Solution: We are given $ p + q = 5 $ and $ pq = 6 $. We use the identity:
["Solving $ p + q = 5 $ and $ pq = 6 $: A Powerful Algebraic Identity Explained", "When faced with a system of equations like\n$$\np + q = 5 \quad \ ext{and} \quad pq = 6,\n$$\nmany students wonder how to efficiently find the values of $ p $ and $ q $ — or better yet, how to understand and apply a key algebraic identity that simplifies the solution. This article explores the elegant mathematical method behind solving for $ p $ and $ q $ using Vieta’s formulas and a fundamental identity.", "---", "### Understanding the Problem", "We are given:", "$$\np + q = 5 \quad \ ext{(sum)} \\npq = 6 \quad \ ext{(product)}\n$$", "These are symmetric expressions in $ p $ and $ q $, which naturally connects to the concept of quadratic roots. In fact, $ p $ and $ q $ are the roots of the quadratic equation:", "$$\nx^2 - (p + q)x + pq = 0\n$$", "Substituting the known values:", "$$\nx^2 - 5x + 6 = 0\n$$", "---", "### Using Factoring: A Classic Identity", "The next step leverages a well-known factoring identity:", "$$\nx^2 - (p + q)x + pq = (x - p)(x - q)\n$$", "This identity is grounded in polynomial algebra and symmetry in roots. It tells us that any quadratic equation with given sum and product factors neatly.", "Factoring $ x^2 - 5x + 6 $:", "$$\nx^2 - 5x + 6 = (x - 2)(x - 3)\n$$", "So the roots are $ x = 2 $ and $ x = 3 $, meaning:", "$$\np = 2, \quad q = 3 \quad \ ext{or vice versa}\n$$", "---", "### The Key Identity: A Deeper Dive", "The identity used here — combining sum and product — is rooted in Vieta’s formulas, which link the coefficients of a polynomial to sums and products of its roots. For a quadratic $ x^2 - Sx + P = 0 $, where $ S = p + q $ and $ P = pq $, the roots are always:", "$$\n\frac{S \pm \sqrt{S^2 - 4P}}{2}\n$$", "Let’s verify this with our values:", "$$\nS = 5, \quad P = 6 \Rightarrow \sqrt{5^2 - 4 \cdot 6} = \sqrt{25 - 24} = \sqrt{1} = 1\n$$", "So,", "$$\np, q = \frac{5 \pm 1}{2} \Rightarrow x = \frac{6}{2} = 3, \quad x = \frac{4}{2} = 2\n$$", "This confirms the identity’s power and consistency.", "---", "### Why This Matters for Problem Solving", "Understanding and applying this identity streamlines solving symmetric equations:", "- Instead of trial and error, you recognize the root-factor form.\n- You automate the process using algebra, not guesswork.\n- It serves as a foundational tool in quadratic equations, systems of equations, and polynomial roots.", "---", "### Conclusion", "Given $ p + q = 5 $ and $ pq = 6 $, the solution isn't just $ p = 2, q = 3 $ — it’s a case study in elegant algebraic identity. By applying the quadratic formation and factoring identity $ x^2 - Sx + P = 0 $, we efficiently uncover the values of $ p $ and $ q $. This method exemplifies how recognizing symmetry and leveraging core algebraic identities can transform complex problems into manageable steps.", "Whether you're solving for variables, tackling Olympiad-style problems, or building deeper math intuition, mastering this identity is essential for anyone advancing in algebra and beyond.", "---", "Keywords: solve $ p + q = 5 $, solve $ pq = 6, $ algebraic identity $ p^2 - (p+q)x + pq = 0 $, Vieta’s formulas, factoring quadratics, sum and product of roots, polynomial roots identity, algebra problem solving.\nMeta Description: Solve $ p + q = 5 $ and $ pq = 6 $ using the identity $ x^2 - 5x + 6 = 0 $. Learn how Vieta’s formulas enable efficient root identification through algebraic factoring."]
