Can you Pass Pure Mathematics Entrance Exam from Oxford University ?

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  • Опубліковано 30 жов 2024

КОМЕНТАРІ • 9

  • @kareolaussen819
    @kareolaussen819 День тому +4

    First search for integer solutions. Since none is found try factorization into two quadratic terms with integer coefficients,
    (x^2 + px + q)(x^2 - px + r) =
    x^4 + (q+r-p^2)x^2 - (q-r)px + qr =
    x^4 -12x -5.
    We must have qr = -5 with q+r positive (equal to p^2). The only integer solution is q=5, r=-1, which gives p^2=4. This matches the condition that (q-r)p=12 when p=2. Hence,
    (x^2 + 2x +5)(x^2 -2x-1) = x^4 -12x -5.

  • @ShriH-d1o
    @ShriH-d1o 14 годин тому +1

    0:59 Let X^4 - 12X - 5 =(X^2 +aX+ p)(X^2 +bX + q) so
    a+b=0 ; pq= -5;
    p + q + ab = 0; aq + pb = - 12
    solving a = 2; b = - 2 ; p = 5; q = - 1 ...

  • @jorgevilas1603
    @jorgevilas1603 2 дні тому +2

    Beautiful exercise and very well resolved, thank you.

    • @superacademy247
      @superacademy247  2 дні тому

      Thanks for watching! 🙏😊I’m glad you found it helpful! 💯💖

  • @key_board_x
    @key_board_x 2 дні тому +1

    x⁴ - 12x - 5 = 0 ← it would be interesting to have 2 squares on the left side (because power 4)
    Let's rewrite the equation by introducing a variable "λ" which, if carefully chosen, will produce a square on the left side
    Let's tinker a bit with x⁴ as the beginning of a square: x⁴ = (x² + λ)² - 2λx² - λ²
    x⁴ - 12x - 5 = 0 → where: x⁴ = (x² + λ)² - 2λx² - λ²
    (x² + λ)² - 2λx² - λ² - 12x - 5 = 0
    (x² + λ)² - [2λx² + λ² + 12x + 5] = 0 → let's try to get a second member as a square
    (x² + λ)² - [(2λ).x² + 12x + (λ² + 5)] = 0 → a square into […] means that Δ = 0 → let's calculate Δ
    Δ = 12² - 4.[2λ * (λ² + 5)] → then, Δ = 0
    12² - 8λ.(λ² + 5) = 0
    8λ.(λ² + 5) = 12²
    λ.(λ² + 5) = 18 → we can observ an obvious solution
    λ = 2
    Restart
    (x² + λ)² - [(2λ).x² + 12x + (λ² + 5)] = 0 → where: λ = 2
    (x² + 2)² - [4x² + 12x + 9] = 0 ← we can see a square inside […]
    (x² + 2)² - (2x + 3)² = 0 → a² - b² = (a + b).(a - b)
    [(x² + 2) + (2x + 3)].[(x² + 2) - (2x + 3)] = 0
    (x² + 2 + 2x + 3).(x² + 2 - 2x - 3) = 0
    (x² + 2x + 5).(x² - 2x - 1) = 0
    First case: (x² + 2x + 5) = 0
    x² + 2x + 5 = 0
    Δ = 2² - (4 * 5) = 4 - 20 = - 16 = 16i²
    x = (- 2 ± 4i)/2
    → x = - 1 ± 2i
    Second case: (x² - 2x - 1) = 0
    x² - 2x - 1 = 0
    Δ = (- 2)² - (4 * - 1) = 4 + 4 = 8
    x = (2 ± √8)/2
    x = (2 ± 2√2)/2
    → x = 1 ± √2

  • @9허공
    @9허공 День тому

    Since there is no x^3, x^2 terms in the given equation,
    x^4 - 12x - 5 = (x^2 + a/2)^2 - a(x + b/2)^2 = x^4 - abx + (a^2 - ab^2)/4
    => -ab = -12 & (a^2 - ab^2)/4 = -5 => a^2 - ab^2 = a^2 - a(12/a)^2 = -20
    => a^3 + 20a - 144 = (a - 4)(a^2 + 4a + 36) = 0 => a = 4 => b = 3
    => x^4 - 12x - 5 = (x^2 + 2)^2 - 4(x + 3/2)^2 = (x^2 + 2)^2 - (2x + 3)^2 = (x^2 + 2x + 5)(x^2 -2x - 1)

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 2 дні тому

    (x^4)^2 ➖( 12)^2 ➖ 5={x^16 ➖ 144} ➖ 5=128 ➖ (5)^2={128 ➖ 25}=103 10^10^3.2^5^2^5^3 1^1^2^1^3.23 (x ➖ 3x+2).

  • @sorinescu123456789
    @sorinescu123456789 День тому

    Taking super small steps and writing every single detail and repeating the writing of everything makes it boring and unnerving to the highest degree. I'd rather give up math altogether than see another UA-cam clip by you. Not that you weren't correct. Just maddening...

    • @superacademy247
      @superacademy247  День тому +1

      When I move with speed avoiding trivial steps some viewers complain I'm too fast for them to understand . I think I need to strike a balance of all my viewership.