Two roots of a cubic equation

Question

Question:

$x^3+px+q$ has two distinct zeroes. Then prove that $4pq+27q^2=0$.

---

My attempt:

Let the roots be $\alpha$ and $\beta$.

Then the equation $x^3+px+q=(x-\alpha)^2(x-\beta)$.

This gives

$2\alpha+\beta=0$

$\alpha^2+2\alpha\beta=p$

$\alpha^2\beta=q$

Substituting $\beta=-2\alpha$ gives

$-3\alpha^2=p$

$-2\alpha^3=q$

Now, if $4pq+27q^2=0$, then either $q=0$, which is trivial, or $4p+27q=0$.

Substituting values,

$-12\alpha^2-27\alpha^3=0$, which gives $\alpha=\frac49$.

---

This feels wrong, and I can't properly prove it anyway.

Please help.

---

Edit: I am told that the correct formula is $4p^3+27q^2=0$. How can this be proved?

Edit 2: I have proved this for $p^3$ instead of $pq$. Thanks, everyone!!!

Answer 1

The polynomial has a double root, so that the first derivative cancels at one of them and

$$3x^2+p=0,$$

$$x=\pm\sqrt{-\frac p3}.$$

When we plug this value in the polynomial,

$$\mp\sqrt{-\frac p3}\frac p3\pm p\sqrt{-\frac p3}+q=\pm\frac23p\sqrt{-\frac p3}+q=0$$

or

$$-\frac4{27}p^3=q^2.$$