- 4y^2 + 12y - 8 \geq 0

["# Understanding and Solving the Inequality: 4y² + 12y – 8 ≥ 0", "When tackling quadratic inequalities, understanding how to find solution intervals is essential for both academic success and real-world problem-solving. In this article, we’ll explore how to solve and interpret the inequality:", "[\n4y^2 + 12y – 8 \geq 0\n]", "## What Is the Inequality?", "The inequality (4y^2 + 12y - 8 \geq 0) asks: For which values of (y) does the quadratic expression (4y^2 + 12y - 8) become zero or positive?", "This type of inequality is common in algebra courses and helps build intuition about parabolas, zero product methods, and sign analysis.", "---", "## Step 1: Simplify the Inequality", "First, simplify the quadratic expression by dividing through by the greatest common divisor (GCD) of the coefficients. Here, the coefficients 4, 12, and -8 have a GCD of 4:", "[\n4y^2 + 12y - 8 \geq 0\n\Rightarrow\ny^2 + 3y - 2 \geq 0 \quad \ ext{(dividing all terms by 4)}\n]", "This simplification preserves the solution set since dividing/ multiplying by a positive number does not change inequality direction.", "---", "## Step 2: Find the Roots Using the Quadratic Formula", "To solve (y^2 + 3y - 2 \geq 0), start by finding the roots of the corresponding equation:", "[\ny^2 + 3y - 2 = 0\n]", "Use the quadratic formula:", "[\ny = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad \ ext{where } a = 1, b = 3, c = -2\n]", "Calculate the discriminant:", "[\n\Delta = 3^2 - 4(1)(-2) = 9 + 8 = 17\n]", "So,", "[\ny = \frac{-3 \pm \sqrt{17}}{2}\n]", "Approximate values:", "[\n\sqrt{17} \approx 4.123\n\Rightarrow\ny_1 = \frac{-3 - 4.123}{2} \approx -3.561, \quad\ny_2 = \frac{-3 + 4.123}{2} \approx 0.561\n]", "The exact roots are:", "[\ny = \frac{-3 \pm \sqrt{17}}{2}\n]", "---", "## Step 3: Analyze the Parabola and Sign Changes", "The quadratic (y^2 + 3y - 2) has a positive leading coefficient (1), meaning the parabola opens upward. This means:", "- The expression is positive outside the interval between the roots\n- The expression equals zero at the roots", "Thus,", "[\ny^2 + 3y - 2 \geq 0 \quad \ ext{when} \quad y \leq \frac{-3 - \sqrt{17}}{2} \quad \ ext{or} \quad y \geq \frac{-3 + \sqrt{17}}{2}\n]", "---", "## Step 4: Write the Solution in Interval Notation", "Using the approximate roots for clarity:", "[\ny \in (-\infty, \frac{-3 - \sqrt{17}}{2}] \cup [\frac{-3 + \sqrt{17}}{2}, \infty)\n]", "This means the inequality holds for all real numbers (y) less than or equal to approximately (-3.561), or greater than or equal to approximately (0.561).", "---", "## Step 5: Graphical Interpretation (Optional Insight)", "Plotting (f(y) = y^2 + 3y - 2) shows a U-shaped curve crossing the y-axis below the x-axis, dipping low enough between the roots to be negative, and rising again beyond those roots — confirming that the function is non-negative outside the interval ((-3.561, 0.561)).", "---", "## Why This Matters in Real Life", "Solving inequalities like (4y^2 + 12y - 8 \geq 0) builds foundational skills for engineering, economics, and science, where constraints and thresholds are modeled by quadratic relationships. Knowing where a function is positive or non-negative helps predict valid ranges for variables in real-world systems.", "---", "## Summary", "To solve (4y^2 + 12y – 8 \geq 0):", "1. Simplify to (y^2 + 3y - 2 \geq 0)\n2. Find roots: (y = \frac{-3 \pm \sqrt{17}}{2})\n3. As the parabola opens upward, the solution is\n [\n y \leq \frac{-3 - \sqrt{17}}{2} \quad \ ext{or} \quad y \geq \frac{-3 + \sqrt{17}}{2}\n ]\n4. In interval notation:\n [\n (-\infty, \frac{-3 - \sqrt{17}}{2}] \cup [\frac{-3 + \sqrt{17}}{2}, \infty)\n ]", "Understanding how to analyze and solve such inequalities empowers deeper problem-solving in math and applied fields.", "---", "Keywords: 4y² + 12y – 8 ≥ 0, quadratic inequality, solving quadratics, parabola analysis, real roots, algebraic methods, math tutorial, inequality solutions, step-by-step math", "---", "Have more quadratic inequalities to solve? Try simplifying, finding roots, analyzing the parabola, and testing intervals—your algebra confidence will grow!"]