Given one root is 3, the other root \( r \) is:

["# Given One Root is 3, the Other Root ( r ) Is: How to Find the Missing Root in Quadratic Equations", "When working with quadratic equations, understanding the relationship between a root and other roots can simplify solving equations and uncovering vital properties like the sum and product of roots. A fundamental idea to master is: if one root of a quadratic equation is known, how can you find the other root?", "## Understanding Roots in Quadratic Equations", "For a standard quadratic equation of the form\n[\nax^2 + bx + c = 0\n]\nthe roots can be found using the quadratic formula:\n[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "The sum and product of the roots are given by Vieta’s formulas:\n- Sum of roots: ( r_1 + r_2 = -\frac{b}{a} )\n- Product of roots: ( r_1 \ imes r_2 = \frac{c}{a} )", "## Given Root = 3: Finding the Other Root", "Suppose one root ( r_1 = 3 ), and the quadratic equation has real coefficients. Then, to find the other root ( r_2 ), we use Vieta’s formula for the product of roots:\n[\nr_1 \ imes r_2 = \frac{c}{a}\n]\nHowever, we don’t need to solve the entire equation explicitly. Since the product of the roots is determined by coefficients, and you know one root, the product of the roots ( r_1 \cdot r_2 ) equals ( \frac{c}{a} ), this helps relate the two roots without full coefficients.", "Alternatively, using the sum:\n[\nr_1 + r_2 = -\frac{b}{a} \quad \Rightarrow \quad 3 + r_2 = -\frac{b}{a}\n]", "But here’s a key shortcut:\nIf you know one root and the equation is monic (i.e., ( a = 1 )), the quadratic can be written as:\n[\n(x - 3)(x - r) = 0\n]\nExpanding this gives:\n[\nx^2 - (3 + r)x + 3r = 0\n]\nComparing coefficients with ( ax^2 + bx + c = 0 ):\n- ( b = -(3 + r) )\n- ( c = 3r )", "Therefore, from the known product of roots ( r_1 \cdot r_2 = r \cdot 3 = \frac{c}{a} ), and assuming ( a = 1 ),\n[\nr = 3r = c \quad \Rightarrow \quad r = \frac{c}{3}\n]", "But even without full coefficients, since the product of the roots must match ( r \cdot 3 ), transferring knowns gives:\nThe other root ( r ) multiplies with 3 to yield the constant term divided by ( a ), simplifying understanding.", "### In simple terms:\nSince one root is 3, and assuming ( a = 1 ), the other root ( r ) satisfies:\n[\nr = \frac{c}{3} \quad \ ext{(if equation is monic: } x^2 - (3 + r)x + 3r = 0 \ ext{)}\n]\nBut more directly:\nThe product of the roots is ( 3r ), so knowing one root, the other root scales accordingly relative to coefficients.", "### Final Answer\nGiven one root is 3, the other root ( r ) depends on the quadratic equation’s coefficients, but:\nThe key relationship is:\n[\nr = \frac{c}{3a} \quad \ ext{(from } r_1 r_2 = \frac{c}{a} \ ext{ with } r_1 = 3\ ext{)}\n]\nSo, to find ( r ), determine ( c ) and ( a ), then compute ( r = \frac{c}{3a} ).\nAlternatively, if the quadratic is monic (( a = 1 )) and written as ( (x - 3)(x - r) = x^2 - (3 + r)x + 3r ), then the other root is any number ( r ) such that the equation holds—often found directly via sum or product.", "---", "### Why This Matters in Algebra", "Knowing how to relate roots protects against comm pared and conceptual errors in equation solving. It empowers students and mathematicians alike to efficiently determine missing values when partial information is available.", "Mastering this principle strengthens foundations in algebra and prepares learners to explore roots of higher-degree polynomials with confidence.", "---", "Explore further:\n- How to use Vieta’s formulas in real-world problem solving\n- Solving quadratics with one known root\n- Applications of roots in geometry and optimization", "Keywords: quadratic roots, Vieta’s formulas, unknown root calculation, sum and product of roots, solving quadratics, algebra tips, math help"]