\( x = \frac{35 \pm \sqrt{1225 - 240}}{4} = \frac{35 \pm \sqrt{985}}{4} \).
["Solving the Quadratic Equation: A Step-by-Step Guide to ( x = \frac{35 \pm \sqrt{1225 - 240}}{4} = \frac{35 \pm \sqrt{985}}{4} )", "Quadratic equations are fundamental in algebra, with applications ranging from physics to engineering and economics. One such equation—( x = \frac{35 \pm \sqrt{1225 - 240}}{4} )—can be simplified and solved using clear algebraic steps. This article breaks down the process of solving this expression, walks through the meaning of each component, and explains how to find the exact and approximate values of ( x ).", "---", "### Understanding the General Form", "Quadratic equations take the standard form:\n[\nax^2 + bx + c = 0\n]\nThe quadratic formula provides solutions for ( x ):\n[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]\nIn our case, comparing the expression ( x = \frac{35 \pm \sqrt{1225 - 240}}{4} ) with the general form reveals:\n- ( 2a = 4 ) → ( a = 2 )\n- ( b = 35 )\n- The discriminant ( D = b^2 - 4ac = 1225 - 240 = 985 )", "---", "### Step 1: Simplify the Discriminant", "The discriminant is key to determining the nature of the roots:\n[\nD = 1225 - 240 = 985\n]\nSince ( D > 0 ) and not a perfect square, the equation has two distinct real irrational roots.", "---", "### Step 2: Apply the Quadratic Formula", "Substituting ( a = 2 ), ( b = 35 ), and ( D = 985 ) into the quadratic formula:\n[\nx = \frac{ -35 \pm \sqrt{985} }{ 2 \cdot 4 } = \frac{35 \pm \sqrt{985}}{4}\n]\nThe ( \mp ) in the original expression reflects the ( \pm ) in the formula, signaling both roots: one using the plus and one using the minus.", "---", "### Step 3: Calculate Approximate Values", "While exact form is preferred mathematically, understanding decimal approximations aids interpretation:\n[\n\sqrt{985} \approx 31.3847\n]\nThus,\n[\nx = \frac{35 \pm 31.3847}{4}\n]\nCompute both solutions:\n- Positive root: ( x_+ = \frac{35 + 31.3847}{4} = \frac{66.3847}{4} \approx 16.596 )\n- Negative root: ( x_- = \frac{35 - 31.3847}{4} = \frac{3.6153}{4} \approx 0.904 )", "---", "### Step 4: Interpret the Solution Set", "The solutions:\n[\nx = \frac{35 \pm \sqrt{985}}{4}\n]\nrepresent two real and distinct values of ( x )—164.6% and 0.9%—depending on the sign used. These can model real-world scenarios such as break-even points, projectile trajectories, or optimization problems where positive solutions often indicate feasible outcomes.", "---", "### Why This Equation Matters", "Understanding this form and its derivation strengthens skills in:\n- Simplifying radicals\n- Applying the quadratic formula\n- Interpreting discriminants\n- Converting symbolic expressions into numerical approximations", "Such manipulations are essential in academic settings and technical applications across STEM fields.", "---", "### Summary", "Solving ( x = \frac{35 \pm \sqrt{1225 - 240}}{4} ) leads directly to ( x = \frac{35 \pm \sqrt{985}}{4} ), revealing two distinct real roots. Mastery of this process empowers precise analysis and application in equations modeling real-world phenomena.", "---", "### Frequently Asked Questions (FAQ)", "Q: Why do we subtract 240 from 1225 inside the square root?\nA: Because ( b^2 = 35^2 = 1225 ), and ( 4ac = 4(2)(c) = 4 \cdot 2 \cdot 85 = 240 ), so ( b^2 - 4ac = 1225 - 240 = 985 ).", "Q: Does this equation have exact or approximate solutions?\nA: The exact solutions are ( \frac{35 \pm \sqrt{985}}{4} ); numerical approximation yields roughly 16.6 and 0.9.", "Q: What does the discriminant tell us?\nA: Since ( 985 > 0 ) and not a perfect square, there are two distinct real (irrational) solutions.", "---", "### Final Note", "Whether you're a student mastering algebra or a professional applying math in practice, understanding how to simplify and solve quadratic expressions like ( x = \frac{35 \pm \sqrt{985}}{4} ) equips you with tools for precise, confident problem-solving. Explore the roots, verify with graphing tools, and apply them confidently across disciplines!"]
