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A very tricky algebra problem from Stanford university admission exam
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Can you solve this? | iota maths problem | Oxford entrance exam questions
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Can you solve this? | iota maths problem | Oxford entrance exam question
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Nice exponential equation | Math Olympiad Question
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The Hardest Exam Question | Only 6% of students solved it correctly | Oxford entrance exam question
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Can you solve this? | iota maths problem | Oxford entrance exam question
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Can you solve this? | iota maths problem | Oxford entrance exam question
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The Hardest Exam Question | Only 6% of students solved it correctly
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Finding the value of x using two methods | Lambert's W function | Oxford exam question
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Math Olympiad Question | Find the value of x
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The Hardest Exam Question | Only 6% of students solved it correctly
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Nice Math Problem | Oxford entrance exam question
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A Very Interesting Harvard Entrance Exam Algebra Equation | How to solve for a,b & c
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A very tricky question from Oxford University Entrance Aptitude Test
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The Hardest Exam Question | Only 6% of students solved it correctly
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Hello my Wonderful family 😍😍😍 Trust you're doing fine 😊. •If you like this video about Math Olympiad Problem Solving. ~Please like and Subscribe to my channel. It helps me a lot. Thanks 🙏 •Oxford University Entrance Exam Question •Harvard University Entrance Examination •International Math Olympiad •Math Olympiad Questions •Mathematics Education •Math Problem Solving •Advanced Math Concepts •Ch...
Nice Algebra Equation | Math Olympiad Question
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Nice Algebra Equation | Math Olympiad Question
Nice algebra problem | Harvard entrance interview question
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Nice algebra problem | Harvard entrance interview question
Can you solve this? | iota maths problem | Oxford entrance exam question
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Can you solve this? | iota maths problem | Oxford entrance exam question
A very tricky algebra question from Oxford Entrance Aptitude Test
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A very tricky algebra question from Oxford Entrance Aptitude Test
A very tricky problem from Harvard Entrance Exam | How to find x
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A very tricky problem from Harvard Entrance Exam | How to find x
Nice Exponential Problem | Math Olympiad Question
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Nice Exponential Problem | Math Olympiad Question
Harvard University Admission Interview Tricks
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Harvard University Admission Interview Tricks
Oxford Entrance Exam Algebra Equation | Find k
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Oxford Entrance Exam Algebra Equation | Find k
A Very Interesting Exponential Math Olympiad Question | Find x
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A Very Interesting Exponential Math Olympiad Question | Find x
Nice Exponential Problem | cube roots of unity | Math Olympiad Question
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Nice Exponential Problem | cube roots of unity | Math Olympiad Question
A Nice Olympiad Trigonometric Exponential Equation | Find x
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A Nice Olympiad Trigonometric Exponential Equation | Find x
Entrance Exam | A tricky exponential problem
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Entrance Exam | A tricky exponential problem
Nice Olympiad Exponential Equation | Find "a"
Переглядів 1,3 тис.28 днів тому
Nice Olympiad Exponential Equation | Find "a"
Can you solve this? | iota maths problem | Oxford entrance exam question
Переглядів 2,2 тис.28 днів тому
Can you solve this? | iota maths problem | Oxford entrance exam question
A Nice Math Olympiad Exponential Equation
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A Nice Math Olympiad Exponential Equation

КОМЕНТАРІ

  • @segaranpillay7154
    @segaranpillay7154 5 годин тому

    stupid music always slows my thought

  • @popodori
    @popodori 6 годин тому

    Answer = 1-i, (1+i)^4 = -4, (1-i)^4 = -4i, (1+i)^2024 / (1-i)^2023 = (-4)^506*(1-i)/(-4i)^506 = (1-i)

  • @에스피-z2g
    @에스피-z2g 7 годин тому

    P=given formula P=[(1+i)/(1-i)]^2024(1-i) (1+i)/(1-i)=i i^2024=1 P=1-i

  • @ghamoz
    @ghamoz 8 годин тому

    Ma perché scrivete sempre i problemi ditti notevoli? Dovrebbero essere contenti sciuti se fai la n test d'entrata e se non li sai è giusto che stai fuori

