To find the value of $p$ so that a pair of quadratic equations has a common root, let's generalize the problem. Assume the two quadratic equations are:

1. $ax^2 + bx + c = 0$
2. $px^2 + qx + r = 0$

Let the common root of both equations be $\alpha$. This means that $\alpha$ satisfies both equations. Therefore, we have:

$a\alpha^2 + b\alpha + c = 0 \quad \text{(Equation 1)}$ $p\alpha^2 + q\alpha + r = 0 \quad \text{(Equation 2)}$

To solve this problem, follow these steps:

Step 1: Express one root of each equation using the quadratic formula.

Let the roots of the first equation be $\alpha$ and $\beta$. The quadratic formula gives: $\alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Similarly, let the roots of the second equation be $\alpha$ and $\gamma$. The quadratic formula for the second equation is: $\alpha, \gamma = \frac{-q \pm \sqrt{q^2 - 4pr}}{2p}$

Step 2: Set up the relationship between the roots.

Since both equations share a common root $\alpha$, let's assume that the sum and product of the roots for both equations are related. Using Vieta's formulas:

• For the first equation $ax^2 + bx + c = 0$:

  • Sum of the roots: $\alpha + \beta = -\frac{b}{a}$
  • Product of the roots: $\alpha \beta = \frac{c}{a}$

• For the second equation $px^2 + qx + r = 0$:

  • Sum of the roots: $\alpha + \gamma = -\frac{q}{p}$
  • Product of the roots: $\alpha \gamma = \frac{r}{p}$

Step 3: Compare the equations and find the value of $p$.

If $\alpha$ is a common root, we can use the following strategy: eliminate $\alpha$ from the equations by subtracting one from the other. This will allow us to determine a relationship between $a$, $b$, $c$, $p$, $q$, and $r$.

However, without specific forms for the equations or additional constraints, this becomes a system of equations problem, typically solved with the help of simultaneous equations.

To assist further, could you please provide the specific pair of quadratic equations you're working with?

Once the specific equations are given, we can determine the value of $p$ precisely.