y^2 - 2xy - 16 = 0
y = ??
Can't seem to express y in terms of x
y^2 - 2xy - 16 = y^2 - 2xy + (x^2 - x^2) - 16
= (y^2 - 2xy + x^2) - x^2 - 16
= (y-x)^2 - x^2 - 16 = 0
Adding x^2+16 to both sides, you get:
(y-x)^2 = x^2 + 16
Taking the square root of both sides, you get:
y-x = +/- SQRT(x^2 + 16)
Adding x to both sides, you get:
y = x +/- SQRT(x^2+16)
Yes. The text uses it here but I was still confused by the reversal of the common variables and also the way x figures into the quadratic. Before I worked it though, I decided I ought to be able to manipulate the equation algebraicly so that y is expressed in terms of x...the first answer here does that nicely.