Alice and Bob were asked to solve the equation x^2+px+q=0. Alice obtai

niks18 's solution is characteristically elegant. This way is more work, but very doable.

Alice's and Bob's solutions were incorrect, but the equations resulting from the solutions do have a coefficient or constant that matches what we need for the original.

Key variables are capitalized, thus:
\(x^2+px+q=0\)
\(x^2+Px+Q=0\)

As solutions to the above, Alice incorrectly found \(x=1\) and \(x=4\).
She obtained the solution set of the equation \(x^2+Px+s=0\)
Her solutions' resultant equation nonetheless gives us coefficient P for the original equation.
(Her constant, \(s\), is wrong. That letter must be \(q\).)

Solutions of \(x=1\) and \(x=4\) mean equation is
\((x-1)(x-4) = x^2 -5x + 4\)
That correlates with \(x^2+Px+s=0\)
\(P = (p = -5)\) in original equation

Bob's solution set \(x=-2\) and \(-3\) was also incorrect. But the solutions' resultant equation did yield constant Q.

\(x^2+tx+Q=0\)
(His [t] coefficient is wrong. It is not \(p\).)

Solutions \(x=-2\) and \(x= -3\) yield equation
\((x + 2)(x + 3) = x^2 + 5x + 6\)
That correlates with \(x^2+tx+Q=0\)
\(Q = (q = 6)\) in original equation

Plug the correct \(p\) and \(q\) values into the original equation. Solve.

\(x^2 - 5x + 6 = 0\)

\((x - 2)(x - 3) = 0\)

\(x = 2\) and \(x = 3\)
are the solutions of the original equation

The answer is C (2,3)