The case of the vanishing discriminant

Problem

Show that in $y^3+py+q=0$ with $D=0$ the equation has a double root and a simple root, and express both as a function of $q$.

Solution

In the depressed form the second-degree term is missing, so the sum of the three roots is $0$. If the double root is $s$ and the simple one is $t$, then $2s+t=0 \implies t=-2s.$ The product of the roots of $y^3+py+q$ equals $-q$, hence $s\cdot s\cdot t = -q \implies s^2(-2s)=-q \implies s^3=\frac{q}{2} \implies s=\sqrt[3]{\frac{q}{2}}.$ So the double root is $\boxed{y_1=y_2=\sqrt[3]{q/2}}$ and the simple one is $\boxed{y_3=-2\sqrt[3]{q/2}}$.

Consistency with Cardano. When $D=0$ the two radicands coincide: $-\dfrac q2\pm\sqrt D=-\dfrac q2$. The formula then gives $y = 2\sqrt[3]{-\frac q2} = -2\sqrt[3]{\frac q2} = y_3,$ that is, the principal value of Cardano’s formula returns precisely the simple root. Check on the example $y^3-3y+2=0$ (where $p=-3,q=2,D=0$): we have $s=\sqrt[3]{1}=1$ double and $t=-2$ simple, in agreement with $(y-1)^2(y+2)=0$. ∎

Links

Topics: Quadratic equations
Concepts: Cubic discriminant · Cubic equation
Methods: Cardano cubic
Skills: Proving
Exercise type: Proof