Oxford University Admission Interview Tricks

[@pierrettebalazut9407]

Enfin j'ai compris !
Merci

[@shmuelzehavi4940]

Another approach:
The original equation is:
x + y = 6 √xy
y can't be zero. Therefore, by dividing this equation by y we obtain:
x/y + 1 = 6 √(x/y)
We denote z = √(x/y) and obtain the following quadratic equation:
z^2 - 6z + 1 = 0
Or:
z^2 - 6z + 9 = (z - 3)^2 = 8
Therefore we obtain 2 real roots:
z = 3 ± 2 √2 = √(x/y)
By squaring this result we obtain:
x/y = 17 ± 12 √2

[@aapkiawaz605]

Good one.

[@shmuelzehavi4940]

Following the solution presented in this video clip, the quadratic equation to be solved is:
u^2 - 34u + 1 = 0
The classic solution is:
u = (34 ± √(34^2 - 4)) / 2
Therefore,
u = (34 ± √(34^2 - 2^2 )) / 2
= ( 34 ± √( (34 + 2) (34 - 2) ) ) / 2
= (34 ± √(36×32)) / 2
= (34 ± √(2 (6×4)^2) ) / 2
= (34 ± √(2×24^2)) / 2
= (34 ± 24√2) / 2
= 17 ± 12√2

[@RyanLewis-Johnson-wq6xs]

x+y=6Sqrt[xy] x/y=17±12Sqrt[2]

[@nasrullahhusnan2289]

x+y=6sqrt(xy)
Divide by sqrt(xy):
sqrt(x/y)+sqrt(y/x)=6
Let t=sqrt(x/y) --> sqrt(y/x)=1/t
Thus t+1/t=6 --> t²-6t+1=0
t=½[6±sqrt(32)]
=3±2sqrt(2)
x/y=t²
=17±12sqrt(2)