\( x = \frac{1 \pm \sqrt{1 + 32}}{2} = \frac{1 \pm \sqrt{33}}{2} \)

["# Solving the Quadratic Equation: ( x = \frac{1 \pm \sqrt{1 + 32}}{2} ) – A Step-by-Step Guide", "When faced with a quadratic equation, one powerful algebraic technique is test. A notable example arises from solving the quadratic in the form:", "[\nx = \frac{1 \pm \sqrt{1 + 32}}{2}\n]", "This expression leads directly to the solution of ( x = \frac{1 \pm \sqrt{33}}{2} ). In this article, we explore how this equation is derived, simplified, and interpreted—key insights every student of algebra, math enthusiast, and problem solver should understand.", "## Understanding the Formula", "The formula ( x = \frac{1 \pm \sqrt{1 + 32}}{2} ) stems from the quadratic formula applied to a specific quadratic equation:", "[\nx^2 - x - 32 = 0\n]", "Recall the standard quadratic equation:", "[\nax^2 + bx + c = 0 \quad \Rightarrow \quad x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "For our equation:\n- ( a = 1 )\n- ( b = -1 )\n- ( c = -32 )", "Substituting into the formula:", "[\nx = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-32)}}{2(1)} = \frac{1 \pm \sqrt{1 + 128}}{2} = \frac{1 \pm \sqrt{129}}{2}\n]", "Wait — this conflicts with the earlier expression ( \sqrt{1 + 32} ). That’s because the question simplified it as ( \sqrt{1 + 32} = \sqrt{33} ). However, ( 1 + 32 = 33 ), so ( \sqrt{33} ) is indeed correct. Therefore, correcting the equation:", "To match the original:", "[\nx = \frac{1 \pm \sqrt{1 + 4 \cdot 1 \cdot 32}}{2} = \frac{1 \pm \sqrt{129}}{2} \quad \ ext{(Correct precise value)}\n]", "But the expression ( x = \frac{1 \pm \sqrt{33}}{2} ) suggests a possible typo—unless the constant is misinterpreted. Let’s recheck inputs.", "Wait: ( 4ac = 4 \ imes 1 \ imes (-32) = -128 ), so discriminant is ( b^2 - 4ac = 1 - (-128) = 129 ), hence:", "[\nx = \frac{1 \pm \sqrt{129}}{2}\n]", "So this refined equation is accurate—but the original question’s discriminant ( 1 + 32 = 33 ) seems off. Unless:", "> Possibly, the intended equation was ( x^2 - x - 32 = 0 ), so discriminant ( \Delta = 1 + 128 = 129 ), not 33.", "But — hypothetically — if ( c = -8 ), then ( b^2 - 4ac = 1 + 32 = 33 ), so ( \sqrt{33} ) fits.", "Perhaps the equation was ( x^2 - x - 8.25 = 0 ), but no — the key insight is understanding the method, not just the number.", "Assuming instead the problem aims to illustrate discriminant logic, we proceed.", "But let’s clarify:", "If the equation was ( x^2 - x - 32 = 0 ), then indeed:", "[\nx = \frac{1 \pm \sqrt{1 + 128}}{2} = \frac{1 \pm \sqrt{129}}{2}\n]", "Yet the given expression simplifies to ( \frac{1 \pm \sqrt{33}}{2} ), so a contradiction. To resolve this, this article corrects the discriminant:", "> If solving ( x^2 - x - 8.25 = 0 ), then discriminant is ( 1 + 33 = 34 ), still not 33.", "Wait — suppose the equation is instead ( x^2 - x - 8.5 = 0 )? Then:", "[\n\Delta = 1 + 4 \cdot 8.5 = 1 + 34 = 35\n]", "No. Only if ( 4ac = 32 ) — i.e., ( c = -8 ), then ( b^2 - 4ac = 1 + 32 = 33 ) exactly.", "Therefore: Let’s assume the intended quadratic was ( x^2 - x - 8 = 0 ) (simplified), then:", "[\n\Delta = (-1)^2 - 4(1)(-8) = 1 + 32 = 33\n]", "Then:", "[\nx = \frac{1 \pm \sqrt{33}}{2}\n]", "This matches the given expression.", "Conclusion: The equation ( x = \frac{1 \pm \sqrt{33}}{2} ) most logically comes from ( x^2 - x - 8 = 0 ), not ( x^2 - x - 32 = 0 ). This is likely a typo in the prompt—correcting it reveals a beautifully solvable quadratic.", "---", "## Step-by-Step Solution", "Let’s solve ( x^2 - x - 8 = 0 ) step-by-step.", "### Step 1: Identify coefficients", "From ( ax^2 + bx + c = 0 ):", "- ( a = 1 )\n- ( b = -1 )\n- ( c = -8 )", "### Step 2: Apply the quadratic formula", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "Substitute values:", "[\nx = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-8)}}{2(1)} = \frac{1 \pm \sqrt{1 + 32}}{2} = \frac{1 \pm \sqrt{33}}{2}\n]", "### Step 3: Simplify and state the solution", "[\nx = \frac{1 + \sqrt{33}}{2} \quad \ ext{or} \quad x = \frac{1 - \sqrt{33}}{2}\n]", "---", "## Why This Matters", "The quadratic formula allows us to solve any quadratic equation without factoring — critical in physics, engineering, economics, and computer science. Recognizing when the discriminant (( \Delta = b^2 - 4ac )) yields real, imaginary, or repeated roots helps predict behavior in optimization, motion, and modeling.", "Here, ( \sqrt{33} ) is irrational (~5.744), so both roots are real and irrational — reflecting the nature of solutions when discriminant > 0 but not perfect squares.", "---", "## Real-World Applications", "1. Projectile Motion:\n When modeling height ( h(t) = -\frac{1}{2}gt^2 + v_0t + h_0 ), solving for time of flight feeds into ( x = \frac{1 \pm \sqrt{D}}{2a} ) forms.", "2. Profit Maximization:\n Quadratic profit functions ( P(x) = -ax^2 + bx - c ) yield maximum revenue at ( x = \frac{b}{2a} ), derived via vertex formula linked to square roots.", "3. Geometry:\n Finding intersection points of parabolas often results in quadratic equations solvable via this method.", "---", "## Final Summary", "Although the original expression ( x = \frac{1 \pm \sqrt{1 + 32}}{2} ) implies ( \sqrt{33} ), standard algebra confirms:", "[\nx = \frac{1 \pm \sqrt{33}}{2} \quad \ ext{comes from solving} \quad x^2 - x - 8 = 0\n]", "But the method remains universal. Mastering this — recognizing discriminants, applying formulas, checking solutions — is essential for problem-solving across STEM fields.", "Whether dealing with simple quadratics or complex models, understanding how to derive and interpret solutions like ( x = \frac{1 \pm \sqrt{33}}{2} ) empowers confident mathematical reasoning.", "---", "## Further Reading", "- Quadratic Formula Derivation and Proof\n- Discriminant Analysis and Root Nature\n- Applications of Quadratics in Physics and Economics\n- Solving Quadratics Graphically and Algebraically", "---", "Keywords: quadratic formula, solve quadratic equation, ( x = \frac{1 \pm \sqrt{33}}{2} ), discriminant, rational and irrational roots, algebra tutorial, math equations, quadratic solutions", "---", "Download a printable cheat sheet for solving quadratics using the formula ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ) and practice with multiple-choice and word problems."]