  • @paulortega5317
    @paulortega5317 12 годин тому

    [(1 + i)/√2]⁸ = [(1 - i)/√2]⁸ = [(1 + i)/(1 - i)]⁸ = 1 so (1 + i)²⁰²⁴/(1 - i)²⁰²³ = (1 - i)•(1 + i)²⁰²⁴/(1 - i)²⁰²⁴ = (1 - i)•[[(1 + i)/(1 - i)]⁸]²⁵³ = (1 - i)•(1)²⁵³ = 1 - i

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 14 годин тому

    (2024i+^2024)/(2023 ➖i^2023)=(2024i^2024/(0+0 ➖ {i^0+i^0 ➖}=(2024i^2024)/(1i^1) (1^1^2^2i^1^2^12)/(1i^1) (1^1^i^2^2^6)/(1i^1) (i^1^12^3)/(1i^1) (i^2^3)/(1i^1) (i^2^3 ) (x ➖ 3ix+2i). I enjoyed avery education mathematics thank you.

  • @walterwen2975
    @walterwen2975 16 годин тому

    Oxford entrance exam question: [(1 + i)²⁰²⁴]/[(1 - i)²⁰²³] =? (1 + i)² = 1 + 2i + i² = 1 + 2i - 1 = 2i, (1 - i)² = 1 - 2i + i² = 1 - 2i - 1 = - 2i [(1 + i)²⁰²⁴]/[(1 - i)²⁰²³] = [(1 + i)²(1 + i)²⁰²²]/[(1 - i)(1 - i)²⁰²²] =[(1 + i)²/(1 - i)]{[(1 + i)/(1 - i)]²⁰²²} = [(1 + i)²/(1 - i)]{[(1 + i)²/(1 - i)²]¹⁰¹¹} = [(2i)/(1 - i)]{[(2i)/(- 2i)]¹⁰¹¹} = [(2i)/(1 - i)][(- 1)¹⁰¹¹] = (- 2i)/(1 - i) = [(- 2i)(1 + i)]/[(1 - i)(1 + i)] = [(- 2i)(1 + i)]/(1 - i²) = [(- 2i)(1 + i)]/2 = - i(1 + i) = - i - i² = 1 - i

  • @user-오리지날
    @user-오리지날 18 годин тому

    풀이가 틀렸다. 답이 1+¿될려면 분모가 1+¿가 되어야 한다.

  • @ahsgdf1
    @ahsgdf1 20 годин тому

    Nice, but I don't see an advantage in using the "trick" presented here over the direct calculation as follows: x^2 = (sqrt(2)-1)^2 = 2 - 2 sqrt(2) + 1 = 3-2 sqrt(2); x^4 = (x^2)^2 = (3-2 sqrt(2))^2 = 9 + 8 - 12 sqrt(2) = 17-12 sqrt(2); x^6 = x^2*x^4 = (3-2 sqrt(2))*(17-12 sqrt(2))= 99-70 sqrt(2), hence x^12=(x^6)^2 = (99-70 sqrt(2))^2 = 99^2+2*70^2 - 2*70*99 sqrt(2) = 19601-13860 sqrt(2) # Or, even shorter, x^12 = ((x^3)^2)^2.

  • @walterwen2975
    @walterwen2975 21 годину тому

    Oxford entrance exam question: √(- i) =? - i = (- 2i)/2 = (1 - 2i - 1)/2 = (1 - 2i + i²)/2 = [(1 - i)²]/2 = [(1 - i)/√2]² √(- i) = √{[(1 - i)/√2]²} = ± (1 - i)/√2 = ± (√2 - i√2)/2 Final answer: √(- i) = (√2 - i√2)/2 or √(- i) = (- √2 + i√2)/2

  • @Tommy_007
    @Tommy_007 22 години тому

    98% just like the problem and don't care how many failed.

  • @phillipmaxwellastrology2978
    @phillipmaxwellastrology2978 23 години тому

    Elegant! I wanted to multiply out top and bottom completely with a similar pairing trick to what you used but you showed that that would be cumbersome and unnecessary extra work!

  • @ianboard544
    @ianboard544 23 години тому

    If you think of -i in polar form it's a hell of a lot easier.

  • @NewsTopThe
    @NewsTopThe День тому

    Good solution 4:39

  • @JUGNU.C.MEHROTRA
    @JUGNU.C.MEHROTRA День тому

    I used method 2 to calculate this is my mind & I did it in approx . 2 min 🎉

  • @andreaslarsson4076
    @andreaslarsson4076 День тому

    Well, just visualize it. Angles are added. 270/2 = 135 = 90+45. So that’s -1+i right there. Lengths are multiplied so scale by sqrt(2). (-1+i)/sqrt(2). The other solution is just a mirror. I.e. -45 degrees.

