If 2x^2+4xy+y^2-12x-8y+15=0, show that x cannot lie between 1-(1/√2) and 1+(1/√2)

And that y cannot lie between 1 and 3. x and y are real.

Solve the quadratic in each of x and y separately.
In y:
y^2 +4(x-2) + 2x^2-12x+15=0
For y to exist, the discriminant must be greater than or equal to 0.
D = b^2 - 4ac
= 16(x-2)^2 - 4(2x^2-12x+15)
= 8(x^2 - 2x + 0.5) ≥ 0
x ≤ 1-(1/√2) and x ≥ 1+(1/√2)
If 1-(1/√2) < x < 1+(1/√2), D<0 and y doesn't exist.

Do the same for the quadratic in x:
2x^2+4(y-3)+y^2-8y+15=0
D = b^2-4ac
= 16(y-3)^2 - 8(y^2-8y+15)
= 8(y^2-4y+3) ≥ 0
y ≤ 1, y ≥ 3

homework?:p

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