Harvard University Admission Interview Tricks .✍️🖋️📘💙

[@key_board_x]

(x² - 2)² = x + 2
x⁴ - 4x² + 4 = x + 2
x⁴ - 4x² - x + 2 = 0 ← it would be interesting to have squares on the left side (because power 4 and power 2).
Let's rewrite the equation by introducing a variable "λ" which, if carefully chosen, will produce squares on the left side.
Let's tinker a bit with z⁴ as the beginning of a square: x⁴ = (x² + λ)² - 2λx² - λ²
x⁴ - 4x² - x + 2 = 0 → where: x⁴ = (x² + λ)² - 2λx² - λ²
(x² + λ)² - 2λx² - λ² - 4x² - x + 2 = 0
(x² + λ)² - [2λx² + λ² + 4x² + x - 2] = 0 → let's try to get a second member as a square
(x² + λ)² - [(2λ + 4).x² + x + (λ² - 2)] = 0 → a square into […] means that Δ = 0 → let's calculate Δ
Δ = (1)² - 4.[(2λ + 4).(λ² - 2)] → then, Δ = 0
(1)² - 4.[(2λ + 4).(λ² - 2)] = 0
4.[(2λ + 4).(λ² - 2)] = 1
8.[(λ + 2).(λ² - 2)] = 1
(λ + 2).(λ² - 2) = 1/8
λ³ - 2λ + 2λ² - 4 = 1/8
λ³ + 2λ² - 2λ - 4 - (1/8) = 0
λ³ + 2λ² - 2λ - (33/8) = 0
8λ³ + 16λ² - 16λ - 33 = 0
8λ³ + (4λ² + 12λ²) - 16λ - 33 = 0
8λ³ + 4λ² + 12λ² - 16λ - 33 = 0
(8λ³ + 4λ²) + 12λ² - 16λ - 33 = 0
2λ.(4λ² + 2λ) + 12λ² - 16λ - 33 = 0
2λ.(4λ² + 2λ) + 3.(4λ²) - 16λ - 33 = 0
2λ.(4λ² + 2λ) + 3.(4λ² + 2λ - 2λ) - 16λ - 33 = 0
2λ.(4λ² + 2λ) + 3.(4λ² + 2λ) - 6λ - 16λ - 33 = 0
2λ.(4λ² + 2λ) + 3.(4λ² + 2λ) - 22λ - 33 = 0
2λ.(4λ² + 2λ) - 22λ + 3.(4λ² + 2λ) - 33 = 0
2λ.(4λ² + 2λ - 11) + 3.(4λ² + 2λ - 11) = 0
(2λ + 3).(4λ² + 2λ - 11) = 0
First case: (2λ + 3) = 0 → 2λ + 3 = 0 → 2λ = - 3 → λ = - 3/2
Second case: (4λ² + 2λ - 11) = 0 ← we have already a value (λ = - 3/2) above, so we stop here
Restart
(x² + λ)² - [x².(2λ + 4) - x + (λ² - 2)] = 0 → where: λ = - 3/2
[x² + (- 3/2)]² - [x².(2.({- 3/2} + 4) - x + ({- 3/2}² - 2)] = 0
[x² - (3/2)]² - [x².(- 3 + 4) - x + ({9/4} - 2)] = 0
[x² - (3/2)]² - [x² - x + (1/4)] = 0
[x² - (3/2)]² - [x - (1/2)]² = 0 ← here's the second square → by applying: a² - b² = (a + b).(a - b)
[x² - (3/2) + x - (1/2)].[x² - (3/2) - x + (1/2)] = 0
[x² + x - (4/2)].[x² - x - (2/2)] = 0
(x² + x - 2).(x² - x - 1) = 0
First case: (x² + x - 2) = 0
x² + x - 2 = 0
Δ = (1)² - (4 * - 2) = 1 + 8 = 9
x = (- 1 ± 3)/2
First possibility: x = (- 1 + 3)/2 = 1
Second possibility: x = (- 1 - 3)/2 = - 2
Second case: (x² - x - 1) = 0
x² - x - 1 = 0
Δ = (- 1)² - (4 * - 1) = 1 + 4 = 5
Third possibility: x = (1 ± √5)/2

[@superacademy247]

Thanks for sharing your feedback and perspective. I appreciate your detailed explanation 🙏💕💯✅💪

[@prollysine]

(x+1)(x-2)(x^2+x-1)=0 ,