Solution: Let $ x = \sqrt{7 + 2\sqrt{10}} + \sqrt{7 - 2\sqrt{10}} $. We compute $ x^2 $:
["Title: Simplify $ x = \sqrt{7 + 2\sqrt{10}} + \sqrt{7 - 2\sqrt{10}} $: Compute $ x^2 $ — A Step-by-Step Solution", "---", "### Introduction", "Mathematics often invites us to simplify complex expressions, and this particular expression—defined as $ x = \sqrt{7 + 2\sqrt{10}} + \sqrt{7 - 2\sqrt{10}} $—presents a beautiful opportunity to explore algebraic identities, symmetry, and square root simplifications. In this article, we’ll compute $ x^2 $ step by step, uncovering how this elegant expression resolves into a clean integer value. If you're exploring algebraic identities or preparing for math competitions, this solution offers both clarity and insight.", "---", "### Step 1: Set Up the Expression", "We begin with:", "$$\nx = \sqrt{7 + 2\sqrt{10}} + \sqrt{7 - 2\sqrt{10}}\n$$", "Our goal is to compute $ x^2 $. Recall the identity:", "$$\nx^2 = \left( a + b \right)^2 = a^2 + b^2 + 2ab\n$$", "Here, $ a = \sqrt{7 + 2\sqrt{10}} $ and $ b = \sqrt{7 - 2\sqrt{10}} $. Applying the identity:", "$$\nx^2 = \left( \sqrt{7 + 2\sqrt{10}} \right)^2 + \left( \sqrt{7 - 2\sqrt{10}} \right)^2 + 2 \cdot \sqrt{7 + 2\sqrt{10}} \cdot \sqrt{7 - 2\sqrt{10}}\n$$", "---", "### Step 2: Simplify $ a^2 + b^2 $", "First, compute:", "$$\na^2 + b^2 = (7 + 2\sqrt{10}) + (7 - 2\sqrt{10}) = 7 + 7 + 2\sqrt{10} - 2\sqrt{10} = 14\n$$", "The irrational parts cancel, leaving a simple sum.", "---", "### Step 3: Simplify the Product Term", "Now evaluate:", "$$\n2ab = 2 \cdot \sqrt{7 + 2\sqrt{10}} \cdot \sqrt{7 - 2\sqrt{10}}\n$$", "Use the identity $ \sqrt{a} \cdot \sqrt{b} = \sqrt{ab} $:", "$$\n2ab = 2 \cdot \sqrt{(7 + 2\sqrt{10})(7 - 2\sqrt{10})}\n$$", "This is a difference of squares:", "$$\n(7 + 2\sqrt{10})(7 - 2\sqrt{10}) = 7^2 - (2\sqrt{10})^2 = 49 - 4 \cdot 10 = 49 - 40 = 9\n$$", "So:", "$$\n2ab = 2 \cdot \sqrt{9} = 2 \cdot 3 = 6\n$$", "---", "### Step 4: Combine All Parts to Compute $ x^2 $", "Now sum the components:", "$$\nx^2 = a^2 + b^2 + 2ab = 14 + 6 = 20\n$$", "Thus:", "$$\nx = \sqrt{20} = 2\sqrt{5}\n$$", "But more importantly for this problem, we found:", "$$\nx^2 = 20\n$$", "---", "### Why This Solution Matters", "This problem exemplifies how nested radicals—seemingly complicated expressions—can be simplified using algebraic identities and careful manipulation. Computing $ x^2 $ not only reveals the square of the expression but also validates simplification techniques used widely in algebra, number theory, and even physics.", "Whether you're solving for exact values or preparing for advanced problem-solving, mastering such steps builds a strong foundation in mathematical reasoning.", "---", "### Key Takeaways", "- Use the identity $ (a + b)^2 = a^2 + b^2 + 2ab $ to expand radical expressions.\n- Simplify products inside square roots using the difference of squares: $ (a + b)(a - b) = a^2 - b^2 $.\n- Recognize when irrational parts cancel, simplifying the overall expression.\n- The result $ x^2 = 20 $ provides a clean integer output from an initially complex expression—ideal for verification and deeper insight.", "---", "### Final Thoughts", "Understanding how to compute $ x^2 $ symbolically opens doors to solving complex equations, verifying identities, and appreciating the elegance within algebra. Keep practicing such techniques—each solved expression strengthens your mathematical toolkit.", "---", "Stay curious. Keep simplifying. Your next breakthrough might be just one equation away.", "---", "Keywords:\nLet ( x = \sqrt{7 + 2\sqrt{10}} + \sqrt{7 - 2\sqrt{10}} ), compute ( x^2 ), algebra identity, simplify radicals, nested radicals simplification, mathematical identity proof, competition math solution, algebra AMC level problem", "Meta Description:\nCompute ( x = \sqrt{7 + 2\sqrt{10}} + \sqrt{7 - 2\sqrt{10}} ) and find ( x^2 ). Step-by-step explanation with key algebraic identities and simplification techniques. Ideal for math learners and competition prep.", "---", "Disclaimer: This article is for educational purposes and does not constitute professional mathematical advice."]
