/x²¹ = -x³ + 1/x³ = -x³ -x³ = -2x³ = ±2i

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  @ChidexOnlineMathClass015 күн бұрын

  @@restablex Very nice

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  @muratkaradag37035 күн бұрын

  Nice solution, even nicer to come up with such a problem-

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  @ChidexOnlineMathClass015 күн бұрын

  @@muratkaradag3703 Thanks so much

• 
  @tensor1316 күн бұрын

  very nice. I did it by a slightly different (but no better) method: solving the first equation gives x = cis(+-pi/6), where cis is abbreviation for cos + isin Hence x^21=cis(+-21pi/6)=cis(-+pi/2) = -+i; similarly x^-21=cis(-+21pi/6) = +-i, and thus +-i2i as you obtained. My method is very direct but yours requires less knowledge (e.g. how the cis operator behaves) so that, imo, makes it more elegant than mine. 21 "works" because 21 = -3(mod6). In fact the same answer would follow for any n satisfying n=+-3(mod6), so 21 could be substituted by: 3, 9, 15, (21), 27, 33 ...... with the same result.

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  @ChidexOnlineMathClass016 күн бұрын

  @@tensor131 Your method is also awesome. Thanks for your comments

• 
  @davestillhere41696 күн бұрын

  thanks ....good use of identity.

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  @ChidexOnlineMathClass016 күн бұрын

  @@davestillhere4169 You are welcome

• 
  @Faoler6 күн бұрын

  Good job friend 🎉❤

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  @ChidexOnlineMathClass016 күн бұрын

  @@Faoler Thanks so much sir

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  @ChidexOnlineMathClass016 күн бұрын

  @@Faoler Please subscribe to my channel and also share. Thanks once again sir

• 
  @comdo7776 күн бұрын

  asnwer=2 isit

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  @comdo7776 күн бұрын

  asnwer=81 isit

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  @comdo7776 күн бұрын

  asnwer=1 or 4 isit

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  @ChidexOnlineMathClass016 күн бұрын

  @@comdo777 No, the answer is in the video

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  @1taraxaseeks16 күн бұрын

  Nicely done. Fun problem.

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  @ChidexOnlineMathClass016 күн бұрын

  @@1taraxaseeks1 Thanks so much

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  @davestillhere41696 күн бұрын

  Nicely explained factoring.

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  @ChidexOnlineMathClass016 күн бұрын

  @@davestillhere4169 Thanks so much

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  @runnow26557 күн бұрын

  Shouldn't there be 6 because it's a degree 6? Or does the a^x change that?

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  @runnow26557 күн бұрын

  Oh you said real

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  @ChidexOnlineMathClass017 күн бұрын

  @@runnow2655 Yeah

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  @Viper-_-_-gaming1017 күн бұрын

  You said cute instead of cube 😂😂🎉

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  @ChidexOnlineMathClass017 күн бұрын

  @@Viper-_-_-gaming101 Really ?? 😂🤣

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  @Viper-_-_-gaming1017 күн бұрын

  @@ChidexOnlineMathClass01 yeaa🤣🤣 but. keep it up 💪👍

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  @davestillhere41697 күн бұрын

  well done good to watch .

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  @ChidexOnlineMathClass017 күн бұрын

  @@davestillhere4169 Thanks so much

• 
  @davestillhere41697 күн бұрын

  great lesson in factorizing.

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  @ChidexOnlineMathClass017 күн бұрын

  @@davestillhere4169 Thanks so much

• 
  @davestillhere41697 күн бұрын

  this is fun.

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  @ChidexOnlineMathClass017 күн бұрын

  @@davestillhere4169 Yes exactly and thank you

• 
  @phillipkoci51007 күн бұрын

  Hello, sorry, I don't understand how you go from 2-(1-sqrt2) to = +1 ! Thank you !

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  @ChidexOnlineMathClass017 күн бұрын

  @@phillipkoci5100 Please I will be glad to answer your questions if you write well please. I don't understand what you wrote

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  @phillipkoci51007 күн бұрын

  @@ChidexOnlineMathClass01 : you write ((k-sqrt(2))*(k²+sqrt(2)k+2)-1(k-sqrt(2))=0. That's OK ; Then you say that is is equal to (k-sqrt(2))(k²+sqrt(2)k+1 ; I don't understand that last step, how you simplify "2)-1(k-sqrt(2))" in "+1" Thanks ! (BTW I love your videos ! ;-))

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  @ChidexOnlineMathClass017 күн бұрын

  @@phillipkoci5100 OK. I understand now. Here is what l did. We have: (k - √2)(k² + √2k + 2) - 1(k - √2) Factor out k - √2, we have ( k - √2)[(k² + √2k +2) - 1] ( k - √2)(k² + √2k + 2 - 1) (k - √2)(k² + √2k + 1) So, that was how l came up with the quadratic equation.

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  @phillipkoci51007 күн бұрын

  @@ChidexOnlineMathClass01 Wow, thanks for your answer ! I did'nt see that !

• 
  @ChidexOnlineMathClass017 күн бұрын

  @@phillipkoci5100 You are welcome

• 
  @user-xh3ih4ks9y7 күн бұрын

  令原式為X 則X^2=3X 則X=3

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  @ChidexOnlineMathClass017 күн бұрын

  @@user-xh3ih4ks9y Exactly

• 
  @daveshub7 күн бұрын

  Nice video, Your voice is funny though 😂

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  @ChidexOnlineMathClass017 күн бұрын

  @@daveshub 🤩🤩 Thanks

• 
  @daveshub6 күн бұрын

  @@ChidexOnlineMathClass01 you're welcome 😊

• 
  @Haallxx7 күн бұрын

  Nice video

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  @ChidexOnlineMathClass017 күн бұрын

  @@Haallxx Thanks so much

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  @charleskosina47998 күн бұрын

  This is an engineering approach rather than a mathematical one. Using Excel do a plot of x from 0 to 0.5. Gives value for X of 0.34583 which calculates to 3.000032. Close enough! Yes, not pure maths but gets answer much faster than the long winded way.

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  @ChidexOnlineMathClass018 күн бұрын

  @@charleskosina4799 The long winded way you speak is best for this kind of problem. The Lambert W Function is a mathematical one indeed. You can make researches about it

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  @davestillhere41698 күн бұрын

  Thank you very good.

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  @ChidexOnlineMathClass018 күн бұрын

  @@davestillhere4169 You are welcome. I promise to soon make the video solution of root of i

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  @benjyperez56068 күн бұрын

  Prime factors of 360 = 2x2x2x3x3x5 thus only way to get 4 consecutive factors from combinations of the 6 factors above are: 3x2x2x5x2x3 = 3x4x5x6

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  @ChidexOnlineMathClass018 күн бұрын

  @@benjyperez5606 Yeah

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  @davestillhere41699 күн бұрын

  I m looking forward to the explanation of the root of i :)

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  @ChidexOnlineMathClass019 күн бұрын

  No problem

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  @ChidexOnlineMathClass019 күн бұрын

  Please l am sorry for the statement " all solutions for x" Obviously x has infinitely many number of solutions.