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u and v are interchangeable, obviously. This suggests two solutions for each with u₁ = v₂ and u₂ = v₁. So we expect a quadratic function:
u + v = 8
uv = 8

u + v = uv = 8
u = 8 − v
(8 − v) v = 8
−v² + 8v − 8 = 0
v² − 8v + 8 = 0
v₁,₂ = 4 ± √(16 − 8)
v₁ = 4 − 2√2 ∨ v₂ = 4 + 2√2
u₁ = v₂ = 4 + 2√2 ∨ u₂ = v₁ = 4 − 2√2

Check:
u₁ + v₁ = 4 + 2√2 + 4 − 2√2 = 8
u₁v₁ = (4 + 2√2) (4 − 2√2) = 4² − (2√2)² = 16 − 8 = 8

𝕃 = {(4 − 2√2, 4 + 2√2), (4 + 2√2, 4 − 2√2)}

Good video. This was real helpful.

by Vieta theorem，
it's able to directly conclude that u and v is two roots of the equation z^2 - 8z + 8 = 0.

One thing I like to tell you , whenever you are teaching math put the division line where equal sign is there. I am teaching math since 1962. I have seen that most of the teacher are unaware of it.

How do b and c enter the equation?

1+7 or2+6 or 3+5 or 4+4 or 5+3 or 6+2 or 7+1 or 9_1 like this many.