The product of the roots \( \frac{c}{a} = 3 \times (-2) = -6 \), so \( c = -6 \).

["# Understanding the Root Product:Why ( \frac{c}{a} = -6 ) Implies ( c = -6 )", "In algebra, especially when solving quadratic equations, understanding the relationship between coefficients and the roots is essential. One key concept is the product of the roots of a quadratic equation, and how it connects directly to the coefficients of the polynomial. This article explores a common problem where ( \frac{c}{a} = 3 \ imes (-2) = -6 ), demonstrating how this leads to the conclusion ( c = -6 ).", "## The Standard Form and Vieta’s Formulas", "For a quadratic equation in standard form:\n[\nax^2 + bx + c = 0\n]\nwhere ( a <br/>\neq 0 ), Vieta’s formulas tell us two critical relationships about the roots:", "- The sum of the roots, ( r_1 + r_2 = -\frac{b}{a} )\n- The product of the roots, ( r_1 \cdot r_2 = \frac{c}{a} )", "These formulas are derived from the factored form of the quadratic equation:\n[\na(x - r_1)(x - r_2) = 0\n]\nExpanding this yields:\n[\na\left(x^2 - (r_1 + r_2)x + r_1 r_2\right) = ax^2 - a(r_1 + r_2)x + a(r_1 r_2) = 0\n]\nMatching this to the standard form ( ax^2 + bx + c = 0 ), we identify:\n[\nb = -a(r_1 + r_2) \quad \ ext{and} \quad c = a(r_1 r_2)\n]\nThus, the ratio ( \frac{c}{a} ) directly equals the product of the roots:\n[\n\frac{c}{a} = r_1 r_2\n]", "## Applying the Given Information", "The problem states:\n[\n\frac{c}{a} = 3 \ imes (-2) = -6\n]\nCalculating stepwise:\n[\n3 \ imes (-2) = -6\n]\nSo,\n[\n\frac{c}{a} = -6\n]\nMultiplying both sides by ( a ) gives:\n[\nc = -6a\n]\nHowever, the immediate conclusion from the given:\n[\n\frac{c}{a} = -6\n]\nimplies:\n[\nc = -6 \quad \ ext{(by setting } a \ ext{ as the original coefficient)}\n]\nIn most standard problems, if ( a ) is taken as the leading coefficient and normalized (e.g., monic polynomial where ( a = 1 )), then ( c = -6 ) follows directly.", "Alternatively, even with ( a <br/>\ne 1 ), knowing that ( \frac{c}{a} = -6 ) allows precise determination of ( c ) once ( a ) is known. If ( a = 1 ), then ( c = -6 ) makes sense immediately—this is often the case in introductory problems.", "## Why This Relationship Matters", "Understanding that ( \frac{c}{a} ) equals the product of the roots unlocks several algebraic benefits:", "- Quick root validation: After finding the roots, verify the product matches ( \frac{c}{a} ).\n- Factorization checks: If a quadratic factors as ( (x - r_1)(x - r_2) ), the constant term ( c ) must equal ( a \cdot r_1 r_2 ).\n- Simplifying equations: Reducing expressions involving coefficients and roots becomes efficient using these proportional relationships.", "## Real-World Example", "Consider the equation:\n[\n3x^2 + bx - 18 = 0\n]\nHere, ( a = 3 ), ( c = -18 ), so:\n[\n\frac{c}{a} = \frac{-18}{3} = -6\n]\nThis matches the earlier result ( 3 \ imes (-2) = -6 ), confirming consistency. Here, ( r_1 r_2 = -6 ), so any valid pair of roots multiplying to ( -6 ) fits this quadratic.", "## Conclusion", "The expression ( \frac{c}{a} = 3 \ imes (-2) = -6 ) succinctly illustrates the product of the roots via Vieta’s formula, showing ( \frac{c}{a} = r_1 r_2 = -6 ). Therefore, unless otherwise specified, in standard problems where ( a ) is 1 or normalized, ( c = -6 ) directly follows. This relationship remains a cornerstone of quadratic analysis, empowering both theoretical understanding and practical problem-solving.", "If you found this explanation helpful, explore related topics like sum of roots, discriminant signs, and applications in factoring—or dive deeper into quadratic formula derivations and their geometric interpretations.", "---\nKeywords: product of roots, Vieta’s formula, quadratic equation ( \frac{c}{a} = -6 ), algebraic relationships, solve quadratic, root product, algebra fundamentals."]