Stanford University Entrance Exam | x^4−12x−5=0

[@ldeans5620]

Awesome

[@superacademy247]

Thanks for visiting my channel 🤩👏😎🙏

[@9허공]

another method
since there are no x^3 and x^2 terms in the given equation,
let 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
=> ab = 12 , a^2 - ab^2 = a^2 - b(ab) = a^2 - 12a = -20
=> a^2 - 12(12/a) + 20 = 0 => 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 + 2x + 5)(x^2 - 2x - 1) = 0
=> x = { -1 ± 2i, 1 ± √2 }

[@superacademy247]

Thanks for your nice method. You've shed light 💡😍🔥🤩💕

[@key_board_x]

x⁴ - 12x - 5 = 0 ← it would be interesting to have a perfect square → let's define x⁴ as the beginning of a square
(x² + λ)² - 2x²λ - λ² - 12x - 5 = 0
(x² + λ)² - [2x²λ + λ² + 12x + 5] = 0
(x² + λ)² - [2x²λ + 12x + λ² + 5] = 0 ← memorize this equation
2x²λ + 12x + λ² + 5 = 0 → if we want to have a square, Δ = 0 → let's define Δ
Δ = (12)² - 4.[2λ.(λ² + 5)] → then Δ = 0
144 - 4.[2λ.(λ² + 5)] = 0
4.[2λ.(λ² + 5)] = 144
2λ.(λ² + 5) = 36
λ.(λ² + 5) = 18
λ³ + 5λ - 18 = 0
λ³ + (2λ² - 2λ²) + (9λ - 4λ) - 18 = 0
λ³ + 2λ² - 2λ² + 9λ - 4λ - 18 = 0
λ³ + 2λ² + 9λ - 2λ² - 4λ - 18 = 0
(λ³ + 2λ² + 9λ) - (2λ² + 4λ + 18) = 0
λ.(λ² + 2λ + 9) - 2.(λ² + 2λ + 9) = 0
(λ² + 2λ + 9).(λ - 2) = 0
First case: (λ² + 2λ + 9) = 0
λ² + 2λ + 9 = 0
λ = (2)² - (4 - 9) = 4 - 36 = - 32 ← no real solution, because negative
Second case (λ - 2) = 0
λ - 2 = 0
λ = 2
Recall the memorized equation
(x² + λ)² - [2x²λ + 12x + λ² + 5] = 0 → now, we know that λ = 2 to get a perfect square
(x² + 2)² - [4x² + 12x + 4 + 5] = 0
(x² + 2)² - [4x² + 12x + 9] = 0
(x² + 2)² - (2x + 3)² = 0 → recall: 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

[@билал-ж2к]

Шедевр!!!

[@mrsmiley631]

Gold star!

[@lananhnguyen5899]

stanford university entrance exam and a secondary student is doing it right now

[@superacademy247]

Thanks for your feedback. I'm glad you found it helpful 🙏🤩🤩🙏😎💕

[@MrRichie444]

z^4 - 12z - 5 = 0 : First of all, this equation is unsolvable using the normal method, so we might need to break the z^4 down, so we get z^4 = (z^2 + 1)^2 -1-2z^2 = z^4 + 1 + 2z^2 - 1 - 2z^2 = z^4 ||rewriting the equation, we get || (z^2 + 1)^2 - 1 - 2z^2 - 12z - 5 = 0 || splitting the equation into two we get (z^2 + 1)^2 - 1 = 0 and 2z^2 + 12z + 5 = 0 ...Now the equation is solvable. I leave it to the kids to decompose it.

[@ruany91719]

Pardon me, how to know that its delta = 0 ? at 4:10 of this video

[@superacademy247]

It's a brute force rule to make the equation a perfect square

[@sergeilyubski852]

agree it is strange .