0
$\begingroup$

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 $y$?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.