Checking a root of the cubic

Problem

Verify that $y^3+3y-4=0$ has $y=1$ as a root. Compute $D$ and confirm using Cardano’s formula.

Solution

Substituting $y=1$: $1+3-4=0$, so $y=1$ is a root.

Here $p=3$, $q=-4$. The discriminant is $D=\Bigl(\frac{q}{2}\Bigr)^2+\Bigl(\frac{p}{3}\Bigr)^3 = 4 + 1 = 5 > 0,$ hence there is a single real root. Cardano’s formula gives $y = \sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}.$ Since $2-\sqrt{5}<0$, the second cube root is negative. Noticing that $\Bigl(\dfrac{1+\sqrt5}{2}\Bigr)^3 = 2+\sqrt5$ and $\Bigl(\dfrac{1-\sqrt5}{2}\Bigr)^3 = 2-\sqrt5$, the sum equals $y = \frac{1+\sqrt5}{2}+\frac{1-\sqrt5}{2} = \boxed{1},$ in agreement with the direct check.

Links

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