Given \( b + c = 12 \) and \( b^2 + c^2 = 74 \), we can find \( bc \) using:

["Finding the Product ( bc ) Given ( b + c = 12 ) and ( b^2 + c^2 = 74 )", "When working with two equations involving two variables, one powerful technique is using the identity that connects the sum, sum of squares, and product of two numbers. Given:", "[\nb + c = 12 \quad \ ext{and} \quad b^2 + c^2 = 74\n]", "there's a well-known algebraic identity that allows us to compute the product ( bc ) efficiently:", "[\n(b + c)^2 = b^2 + c^2 + 2bc\n]", "This identity rearranges to solve for ( bc ):", "[\nbc = \frac{(b + c)^2 - (b^2 + c^2)}{2}\n]", "### Step-by-Step Calculation", "1. Use the given sum:\n [\n b + c = 12 \Rightarrow (b + c)^2 = 12^2 = 144\n ]", "2. Plug in the known sum of squares:\n [\n b^2 + c^2 = 74\n ]", "3. Substitute into the identity:\n [\n bc = \frac{144 - 74}{2} = \frac{70}{2} = 35\n ]", "### Conclusion", "Thus, from the equations ( b + c = 12 ) and ( b^2 + c^2 = 74 ), we find that the product ( bc = 35 ). This method avoids solving complex quadratic equations directly, using only elementary algebraic identities for a fast and accurate result.", "This technique is particularly useful in algebra, geometry, and problem-solving contexts where symmetry in variables simplifies calculations. Always remember:\n[\nbc = \frac{(b + c)^2 - (b^2 + c^2)}{2}\n]\n— your go-to formula for finding the product when sum and sum of squares are known.", "---", "Keywords: find ( bc ), ( b + c = 12 ), ( b^2 + c^2 = 74 ), algebraic identity, sum of squares and products, solve equations, math trick, simplify expressions, algebra problem-solving."]