Solving the quadratic equation: \(x = 12\) or \(x = -14\).

["# Solving the Quadratic Equation: How to Find (x = 12) or (x = -14)", "Quadratic equations are fundamental in algebra and commonly appear in science, engineering, and everyday problem-solving. While many quadratic equations require finding roots through complex methods like factoring or the quadratic formula, some simplify directly to straightforward solutions—like (x = 12) or (x = -14). This comprehensive guide explains how to solve such equations, why these specific values satisfy them, and how to apply this knowledge confidently.", "---", "## What Is a Quadratic Equation?", "A quadratic equation is a second-degree polynomial of the form:", "[\nax^2 + bx + c = 0\n]", "where ( a <br/>\neq 0 ). The solutions (roots) represent the ( x )-values where the graph of the equation intersects the ( x )-axis. For certain cases, these solutions can be found directly without complex calculations.", "---", "## When Does a Quadratic Equation Yield (x = 12) or (x = -14)?", "The solutions (x = 12) and (x = -14) emerge when the quadratic equation is factorable into linear components that produce these exact values. Let’s explore the general form that leads to these roots.", "### Factoring Underground", "If a quadratic equation factors as:", "[\n(x - r_1)(x - r_2) = 0\n]", "then the solutions are (x = r_1) and (x = r_2). For (x = 12) or (x = -14), a simplified factorable form is:", "[\n(x - 12)(x + 14) = 0\n]", "Expanding this confirms the roots:", "[\nx^2 + 14x - 12x - 168 = x^2 + 2x - 168 = 0\n]", "Now substitute (x = 12):", "[\n(12)^2 + 2(12) - 168 = 144 + 24 - 168 = 0\n]", "Similarly, for (x = -14):", "[\n(-14)^2 + 2(-14) - 168 = 196 - 28 - 168 = 0\n]", "Hence, (x = 12) and (x = -14) are valid roots of the equation (x^2 + 2x - 168 = 0).", "---", "## Why Do These Particular Roots Matter?", "These values are not random—they stem from a carefully constructed quadratic that simplifies solving by direct factorization. Understanding this helps in:", "- Recognizing patterns: Recognizing factor patterns saves time compared to applying the quadratic formula.\n- Validating solutions: Plugging roots back confirms correctness.\n- Applying formulas: Tools like the quadratic formula always yield (x = 12) or (x = -14) when the correct (a), (b), and (c) are used.", "---", "## Step-by-Step: Solving a Quadratic Equation to Find (x = 12) or (x = -14)", "### Example:\nSuppose the equation is (x^2 + 2x - 168 = 0). Follow these steps to find the roots:", "### Step 1: Factor the quadratic\nFactor (x^2 + 2x - 168) into two binomials.\nWe know the product is (-168) and sum is (2). The pair (14) and (-12) works:", "[\n(x + 14)(x - 12) = 0\n]", "### Step 2: Set each factor to zero\n[\nx + 14 = 0 \quad \Rightarrow \quad x = -14\n]\n[\nx - 12 = 0 \quad \Rightarrow \quad x = 12\n]", "### Step 3: Verify\nEnsure both values satisfy the original equation by substituting back.", "---", "## How to Create a Quadratic with Roots (x = 12) and (x = -14)", "From roots, build the equation:", "[\n(x - 12)(x + 14) = 0\n]", "Expanding:", "[\nx^2 + 14x - 12x - 168 = x^2 + 2x - 168 = 0\n]", "So, the standard form is:", "[\n\boxed{x^2 + 2x - 168 = 0}\n]", "---", "## Common Tricks: Quick Checks and Solving Shortcuts", "- Factoring by grouping: Useful when coefficients are larger, but not necessary for simple factored forms like these.\n- Quadratic formula: Always gives (x = 12) and (x = -14) when (a = 1), (b = 2), (c = -168).\n [\n x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{4 + 672}}{2} = \frac{-2 \pm \sqrt{676}}{2} = \frac{-2 \pm 26}{2}\n ]\n So,\n [\n x = \frac{24}{2} = 12 \quad \ ext{and} \quad x = \frac{-28}{2} = -14\n ]", "---", "## Practice Problems to Reinforce Learning", "1. Factor and solve: (x^2 - 5x - 84 = 0), should yield (x = 12) or (x = -7)? (Answer: (x^2 - 5x - 84 = (x - 12)(x + 7); x = 12, -7)\n2. Given (a = 1), (b = 2), (c = -168), verify roots by factoring.", "---", "## Conclusion", "Finding (x = 12) and (x = -14) as solutions to a quadratic equation showcases how factoring simplifies equation solving. By recognizing recognizable patterns, expanding binomial products, and applying verification, anyone can confidently solve such equations. Whether you’re a student, educator, or math enthusiast, mastering these roots enhances problem-solving fluency in algebra and beyond.", "---", "### Key Takeaways:", "- Quadratic equations with roots (12) and (-14) factor as ((x - 12)(x + 14) = 0).\n- Factoring reduces solving to simple arithmetic.\n- Expanding back confirms correctness.\n- The quadratic formula always honors these roots when coefficients match.\n- Practice helps recognize and build these equations effortlessly.", "Start using these insights today—solving quadratics has never been clearer!"]