I'm trying to work through the explanation of Ferrari's Solution of the Quartic.

The arbitrary variable: $y$ is introduced to the depressed quartic in the link's step 3, yielding a right side of: $$(\alpha + 2y)u^2 - \beta u + (y^2 + 2\alpha y + \alpha^2 - \gamma)$$

Then the statement is then made that:

As the value of $y$ may be arbitrarily chosen, we will choose it in order to get a perfect square in the right-hand side. This implies that the discriminant in $u$ of this quadratic equation is zero, that is $y$ is a root of the equation

Somehow that yields the equation: $$(-\beta)^2 - 4(2y + \alpha)(y^2 + 2\alpha y + \alpha^2 - \gamma) = 0$$

I need help seeing what was done to achieve that jump. Can someone explain to me how we just decided that this equation could be used to solve for $y$?


We can rewrite the equation as $$(\alpha+2y) \left (u^2 + \frac{-\beta}{\alpha+2y} u + \frac{y^2+2\alpha y + \alpha^2 - \gamma^2}{\alpha+2y}\right ).$$ In order for $$\left (u^2 + \frac{-\beta}{\alpha+2y} u + \frac{y^2+2\alpha y + \alpha^2 - \gamma^2}{\alpha+2y}\right )\qquad(*)$$ to be a perfect square we must have that $$\left(\frac{-\beta}{2(\alpha+2y)}\right)^2 = \frac{y^2+2\alpha y + \alpha^2 - \gamma^2}{\alpha+2y}$$ or equivalently $$(-\beta)^2- 4(\alpha+2y)(y^2+2\alpha y + \alpha^2 - \gamma^2)=0.$$ This means that $(*)$ will be a perfect square whenever $y$ is a solution to this last equation.


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