Stanford University Entrance Interview Tricks | Find x=?

[@yuusufliibaan1380]

❤❤❤ thanks 👍🙏💯😊

[@superacademy247]

💪💪💪😍🥰🤩💡💕🔥

[@musicsubicandcebu1774]

(A²)² - (B²)² = 16² - 4²

[@RealQinnMalloryu4]

(28+4/x^4) ➖ (20+4/x^4)={32/x^4 ➖ 24/x^4 }=8/{x^0+x^0 ➖ }=8/x^1=8 (x ➖ 8x+8).

[@himo3485]

7+1/x=A 5+1/x=B
A⁴ - B⁴ = 240 (A2 + B2)(A + B)(A - B) = 240 = 20 * 6 * 2
A + B = 6 A - B = 2 2A = 8 A = 4 B = 2
7+1/x=4 5+1/x=2 1/x = -3 x = -1/3

[@peterotto712]

Geht es noch umständlicher?

[@prollysine]

let u=5+1/x , (2+u)^4-u^4=240 , we get , (u-2)(u^2+5u+12)=0 , u=2 , 5+1/x = 2 , 1/x = -3 , x= -1/3 ,
u^2+5u+12=0 , u=(-5+/-V(25-48))/2 , u=(-5+/-i*V(23))/2 , x= 2/(-15+i*V23) , 2/(-15-i*V23) ,
test , (7-3)^4-(5-3)^4=256-16 , --> 240 , OK ,

[@kd8opi]

There aren’t specific math questions you need to answer to get into Stanford or any other American University. Admissions are based on high school grades, national aptitude tests like the SAT, letters of recommendation and essays. I don’t know why you keep posting these questions as entrance tests to get into Ivy League schools, especially when half of students enrolled aren’t in science or math related fields.

[@ManjulaBrahmachari]

Mans gotta get his bread somehow

[@kd8opi]

@@ManjulaBrahmachari he can get his bread by doing math problems. Doesn’t have to lie about him and say they’re part of some entry exam when they’re not.

[@key_board_x]

[7 + (1/x)]⁴ - [5 + (1/x)]⁴ = 240 → let: a = 6 + (1/x)
→ 7 + (1/x) = a + 1
→ 5 + (1/x) = a - 1
(a + 1)⁴ - (a - 1)⁴ = 240
[(a + 1)²]² - [(a - 1)²]² = 240 → recall: a² - b² = (a + b).(a - b)
[(a + 1)² + (a - 1)²].[(a + 1)² - (a - 1)²] = 240
[(a + 1)² + (a - 1)²].[(a + 1) + (a - 1)].[(a + 1) - (a - 1)] = 240
[a² + 2a + 1 + a² - 2a + 1].[a + 1 + a - 1].[a + 1 - a + 1] = 240
[2a² + 2].[2a].[2] = 240
2.(a² + 1) * 2a * 2 = 240
2.(a² + 1) * 2a = 120
2.(a² + 1) * a = 60
(a² + 1) * a = 30
a³ + a = 30
a³ + a = 27 + 3
a³ - 27 + a - 3 = 0
(a³ - 3³) + (a - 3) = 0 → recall: p³ - q³ = (p - q).(p² + qp + q²)
(a - 3).(a² + 3a + 9) + (a - 3) = 0
(a - 3).(a² + 3a + 9 + 1) = 0
(a - 3).(a² + 3a + 10) = 0
First case: (a - 3) = 0
→ a = 3
Second case: (a² + 3a + 10) = 0
a² + 3a + 10 = 0
Δ = (3)² - (4 * 10) = 9 - 40 = - 31 = 31i²
→ a = (- 3 ± i√31)/2
First solution: a = 3
Recall: a = 6 + (1/x)
1/x = a - 6
1/x = 3 - 6
1/x = - 3
→ x = - 1/3
Second solution: a = (- 3 + i√31)/2
Recall: a = 6 + (1/x)
1/x = a - 6
1/x = [(- 3 + i√31)/2] - 6
1/x = (- 3 + i√31 - 12)/2
1/x = (- 15 + i√31)/2
x = 2/(- 15 + i√31)
x = 2.(- 15 - i√31)/[(- 15 + i√31).(- 15 - i√31)]
x = 2.(- 15 - i√31)/[225 - 31i²]
x = 2.(- 15 - i√31)/256
→ x = (- 15 - i√31)/128
Third solution: a = (- 3 - i√31)/2
Recall: a = 6 + (1/x)
1/x = a - 6
1/x = [(- 3 - i√31)/2] - 6
1/x = (- 3 - i√31 - 12)/2
1/x = (- 15 - i√31)/2
x = 2/(- 15 - i√31)
x = 2.(- 15 + i√31)/[(- 15 - i√31).(- 15 + i√31)]
x = 2.(- 15 + i√31)/[225 - 31i²]
x = 2.(- 15 + i√31)/256
→ x = (- 15 + i√31)/128

[@superacademy247]

Thanks for detailed and resourceful explanation 💪💪💪

[@Rohmat-oo3pw]

(7+1/x)⁴+(5+1/x)⁴ =240=4⁴-2⁴. 7+1/x=4. 1/x=4-7. 1/x=-3 x=-1/3. 5+1/x=2. 1/x=2-5 1/x=-3. X=-1/3. Ok