A quadratic equation \( x^2 - 5x + 6 = 0 \) has roots \( a \) and \( b \). What is the value of \( a^2 + b^2 \)?

["Understanding Quadratic Roots: Calculating ( a^2 + b^2 ) for ( x^2 - 5x + 6 = 0 )", "Solving quadratic equations is a fundamental skill in algebra, enabling students and professionals alike to uncover critical relationships between variables. One classic example is the equation ( x^2 - 5x + 6 = 0 ), a simple yet powerful illustration of how root properties simplify complex calculations. But what happens when we know the roots—denoted ( a ) and ( b )—and want to find ( a^2 + b^2 )? In this article, we’ll explore the equation, its roots, and the elegant mathematical shortcut to compute ( a^2 + b^2 ) without solving for ( a ) and ( b ) individually.", "---", "### The Quadratic Equation: ( x^2 - 5x + 6 = 0 )", "The standard form of a quadratic equation is:", "[\nax^2 + bx + c = 0\n]", "For our equation, the coefficients are:", "- ( a = 1 )\n- ( b = -5 )\n- ( c = 6 )", "A quadratic equation may have two real roots, and these roots ( a ) and ( b ) follow well-known relationships derived from Vieta’s formulas:", "- Sum of roots: ( a + b = -\frac{b}{a} = -\frac{-5}{1} = 5 )\n- Product of roots: ( ab = \frac{c}{a} = \frac{6}{1} = 6 )", "These relationships are crucial because they link coefficients to root behavior without requiring complex solving techniques like factoring or the quadratic formula.", "---", "### From Roots to ( a^2 + b^2 ): A Mathematical Shortcut", "Rather than factoring the equation directly—though we could easily write ( (x - 2)(x - 3) = 0 ), yielding roots ( x = 2 ) and ( x = 3 )—we can compute ( a^2 + b^2 ) using a smart algebraic identity:", "[\na^2 + b^2 = (a + b)^2 - 2ab\n]", "This formula works for any two real numbers and is especially powerful when you already know ( a + b ) and ( ab ).", "Let’s compute step by step:", "1. From Vieta’s formulas:\n ( a + b = 5 )\n ( ab = 6 )", "2. Square the sum:\n ( (a + b)^2 = 5^2 = 25 )", "3. Multiply the product by 2:\n ( 2ab = 2 \ imes 6 = 12 )", "4. Subtract:\n ( a^2 + b^2 = 25 - 12 = 13 )", "---", "### Verification with Direct Root Calculation", "For clarity, let’s verify by finding the actual roots. Factoring ( x^2 - 5x + 6 = 0 ):", "[\nx^2 - 5x + 6 = (x - 2)(x - 3) = 0\n]", "So, roots are ( x = 2 ) and ( x = 3 ), which we can assign as ( a = 2 ), ( b = 3 ). Compute ( a^2 + b^2 ):", "[\n2^2 + 3^2 = 4 + 9 = 13\n]", "The result matches our earlier calculation, confirming the power of algebraic identities.", "---", "### Why This Approach Matters in Math and Beyond", "Calculating expressions like ( a^2 + b^2 ) quickly is not just an academic exercise. It appears in physics (kinetic energy calculations), engineering (system stability analysis), and data science (variance computations). Understanding how to derive such results efficiently builds both problem-solving flexibility and conceptual depth.", "---", "### Conclusion", "The quadratic equation ( x^2 - 5x + 6 = 0 ) has roots ( a = 2 ) and ( b = 3 ), and their squares sum to:", "[\na^2 + b^2 = 13\n]", "Thanks to Vieta’s formulas and the identity ( a^2 + b^2 = (a + b)^2 - 2ab ), we transformed a potentially tedious root-by-root calculation into a quick, elegant computation. Whether you’re a student mastering algebra or a professional applying math in real-world contexts, mastering these shortcuts empowers smarter, faster problem solving.", "---", "Key Takeaways:", "- Use Vieta’s formulas to find ( a + b ) and ( ab ) quickly.\n- Apply the identity ( a^2 + b^2 = (a + b)^2 - 2ab ) to avoid lengthy root solving.\n- This method applies broadly to quadratic relationships and more complex polynomials.", "Start mastering identities today, and turn complicated equations into confidence!"]