  • @benardolivier6624
    @benardolivier6624 День тому

    I used a third method (basically the method you use to denest radicals for real numbers applied to complex numbers): just set sqrt(3-4i) = sqrt(x)-sqrt(y). When you square both sides you get 3-4i = x+y-2sqrt(xy), so x+y=3 and xy=-4, which gives an easy second degree polynomial to solve: x²-3x-4=0 or (x-4)(x+1)=0. This gives 2 solutions: x=4, y=-1 or x=-1, y=4. Taking the square roots you get sqrt(3-4i)=2-i or i-2.

  • @eamonhannon1103
    @eamonhannon1103 День тому

    You lost me around 6:51 . Where did you get x^2 + y^2 =25 You wrote something in the right that is not visible

  • @satrajitghosh8162
    @satrajitghosh8162 День тому

    √ ( 2 ^2 + i.^2 - 2 * 2 * i) = (2 - i), - (2 - i)

  • @camgere
    @camgere День тому

    Or you could just start from -i = e^i(-pi/2)

  • @petercrem-wl9ev
    @petercrem-wl9ev День тому

    Dragged out and thus tediously boring

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 День тому

    (x ➖ 1ix+1i).

  • @matthewkendall5235
    @matthewkendall5235 День тому

    Once you see the trick to get an x^2 = 1 - 2x relationship - you could just use the binominal theorem to go right to the answer... it's having the insight to get to this point to break the problem down to something that is just straight algebra. Nicely done!

  • @RexxSchneider
    @RexxSchneider День тому

    No. The convention is that √(i) denotes the principal value of the square root of i, which is (1+i)/√2. i = exp(πi/2) so √(i) = exp(πi/4) = cos(π/4) + i.sin(π/4) = 1/√2 + i/√2 = (1+i)/√2. The other square root of i is just -(1+i)/√2. There can only be two values for any square root, not the four that you imply. And please quit the ridiculous click-bait. It's not an "Oxford entrance exam question", whatever you imagine that to be.

  • @adrianojoaquin938
    @adrianojoaquin938 День тому

    I used (a^2 + 2ab + b^2)

  • @鄧貴文-v2t
    @鄧貴文-v2t День тому

    Although the answer is correct, I saw something wrong as x^12=根號2-1 at 9:44

  • @jackardor
    @jackardor День тому

    My question -- this is how I see it: Problem -- a+b=6sqrt(ab) Divide all by b -- a/b + b/b = 6sqrt(ab)/b Reduce -- a/b + 1 = 6sqrt(a/b) All to LHS - a/b -- 6sqrt(a/b) + 1 = 0 Let m = sqrt(a/b) -- m^2 - 6m + 1 = 0 Quadratic -- m = -(-6)/2 +/- sqrt((-6)^2-4(1)(1))/2 Reduce -- m = -3 +/- 2sqrt(2) Equivalence -- sqrt(a/b) = -3 +/- 2sqrt(2) Square both sides => a/b = 17 +/- 12sqrt2 What assumptions (besides that a & b are >0) do I need to make to solve it in this way?

  • @BN-hy1nd
    @BN-hy1nd 2 дні тому

    Brill. Method 2 preferred. Thanks

  • @renesperb
    @renesperb 2 дні тому

    If you consider the complex z-plane the answer is immediately clear : for the 1/3 -power you have to divide the corresponding angle by 3: π/2 * 1/3 = π/6 :hence the answer is cos π/6 +i*sin π/6 =1/2*(√3 + i). In your video you give the solution of z^3 = i, but the third root is uniquely defined, as the solution given here.

  • @borisjeud8464
    @borisjeud8464 2 дні тому

    👍

  • @laogui2425
    @laogui2425 2 дні тому

    using r*cis(θ) works well, I suggest, and gives at once sin^2(θ)=cos^2(θ) and 2r*sinθcosθ=1 so r = 1 and θ=m*pi/4 with m in (1,3,5,7)

    • @RexxSchneider
      @RexxSchneider День тому

      Not well enough, it seems. The expression exp(mπi/4) where m ∈ { 1, 3, 5, 7 } represents four values. Two of these (m=1, 5) represent the two square roots of i. The other two (m=3, 7) represnt the two square roots of -i. Only the one where m=1 represents the principal value of the square root of i denoted by √(i). And exp(πi/4) evaluates to (1+i)/√2.

