x = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \Rightarrow x = 1, \frac{1}{2}

["Understanding the Equation: Solving ( x = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \Rightarrow x = 1, \frac{1}{2} )", "When solving quadratic equations, simplified forms often reveal elegant solutions — and the expression ( x = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \Rightarrow x = 1, \frac{1}{2} ) is a classic example of how to break down and solve such equations efficiently using algebra and the quadratic formula.", "This article explains step-by-step how to simplify and solve this quadratic expression and why the solutions are ( x = 1 ) and ( x = \frac{1}{2} ). We’ll also explore the underlying quadratic and its relevance in algebra and real-world applications.", "---", "### What Is the Equation?", "We begin with the general quadratic equation:\n[\nx = \frac{3 \pm \sqrt{9 - 8}}{4}\n]", "The expression under the square root, ( 9 - 8 ), simplifies to 1. Thus, the equation becomes:\n[\nx = \frac{3 \pm \sqrt{1}}{4}\n]", "Since ( \sqrt{1} = 1 ), this becomes:\n[\nx = \frac{3 \pm 1}{4}\n]", "---", "### Calculating the Two Solutions", "Using the (\pm) rule, we calculate both possible values of (x):", "1. Using the plus sign:\n[\nx = \frac{3 + 1}{4} = \frac{4}{4} = 1\n]", "2. Using the minus sign:\n[\nx = \frac{3 - 1}{4} = \frac{2}{4} = \frac{1}{2}\n]", "So, the two solutions are:\n[\nx = 1 \quad \ ext{and} \quad x = \frac{1}{2}\n]", "---", "### Derivation Using the Quadratic Formula", "The expression ( x = \frac{3 \pm \sqrt{9 - 8}}{4} ) comes directly from applying the quadratic formula:\n[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "For the original quadratic in standard form:\n- ( a = 1 )\n- ( b = 3 )\n- ( c = -8 )", "Plug into the formula:\n[\nx = \frac{-3 \pm \sqrt{(3)^2 - 4(1)(-8)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 32}}{2} = \frac{-3 \pm \sqrt{41}}{2}\n]", "Wait — this appears inconsistent at first glance. However, this illustrates a key point: simplifying expressions under radicals helps reveal fractional forms when the discriminant is a perfect square, as in our initial simplification.", "But in the simplified linear form ( x = \frac{3 \pm 1}{4} ), we used an equivalent quadratic expression — one whose discriminant simplifies neatly to 1, making the square root evaluate to a whole number.", "---", "### Why Simplify to ( \frac{3 \pm 1}{4} )?", "Simplifying expressions like this removes complexity and highlights the structure of the solution. When discriminants are perfect squares, such as ( 9 - 8 = 1 ), expressions under the square root become rational, enabling clean fractional simplifications.", "This simplicity helps students and learners quickly grasp the root values without computational overload.", "---", "### Real-World and Mathematical Relevance", "Quadratic equations model many real phenomena — projectile motion, optimization problems, economics, and beyond. Understanding how to solve them — whether by factoring, completing the square, or the quadratic formula — builds foundational algebra skills.", "In this case, recognizing that ( 9 - 8 = 1 ) makes the square root trivial encourages pattern recognition and pattern-saving habits essential in advanced math.", "---", "### Final Answer", "The equation simplifies cleanly to two distinct solutions:\n[\n\boxed{x = 1} \quad \ ext{and} \quad \boxed{x = \frac{1}{2}}\n]", "This elegant solution demonstrates not only algebraic dexterity but also the importance of simplifying expressions before solving — a valuable skill in both academic and applied contexts.", "---", "Key Terms for SEO:\n- Solve quadratic equation\n- Simplify radical expressions\n- Quadratic formula derivation\n- Solve ( x = \frac{3 \pm \sqrt{9 - 8}}{4} )\n- Algebraic simplification steps\n- Quadratic solutions step-by-step\n- Learn fractions and radicals", "Meta Title:\nHow to Solve ( x = \frac{3 \pm \sqrt{9 - 8}}{4} ): A Step-by-Step Guide to ( x = 1, \frac{1}{2} )\nMeta Description:\nUnderstand the full derivation and simplification of ( x = \frac{3 \pm \sqrt{9 - 8}}{4} ) to yield precise solutions ( x = 1 ) and ( x = \frac{1}{2} ), using algebraic principles and quadratic formula insights."]