\frac{1}{2} \times b \times \sqrt{100 - \frac{b^2}{4}} = 48

["Solving the Equation: (\frac{1}{2} \ imes b \ imes \sqrt{100 - \frac{b^2}{4}} = 48)", "Are you looking to solve a challenging algebraic equation? Consider this equation:", "[\n\frac{1}{2} \ imes b \ imes \sqrt{100 - \frac{b^2}{4}} = 48\n]", "This equation combines linear and square root components, making it ideal for exploring algebraic techniques and real-world applications. In this article, we guide you through solving it step-by-step, interpret its meaning, and explore practical implications.", "---", "### Step 1: Simplify the Equation", "Start by eliminating the fraction and square root:", "Multiply both sides by 2:", "[\nb \ imes \sqrt{100 - \frac{b^2}{4}} = 96\n]", "Let’s isolate the square root:", "[\n\sqrt{100 - \frac{b^2}{4}} = \frac{96}{b}\n]", "---", "### Step 2: Square Both Sides to Eliminate the Square Root", "[\n\left( \sqrt{100 - \frac{b^2}{4}} \right)^2 = \left( \frac{96}{b} \right)^2\n]", "[\n100 - \frac{b^2}{4} = \frac{9216}{b^2}\n]", "---", "### Step 3: Eliminate the Denominator by Multiplying Through by (b^2)", "Multiply every term by (b^2) to eliminate the fraction:", "[\n100b^2 - \frac{b^4}{4} = 9216\n]", "Multiply the entire equation by 4 to clear the denominator:", "[\n400b^2 - b^4 = 36864\n]", "---", "### Step 4: Rearrange into Standard Polynomial Form", "Rewriting terms in descending order:", "[\n-b^4 + 400b^2 - 36864 = 0\n]", "Multiply through by (-1) to make the leading coefficient positive:", "[\nb^4 - 400b^2 + 36864 = 0\n]", "This is a quadratic in disguise — set (x = b^2), so the equation becomes:", "[\nx^2 - 400x + 36864 = 0\n]", "---", "### Step 5: Solve the Quadratic Equation", "Use the quadratic formula (x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}), where (A = 1), (B = -400), (C = 36864):", "[\nx = \frac{400 \pm \sqrt{(-400)^2 - 4 \cdot 1 \cdot 36864}}{2}\n]", "[\nx = \frac{400 \pm \sqrt{160000 - 147456}}{2}\n]", "[\nx = \frac{400 \pm \sqrt{12544}}{2}\n]", "[\n\sqrt{12544} = 112 \quad \ ext{(since (112^2 = 12544))}\n]", "Thus:", "[\nx = \frac{400 \pm 112}{2}\n]", "Calculate the two roots:", "[\nx_1 = \frac{400 + 112}{2} = \frac{512}{2} = 256\n]", "[\nx_2 = \frac{400 - 112}{2} = \frac{288}{2} = 144\n]", "Recall (x = b^2), so:", "[\nb^2 = 256 \Rightarrow b = \pm 16\n]\n[\nb^2 = 144 \Rightarrow b = \pm 12\n]", "---", "### Step 6: Check Which Solutions Satisfy the Original Equation", "Substitute (b = 16):", "[\n\frac{1}{2} \ imes 16 \ imes \sqrt{100 - \frac{16^2}{4}} = 8 \ imes \sqrt{100 - \frac{256}{4}} = 8 \ imes \sqrt{100 - 64} = 8 \ imes \sqrt{36} = 8 \ imes 6 = 48\n]", "✅ Valid.", "Try (b = -16):", "[\n\frac{1}{2} \ imes (-16) \ imes \sqrt{100 - \frac{256}{4}} = -8 \ imes 6 = -48 <br/>\ne 48\n]", "❌ Invalid — square root is non-negative, so the product with negative (b) fails.", "Try (b = 12):", "[\n\frac{1}{2} \ imes 12 \ imes \sqrt{100 - \frac{144}{4}} = 6 \ imes \sqrt{100 - 36} = 6 \ imes \sqrt{64} = 6 \ imes 8 = 48\n]", "✅ Valid.", "Try (b = -12) similarly yields negative result — ❌ invalid.", "---", "### Final Answer", "The real solutions are:", "[\n\boxed{b = 12 \quad \ ext{and} \quad b = 16}\n]", "---", "### Real-World Interpretation", "This equation may model scenarios involving areas and geometry — for example, optimizing dimensions of a shape constrained by certain measurements. The presence of (\sqrt{100 - \frac{b^2}{4}}) suggests a semicircular or curved constraint, such as maximizing area within a fixed radius.", "Solving such equations helps engineers, architects, and data analysts find precise values under constraints — a key skill in applied mathematics and problem-solving.", "---", "### Summary", "- Simplified the original equation step-by-step.\n- Solved via substitution and factoring.\n- Validated solutions to ensure correctness.", "Mastering these techniques opens doors to tackling complex equations across science and engineering.", "---", "Keywords:\n(\frac{1}{2} \ imes b \ imes \sqrt{100 - \frac{b^2}{4}} = 48), solve quadratic, algebra solutions, equation with square root, real applications, (b = 12), (b = 16), algebra tutorial."]