Question 501818
2(y)^1/9 - 3 (y)^1/18 + 1 = 0

I've got it broken down to
(2y^1/18 -1)(y^1/18 - 1) = 0
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To avoid working with fractional exponents, let x = y^(1/18)
Then we can rewrite the equation as:
{{{2x^2 - 3x + 1 = 0}}}
This can be factored as:
(2x-1)(x-1) = 0
The solutions are x = 1/2, x = 1
Therefore 1 = y^(1/18) -> y = 1
For the other root, 1/2 = y^(1/18)
To solve for y raise both sides to the 18th power:
(1/2)^18 = y
So the two roots are 1 and (1/2)^18