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 2 дні тому

    (x^4 ➖ 256)={x^0+x^0+x^0+x^0x^0 ➖ x^0 ➖ 'x^0 ➖ x^0}=x^4 (x ➖ 4x+4).

  • @daniel_feiglin
    @daniel_feiglin 2 дні тому

    Trivial! Draw a picture: sqrt(i) is clearly (1+i)/sqrt(2). It's simple geometry.

  • @confusedmathematican69
    @confusedmathematican69 2 дні тому

    97% failed?bro is capping this shit easy

    • @arthursgarage6550
      @arthursgarage6550 2 дні тому

      That's what I was saying, there certainly is no way that high failed. Perhaps this is a perfidious statement. I'm alright at math, good enough that I can make things work, and I figured it out.

  • @beniocabeleleiraleila5799
    @beniocabeleleiraleila5799 2 дні тому

    you can do it by trigonometric form and moivre

  • @شعرکوتاه-ع7ظ
    @شعرکوتاه-ع7ظ 2 дні тому

    Tanks mr❤❤🎉

  • @شعرکوتاه-ع7ظ
    @شعرکوتاه-ع7ظ 2 дні тому

    Nice❤🎉

  • @RjKingie
    @RjKingie 3 дні тому

    Why make it so complex. It does not take any imagination to see that x=+2 is an answer.

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 3 дні тому

    {4x+4x ➖ }+{x+x ➖ }={8x^2+x^2}=8x^4 2^3x^2^2 1^1^1x1^2 1x^2 (x ➖ 2x+1) .

  • @satyapriyagogula8334
    @satyapriyagogula8334 3 дні тому

    Very good calculation.

  • @Zenadriel
    @Zenadriel 3 дні тому

    The exact result is 1 and I solved it in my head. By the way, I'm not a math genius! You just have to know the power laws and the third binomial formula! (√2-1)^12=(√2-1)^2*(√2-1)^2*(√2-1)^2*(√2-1)^2*(√2-1)^2*(√2-1)^2=1*1*1*1*1*1=1

    • @akaRicoSanchez
      @akaRicoSanchez 3 дні тому

      I have bad news for you: (√2-1)^2 is not the same as (√2-1)(√2+1)

    • @Zenadriel
      @Zenadriel 3 дні тому

      @@akaRicoSanchez Oh, I really got that wrong. The only strange thing is that I also get 1. That's probably why I didn't look so closely.

  • @jesper856
    @jesper856 3 дні тому

    Ugh that loud annoying music!!!

  • @MgtowRubicon
    @MgtowRubicon 3 дні тому

    Stupid music.

  • @ELALFAMentalidad
    @ELALFAMentalidad 3 дні тому

    Could you please lower the volume of the music?

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 3 дні тому

    (3 ➖ 1)^2^6 (3 ➖1)2^2^3 (1 ➖ 1)1^2^3 ()2^3 (x ➖ 3x+2).

  • @ephraimgarrett4727
    @ephraimgarrett4727 3 дні тому

    (1,4142 -1)^12 0.4142^12 I guess Oxford isn't my cup of tea. 😄😄😄

  • @radupopescu9977
    @radupopescu9977 3 дні тому

    I would to like this: ((2)^(1/2)-1)^12=x ((2)^(1/2)-1)=x^12 solve 0.4142...=x^12 which have 2 real and 8 complex solution. Choose the positive real solution. Your way is too complicated. You course you need a calculator, but it is way much is easy.

  • @dd-di3mz
    @dd-di3mz 3 дні тому

    What a complete mess.

  • @renesperb
    @renesperb 4 дні тому

    If you write i = exp[ i π/2] ,then it follows that √ i = exp[ i π/4]= cos π/4 +i sin π/4 = √2/2*(1+i) . Note : there is only one value for √ i .There would be two values if you solve z^2 = i .

  • @Zbigniew-b3u
    @Zbigniew-b3u 4 дні тому

    ... coś tu spierdoliłeś !!! według ciebie (sqrt(2)-1)^12 to około 0,188, przecież to bzdura !!!!!!!!!