If a + b + c = 14 and a^2 + b^2 + c^2 = 96, find the value of ab + bc + ca using the standard identity that links sums and squares.

Introduction / Context:This question targets a staple algebra identity connecting the square of a sum to the sum of squares and pairwise products. It is frequently used to compute ab + bc + ca when you know a + b + c and a^2 + b^2 + c^2. Mastery of this identity saves time in many contest and exam problems.

Given Data / Assumptions:

• a + b + c = 14
• a^2 + b^2 + c^2 = 96
• Real numbers (no special constraints beyond this are required).

Concept / Approach:Use the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca). Rearranging immediately yields ab + bc + ca in one step. This avoids solving for individual variables, which is unnecessary and inefficient here.

Step-by-Step Solution:Compute (a + b + c)^2 = 14^2 = 196.Apply identity: 196 = 96 + 2(ab + bc + ca).Thus 2(ab + bc + ca) = 196 − 96 = 100.Therefore ab + bc + ca = 100 / 2 = 50.

Verification / Alternative check:Sanity check: ab + bc + ca = 50 is consistent with many possible triples; for example, symmetric choices near the mean 14/3 will produce reasonable non-negative pairwise sums.

Why Other Options Are Wrong:

• 51 / 55 / 60 / 65: These are common arithmetic slips (e.g., squaring 14 incorrectly or subtracting 96 from 196 wrongly).

Common Pitfalls:Forgetting the factor 2 in the identity; attempting to solve for a, b, and c individually; mixing up a^2 + b^2 + c^2 with (a + b + c)^2.

Final Answer:50