Cambridge University Entrance Exam Tricks Probably Never Knew Existed!✍️🖋️📘💙

[@9허공]

DOMAIN of x for real solutions is 1/7 ≤ x < 3 or x > 4
from 4:36 f(x) = x^4 - 7x^3 + 12x^2 - 7x + 1,
since f(1) = 0 , f(x) = (x - 1)g(x) = (x - 1)(x^3 - 6x^2 + 6x - 1) using SDM
since g(1) = 0 , f(x) = (x - 1)g(x) = (x - 1)(x -1)(x^2 - 5x + 1) using SDM
x = { 1, (5 + √21)/2, (5 - √21)/2 } within the domain

[@에스피-z2g]

Excellent

[@mmk9456aust]

x⁴-7x²+6x²+6x²+1=0
x²(x-1)(x-6)+(6x-1)(x-1)=0
either x=1 or
x³-6x²+6x-1=0
(x³-1)-6x(x-1)=0
(x-1)(x²+x+1)-6x(x-1)=0
(x-1)(x²-5x+1)=0
either x=1
or x =(5±√21)/2

[@key_board_x]

Condition
[27/(x - 4)] - [20/(x - 3)] ≥ 0
[27.(x - 3) - 20.(x - 4)] / [(x - 4).(x - 3)] ≥ 0
[27x - 81 - 20x + 80] / [x² - 3x - 4x + 12] ≥ 0
(7x - 1) / (x² - 7x + 12) ≥ 0
(7x - 1) ← root: 1/7
x² - 7x + 12 → Δ = 49 - 48 = 1
x = (7 ± 1)/2
x = 4 ← root
x = 3 ← root
x - ∞ 1/7 3 4 +∞
(7x - 1) - 0 + + +
(x² - 7x + 12) + + 0 - 0 +
sign - 0 + II - II +
x ⋲ [1/7 ; 3 [ U ] 4 ; +∞ [
x = √{ [27/(x - 4)] - [20/(x - 3)] }
x = √{ [27.(x - 3) - 20.(x - 4)] / [(x - 4).(x - 3)] }
x = √{ [27x - 81 - 20x + 80] / [x² - 3x - 4x + 12] }
x = √[(7x - 1)/(x² - 7x + 12)]
x² = (7x - 1)/(x² - 7x + 12)
x².(x² - 7x + 12) = 7x - 1
x⁴ - 7x³ + 12x² = 7x - 1
x⁴ - 7x³ + 12x² - 7x + 1 = 0
(x⁴ + 12x² + 1) - (7x³ + 7x) = 0 ← you can see that (x = 1) is an obvious solution
x⁴ - 7x³ + 12x² - 7x + 1 = 0 → so, we can factorize (x - 1)
(x - 1).(x³ + βx² + λx - 1) = 0 → you expand
x⁴ + βx³ + λx² - x - x³ - βx² - λx + 1 = 0 → you group
x⁴ + (β - 1).x³ + (λ - β).x² - x.(1 + λ) + 1 = 0 → we compare with: x⁴ - 7x³ + 12x² - 7x + 1 = 0
For x³ → (β - 1) = - 7 → β = - 6
For x² → (λ - β) = 12 → λ = 12 + β → λ = 6
For x → (1 + λ) = 7 → λ = 6 (of course because above)
(x - 1).(x³ + βx² + λx - 1) = 0 → you it becomes
(x - 1).(x³ - 6x² + 6x - 1) = 0
(x - 1).[x³ - 1 - (6x² - 6x)] = 0
(x - 1).[(x³ - 1³) - 6x.(x - 1)] = 0 → recall: a³ - b³ = (a - b)(a² + ab + b²)
(x - 1).[(x - 1).(x² + x + 1) - 6x.(x - 1)] = 0
(x - 1).[(x - 1).(x² + x + 1 - 6x)] = 0
(x - 1).(x - 1).(x² - 5x + 1) = 0
(x - 1)².(x² - 5x + 1) = 0
First case: (x - 1) = 0
x - 1 = 0
→ x = 1
Second case: (x² - 5x + 1) = 0
x² - 5x + 1 = 0
Δ = (- 5)² - (4 * 1) = 25 - 4 = 21
→ x = (5 ± √21)/2