Use quadratic formula: \( x = \frac{4 \pm \sqrt{16 + 64}}{2} = \frac{4 \pm \sqrt{80}}{2} \)

["Using the Quadratic Formula: Solve ( x = \frac{4 \pm \sqrt{16 + 64}}{2} )", "The quadratic formula is a powerful tool in algebra, enabling precise solutions to quadratic equations of the form ( ax^2 + bx + c = 0 ). One common scenario involves expressions like ( x = \frac{4 \pm \sqrt{16 + 64}}{2} ), which may appear in problems related to solving quadratic equations efficiently. In this SEO-optimized article, we explore how to use the quadratic formula effectively—with step-by-step guidance—including the simplified computation of ( \sqrt{80} ), helping students, educators, and math enthusiasts master this technique.", "---", "### Understanding the Quadratic Formula", "The standard quadratic formula is:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "It solves equations such as ( ax^2 + bx + c = 0 ) by providing exact solutions regardless of whether the roots are rational or irrational.", "---", "### Step-by-Step: Solve ( x = \frac{4 \pm \sqrt{16 + 64}}{2} )", "This expression arises when applying the quadratic formula to a specific equation. Let’s break it down and solve it clearly.", "Step 1: Identify coefficients", "From ( x = \frac{4 \pm \sqrt{16 + 64}}{2} ), we recognize:", "- The numerator follows ( \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), suggesting ( b = -4 ), but here the constant in the radical reflects ( b^2 - 4ac = 80 ), not necessarily tied to ( b = 4 ) directly.", "Step 2: Simplify the radical", "[\n\sqrt{16 + 64} = \sqrt{80}\n]", "Simplify ( \sqrt{80} ):", "[\n\sqrt{80} = \sqrt{16 \ imes 5} = 4\sqrt{5}\n]", "Step 3: Substitute back into the formula", "[\nx = \frac{4 \pm 4\sqrt{5}}{2}\n]", "Step 4: Simplify the expression", "Factor numerator:", "[\nx = \frac{4(1 \pm \sqrt{5})}{2} = 2(1 \pm \sqrt{5})\n]", "Thus, the two solutions are:", "[\nx = 2(1 + \sqrt{5}) \quad \ ext{and} \quad x = 2(1 - \sqrt{5})\n]", "---", "### Why This Formula Works", "The quadratic formula leverages the symmetry of quadratic equations. By evaluating the discriminant ( b^2 - 4ac ), we determine the nature of the roots—real or complex, distinct or repeated. In this case, since ( \sqrt{80} ) is real and positive, we obtained two distinct real roots.", "---", "### Practical Applications", "Understanding and applying the quadratic formula is essential in:", "- Engineering: solving for dimensions, stress points, or resonance frequencies\n- Physics: modeling motion under acceleration\n- Economics: optimizing profit or cost functions\n- Computer Graphics: calculating intersections and curves", "Mastering simplification like transforming ( \sqrt{80} ) to ( 4\sqrt{5} ) enhances clarity and versatility in real-world problem solving.", "---", "### Tips for Using Quadratic Formula Efficiently", "1. Identify coefficients clearly from the equation ( ax^2 + bx + c = 0 ).\n2. Simplify radicals and fractions as much as possible to reduce complexity.\n3. Use tools like √80 = 4√5 to avoid errors in calculations.\n4. Always verify your solutions by substituting back into the original equation.\n5. Practice with word problems to apply formulas in context.", "---", "### Conclusion", "The expression ( x = \frac{4 \pm \sqrt{16 + 64}}{2} ) exemplifies how the quadratic formula transforms complex polynomials into manageable solutions. Knowing how to simplify the radical and reduce the expression ensures accuracy and deeper conceptual understanding. Whether you're solving for math competitions, classroom exams, or practical challenges, mastering the quadratic formula—from coefficients to simplified roots—is key.", "---", "Keywords: quadratic formula, solving quadratic equations, solve ( x = \frac{4 \pm \sqrt{16 + 64}}{2} ), simplify √80, using quadratic formula, step-by-step quadratic solution, math tutorial, simplify radicals, algebra practice", "---", "Backlinks Opportunity: Link to algebra guides, radical simplification tutorials, and quadratic equation practice problems to boost SEO authority."]