We are given the quadratic equation:

$3x^2 + px + 3 = 0$

with the condition that one of the roots is the square of the other. Let's denote the roots of the quadratic equation by $\alpha$ and $\beta$, where we are told that $\alpha = \beta^2$ (i.e., one root is the square of the other).

Step 1: Use Vieta’s Formulas

For any quadratic equation $ax^2 + bx + c = 0$, Vieta's formulas give us the relationships between the coefficients and the roots:

• Sum of the roots: $\alpha + \beta = -\frac{p}{3}$
• Product of the roots: $\alpha \beta = \frac{3}{3} = 1$

Step 2: Substitute the relationship $\alpha = \beta^2$ into the formulas

From the product of the roots, we have:

$\alpha \beta = 1$

Substituting $\alpha = \beta^2$ into this equation:

$\beta^2 \cdot \beta = 1 \implies \beta^3 = 1$

This gives:

$\beta = 1 \quad \text{(since the real root of } \beta^3 = 1 \text{ is } \beta = 1)$

Step 3: Find $\alpha$

Since $\alpha = \beta^2$, we now know:

$\alpha = 1^2 = 1$

Step 4: Use the sum of the roots to find $p$

From the sum of the roots, we have:

$\alpha + \beta = -\frac{p}{3}$

Substituting $\alpha = 1$ and $\beta = 1$ into this equation:

$1 + 1 = -\frac{p}{3} \implies 2 = -\frac{p}{3}$

Multiplying both sides by 3:

$p = -6$

Thus, the value of $p$ is $-6$.