Thus, the sum of the roots is \(\boxed{\frac{9}{4}}\).

["Title: Understanding the Sum of Roots: A Clear Explanation with the Formula (\boxed{\frac{9}{4}})", "Meta Description:\nExplore how to determine the sum of roots in quadratic equations—and why it equals (\boxed{\frac{9}{4}})—with step-by-step reasoning and practical examples.", "---", "### Introduction", "In algebra, understanding the behavior of quadratic equations is fundamental. One key property is the sum of the roots, and today we dive deep into the specific case where the sum is (\boxed{\frac{9}{4}}). Whether you're solving equations for math class, engineering applications, or simply sharpening your skills, this guide will clarify how that elegant fraction emerges from quadratic formulas and factoring.", "---", "### What Does the Sum of Roots Mean?", "For a standard quadratic equation:", "[\nax^2 + bx + c = 0\n]", "If ( r_1 ) and ( r_2 ) are the roots (solutions) of the equation, their sum is given by:", "[\nr_1 + r_2 = -\frac{b}{a}\n]", "This formula makes solving problems efficient: instead of solving for each root individually, you directly compute the sum. When that sum equals (\boxed{\frac{9}{4}}), something key about (a) and (b) must hold. Let’s verify this using a concrete example and algebraic derivation.", "---", "### Step-by-Step Derivation: Why Is the Sum (\frac{9}{4})?", "Suppose your quadratic equation has roots ( r_1 ) and ( r_2 ), with", "[\nr_1 + r_2 = \frac{9}{4}\n\quad \ ext{and} \quad\nr_1 r_2 = c/a\n]", "We want to identify specific values of (a), (b), and (c) that make this true. Let’s assume the simplest scenario: factorable quadratics with rational roots.", "Let’s suppose the sum of roots is derived from:", "[\nr_1 + r_2 = \frac{9}{4}\n]", "From the formula ( r_1 + r_2 = -\frac{b}{a} ), this implies:", "[\n-\frac{b}{a} = \frac{9}{4} \quad \Rightarrow \quad b = -\frac{9}{4}a\n]", "Now, choose a convenient value for (a) to simplify—say, ( a = 4 ) (to eliminate denominators). Then:", "[\nb = -\frac{9}{4} \cdot 4 = -9\n]", "So our quadratic becomes:", "[\n4x^2 - 9x + c = 0\n]", "To find (c), recall that the product ( r_1 r_2 = c/a ), and ( r_1 + r_2 = \frac{9}{4} ). Let the roots be additive to (\frac{9}{4}). Suppose, for simplicity, the roots are ( r_1 = \frac{5}{4} ) and ( r_2 = 1 ), since:", "[\n\frac{5}{4} + 1 = \frac{5}{4} + \frac{4}{4} = \frac{9}{4}\n]", "Then their product is:", "[\nr_1 r_2 = \frac{5}{4} \cdot 1 = \frac{5}{4}\n]", "Now calculate (c):", "[\nc = a \cdot (r_1 r_2) = 4 \cdot \frac{5}{4} = 5\n]", "Thus, one possible quadratic equation is:", "[\n4x^2 - 9x + 5 = 0\n]", "This confirms the structure—rational roots with sum (\frac{9}{4}) and consistent coefficients. Let's now verify the sum using the quadratic formula.", "---", "### Verification Using the Quadratic Formula", "Use the quadratic formula:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "For (4x^2 - 9x + 5 = 0), plug in (a = 4), (b = -9), (c = 5):", "[\nx = \frac{9 \pm \sqrt{(-9)^2 - 4(4)(5)}}{2(4)} = \frac{9 \pm \sqrt{81 - 80}}{8} = \frac{9 \pm \sqrt{1}}{8}\n]", "So,", "[\nx = \frac{9 + 1}{8} = \frac{10}{8} = \frac{5}{4}, \quad x = \frac{9 - 1}{8} = \frac{8}{8} = 1\n]", "Sum of roots:", "[\n\frac{5}{4} + 1 = \frac{9}{4}\n]", "Confirmed! This reinforces that when a quadratic equation has roots summing to (\boxed{\frac{9}{4}}), the coefficient ratio satisfies (b = -\frac{9}{4}a), and choosing appropriate (a, b), and (c) yields a valid equation—here demonstrating both algebraic theory and practical computation.", "---", "### Applications and Why This Formula Matters", "Understanding root sums is essential in:", "- Polynomial construction: Specify sum and product to build equations.\n- Signal processing and control systems: Stability often depends on root locations.\n- Economics and optimization: Find break-even points modeled by quadratics.", "Knowing that the sum equals (\boxed{\frac{9}{4}}) empowers precise problem-solving without brute-force solving.", "---", "### Conclusion", "The sum of the roots of a quadratic equation is a powerful lens into its structure—tied directly to coefficients via (-\frac{b}{a}). When this sum is (\boxed{\frac{9}{4}}), it reflects a clear numerical relationship between linear and constant terms. Armed with this insight, students and professionals alike can decode, analyze, and construct quadratic problems with confidence.", "---", "Want to calculate root sums quickly? Just recall:\n[\n\ ext{Sum of roots} = -\frac{b}{a}\n]\nFor sum (= \frac{9}{4}), ensure ( b = -\frac{9}{4}a ). Then solve for (c) using root product or verify via quadratic formula.", "---", "Tagline: Mastering quadratic roots starts with the simple sum—boxed (\frac{9}{4})—and unlocks deeper algebra mastery.", "---", "Keywords: sum of roots, quadratic roots, algebra formula, (\frac{9}{4}) sum, quadratic equation sum of roots, verify quadratic roots, root sum formula, mathematics education, algebra practice\nSEO Tags: #QuadraticEquations #RootSum #AlgebraTutorial #MathHelp #EquationSolving #SumOfRoots\n---", "Feel free to share this guide to help others visualize how root sums emerge from fundamental quadratic principles!"]