Source: 1976 Euclid Part A Problem 4
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The points $(1,y_1)$ and $(-1,y_2)$ lie on the curve $y=px^2+qx+5$. If $y_1+y_2=14$, then the value of $p$ is

$\textbf{(A) } 2 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 2-q \qquad \textbf{(E) }\text{none of these}$
1. Given the points \((1, y_1)\) and \((-1, y_2)\) lie on the curve \(y = px^2 + qx + 5\), we can substitute these points into the equation to find expressions for \(y_1\) and \(y_2\).
   
   For the point \((1, y_1)\):
   \[
   y_1 = p(1)^2 + q(1) + 5 = p + q + 5
   \]

   For the point \((-1, y_2)\):
   \[
   y_2 = p(-1)^2 + q(-1) + 5 = p - q + 5
   \]

2. We are given that \(y_1 + y_2 = 14\). Substituting the expressions for \(y_1\) and \(y_2\) from step 1, we get:
   \[
   (p + q + 5) + (p - q + 5) = 14
   \]

3. Simplify the equation:
   \[
   p + q + 5 + p - q + 5 = 14
   \]
   \[
   2p + 10 = 14
   \]

4. Solve for \(p\):
   \[
   2p + 10 = 14
   \]
   \[
   2p = 14 - 10
   \]
   \[
   2p = 4
   \]
   \[
   p = 2
   \]

The final answer is \(\boxed{2}\).