What is the value of A such that x2 − 11x + A and x2 − 14x + 2A have a common…

When two quadratic equations in the same variable share a common linear factor, the shared root satisfies both equations simultaneously. Subtracting one equation from the other eliminates the squared term, leaving a linear relation between that common root and the unknown parameter — substituting this relation back into either original quadratic then pins down the parameter's value.

1. Let the common root be k. Then k2 − 11k + A = 0 …(i) and k2 − 14k + 2A = 0 …(ii).

2. Subtract (ii) from (i): (k2 − 11k + A) − (k2 − 14k + 2A) = 3k − A = 0, so k = A/3.

3. Substitute k = A/3 into (i): (A/3)2 − 11(A/3) + A = 0.

4. Multiply through by 9: A2 − 33A + 9A = 0, i.e. A2 − 24A = 0.

5. Factor: A(A − 24) = 0, giving A = 0 or A = 24. A = 0 makes the constant term of both expressions vanish (x2 − 11x and x2 − 14x), so they trivially share the factor x regardless of the specific coefficients 11 and 14 — this doesn't test the coefficient relationship the problem is built on, so the meaningful, non-trivial value is A = 24.

With A = 24, the equations become x2 − 11x + 24 = 0 (roots 3 and 8) and x2 − 14x + 48 = 0 (roots 6 and 8) — both share the root 8, confirming a common factor (x − 8), so A = 24 is correct.