f(3) = (3)^2 - 6(3) + k = 0

["Understanding the Quadratic Equation: Solving f(3) = 3² – 6(3) + k = 0", "When encountering a mathematical expression like f(3) = (3)² – 6(3) + k = 0, it opens the door to understanding quadratic equations, their structure, and how to solve for unknown parameters. In this article, we’ll explore the equation, dissect its components, and learn how to determine the value of k that makes the equation true when x = 3.", "---", "### What Is the Equation f(3) = (3)² – 6(3) + k = 0?", "At first glance, f(3) = 3² – 6(3) + k = 0 might appear as a simple substitution in a quadratic function, but it actually reveals deeper insights into quadratic relationships. Though the standard form of a quadratic is f(x) = ax² + bx + c, here we evaluate f(3) by substituting x = 3 into an expression involving k.", "Let’s compute the left-hand side step by step:", "[\nf(3) = (3)^2 - 6(3) + k = 0\n]", "[\nf(3) = 9 - 18 + k = 0\n]", "[\nf(3) = -9 + k = 0\n]", "---", "### Solving for the Unknown Parameter k", "To satisfy the equation f(3) = 0, we solve:", "[\n-9 + k = 0 \Rightarrow k = 9\n]", "This means that k must equal 9 for x = 3 to be a solution to the quadratic expression defined implicitly by f(3) = 3² – 6(3) + k.", "---", "### Why Finding k Matters in Quadratics", "Finding k ensures the quadratic equation f(x) passes through the point (3, 0) — that is, the graph crosses the x-axis at x = 3. In quadratic functions, this point represents a root of the equation f(x) = 0.", "By fixing k = 9, we satisfy:", "[\nf(3) = 0\n]", "Which implies the quadratic function has a real root at x = 3, or at least touches the x-axis there (depending on multiplicity). If k were any other value, x = 3 would not satisfy the equation unless adjusted.", "---", "### General Form and Roots", "With k = 9, the function becomes:", "[\nf(x) = 3x^2 - 6x + 9\n]", "To find all roots, solve:", "[\n3x^2 - 6x + 9 = 0\n]", "Apply the quadratic formula:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4(3)(9)}}{2(3)}\n]", "[\nx = \frac{6 \pm \sqrt{36 - 108}}{6} = \frac{6 \pm \sqrt{-72}}{6}\n]", "The discriminant is negative, meaning the roots are complex — x = 3 is not a double root, but k = 9 positions 3 as a root only if plugged directly into f(3)=0, as originally posed.", "---", "### Conclusion: The Role of k in Evaluating Quadratic Functions", "The equation f(3) = (3)² – 6(3) + k = 0 serves as a focused way to determine the value of k that ensures x = 3 is a root of the quadratic when f(x) = 3² – 6(3) + k. As shown, k must equal 9.", "Understanding such expressions helps students and learners grasp:", "- How to substitute and simplify expressions in quadratic contexts\n- The relationship between parameters and function roots\n- The introduction of complex solutions when discriminants are negative", "If you're studying quadratics and encountering expressions like this, remember: evaluating at specific values and solving for unknowns is a foundational skill in algebra and equation solving.", "---", "Keywords: quadratic equations, solve for k, f(3) = 0, solve quadratic equations, math tutorial, algebra basics, quadratic functions, find k, discriminant, complex roots, root of quadratic, solve linear equations.", "---", "Related Reading:\n- How to Solve Quadratic Equations Step-by-Step\n- Understanding the Quadratic Formula and Discriminant\n- Real-Life Applications of Quadratic Functions", "---", "For further practice, try substituting other values of x and varying k to see how roots shift — deepening your mastery of quadratic behavior."]