Question:

Step 1: Understand the constraint on irrational roots. Since $p(x)$ and $q(x)$ both have integer coefficients and share an irrational root $r$, we must consider the conjugate root theorem. For any polynomial with integer coefficients, if an irrational number $r$ is a root, then its conjugate $\bar{r}$ must also be a root of that polynomial. Step 2: Determine the common roots of both polynomials. Since both $p(x)$ and $q(x)$ have integer coefficients and share the irrational root $r$, the conjugate $\bar{r}$ must also be a root of both polynomials. Therefore, both polynomials must have the same pair of roots: $r$ and $\bar{r}$. Step 3: Express both polynomials in factored form. We can write: $$p(x) = 51(x - r)(x - \bar{r})$$ $$q(x) = 3(x - r)(x - \bar{r})$$ The leading coefficients (51 and 3) are determined by the given forms of $p(x)$ and $q(x)$. Step 4: Expand the common quadratic factor. Let us denote the common quadratic factor as: $$(x - r)(x - \bar{r}) = x^2 - (r + \bar{r})x + r\bar{r}$$ Since $r$ and $\bar{r}$ are conjugate irrational roots, their sum $r + \bar{r}$ and product $r\bar{r}$ are both rational numbers. Let: $$(x - r)(x - \bar{r}) = x^2 + px + q$$ where $p$ and $q$ are rational numbers. Step 5: Find the constant terms $c$ and $a$. Expanding $p(x)$: $$p(x) = 51(x^2 + px + q) = 51x^2 + 51px + 51q$$ Therefore, $c = 51q$. Expanding $q(x)$: $$q(x) = 3(x^2 + px + q) = 3x^2 + 3px + 3q$$ Therefore, $a = 3q$. Step 6: Calculate the ratio $\dfrac{c}{a}$. $$\dfrac{c}{a} = \dfrac{51q}{3q} = \dfrac{51}{3} = 17$$ The value of $\dfrac{c}{a}$ is **17**. The correct answer is **Option 2: 17**.