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I have the solution, by the way. This is like a challenge. :)

y^2 = 1-sqrt{17/16-y}
(y^2-1)^2 = 17/16-y
solve...

1/2 ?

That's a good heuristic guess. It's correct. Why?

am i on the right track?

i think i accidentally dropped a negative sign

Yes, you did. Haha, I didn't notice. I just saw the polynomials and went, "k"

Not that it makes a difference. A mild algebraic rearrangement leads to the same problem.

cant we just substitute the value in y^2 = 1-sqrt{17/16-y} and see if it is TRUE OR NOT.

As squaring gives some unnecessary roots.

|dw:1349505055344:dw|

Only 1/2 satisfies that loop

That's correct, and answers the question, but I feel like I'm missing a technical detail here.

|dw:1349505240174:dw|

But whatever, if there was subtlety here, I'm still missing it. You got the answer.

\[2^3(2y)-2^3(2y)^2-(2y)^4-(2y)^0=0\]