Solve for x: Harvard University Algebra Problem √(16-x²)+√(9-x²)=5

Problem Overview & Strategy

• 💡 The video presents a Harvard University algebra problem to solve for x in the equation √(16-x²)+√(9-x²)=5.
• 🚫 The speaker advises against directly squaring both sides due to potential complexity.
• 🔑 The recommended approach is the substitution method to simplify the equation.

Initial Substitution & Simplification

• 🎯 The method involves setting a = √(16-x²) and b = √(9-x²).
• ✅ This transforms the original equation into a simpler form: a + b = 5.
• 🔬 By squaring both a and b, the radicals are eliminated, resulting in a² = 16 - x² and b² = 9 - x².

Deriving a Second Equation

• ➕ Subtracting the squared equations yields a² - b² = 7 (since -x² + x² cancels out).
• 🧩 Recognizing a² - b² as a difference of two squares, it's factored into (a + b)(a - b).
• ➡️ Substituting a + b = 5 into the factored form leads to 5(a - b) = 7, which simplifies to a - b = 7/5.

Solving for 'a'

• 📈 A system of two linear equations is formed: a + b = 5 and a - b = 7/5.
• ➕ Adding these two equations together eliminates b, resulting in 2a = 5 + 7/5.
• 🔢 After combining the fractions (25/5 + 7/5), 2a = 32/5, which means a = 16/5.

Finding the Value of 'x'

• 🔄 The value of a is substituted back into the equation a² = 16 - x².
• 🛠️ Rearranging to solve for x², we get x² = 16 - a².
• 📊 Plugging in a = 16/5 gives x² = 16 - (16/5)² = 16 - 256/25.
• ✅ Calculating the difference yields x² = 144/25.
• 💡 Taking the square root of both sides provides the final solution: x = ±12/5.