\(x = rac{-(-8) \pm \sqrt{16}}{4}\)

["# Solving the Quadratic Equation: ( x = \frac{-(-8) \pm \sqrt{16}}{4} )", "Solving quadratic equations is a fundamental skill in algebra, essential for students, educators, and anyone interested in mathematics. One particularly clear example is the equation:", "[\nx = \frac{-(-8) \pm \sqrt{16}}{4}\n]", "This equation arises from applying the quadratic formula to a standard quadratic trinomial. Understanding how to break it down not only helps solve for ( x ) but also deepens your grasp of quadratic expressions and their real-world applications.", "## Understanding the Equation", "The general form of a quadratic equation is:", "[\nax^2 + bx + c = 0\n]", "From the given expression, we can identify:\n- ( a = 1 ) (implied since no ( x^2 ) term written),\n- ( b = -8 ),\n- ( c = 0 ) if inferred from the presence of only linear and constant terms being transformed, but here the constant is already captured by the square root term—more precisely, the discriminant is 16, meaning this comes from solving a quadratic that was simplified during derivation.", "However, let's re-express the equation clearly:", "[\nx = \frac{-(-8) \pm \sqrt{16}}{4}\n]", "Simplifying:", "- ( -(-8) = 8 )\n- ( \sqrt{16} = 4 )", "So the equation becomes:", "[\nx = \frac{8 \pm 4}{4}\n]", "This reveals two solutions:\n- ( x = \frac{8 + 4}{4} = \frac{12}{4} = 3 )\n- ( x = \frac{8 - 4}{4} = \frac{4}{4} = 1 )", "### Therefore, the solutions are:", "[\nx = 3 \quad \ ext{and} \quad x = 1\n]", "## Deriving the Quadratic Equation", "To arrive at this form, solve the corresponding quadratic equation:", "Start with:", "[\nx = \frac{8 \pm 4}{4}\n]", "Multiply both sides by 4:", "[\n4x = 8 \pm 4\n]", "Split into two cases:", "1. ( 4x = 8 + 4 = 12 ) → ( x = 3 )\n2. ( 4x = 8 - 4 = 4 ) → ( x = 1 )", "This traces back directly to the simplified quadratic form—likely derived from factoring, completing the square, or applying the quadratic formula beyond standard order.", "## The Quadratic Formula Connection", "The expression ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ) defines the quadratic formula. In our equation:", "- Sum of solutions from ( x = \frac{8 \pm 4}{4} ) gives:", "[\nx = \frac{-b}{2a} \quad \ ext{(sum of roots)}\n]\n[\n\frac{-b}{2a} = \frac{-(-8)}{2 \cdot 1} = \frac{8}{2} = 4\n]", "This matches the sum: ( 3 + 1 = 4 ).", "Also, product of roots:", "[\n3 \ imes 1 = 3 = \frac{c}{a} \Rightarrow c = 3\n]", "But wait—earlier we noticed no ( c ) term in the initial expression. This indicates that the original equation likely originated from simplifying a depressed quadratic or through specific manipulation where the constant term was absorbed.", "Indeed, verify:", "Using ( a = 1 ), ( b = -8 ), ( c = 3 ), then:", "[\nx = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(3)}}{4} = \frac{8 \pm \sqrt{64 - 12}}{4} = \frac{8 \pm \sqrt{52}}{4}\n]", "But ( \sqrt{52} = 2\sqrt{13} ), not 4. So this does not match unless there was a prior correction or assumption.", "Thus, the given expression ( x = \frac{-(-8) \pm \sqrt{16}}{4} ) assumes the discriminant is 16 — which suggests that ( b^2 - 4ac = 16 ), consistent with simplified quadratic forms.", "Therefore, the correct equation producing this form is likely derived from:", "[\nx^2 + 8x = 0\n]", "(With ( c = 0 )), so:", "[\n\frac{-8 \pm \sqrt{8^2 - 4(1)(0)}}{4} = \frac{-8 \pm \sqrt{64}}{4} = \frac{-8 \pm 8}{4}\n]", "But this gives ( x = 0 ) and ( x = -4 ), which does not match.", "Alternatively, if the discriminant is 16, then:", "[\nb^2 - 4ac = 16\n]", "With ( b = -8 ), then:", "[\n64 - 4ac = 16 \Rightarrow 4ac = 48 \Rightarrow ac = 12\n]", "Choosing ( a = 1 ), ( c = 12 ), we get:", "[\nx = \frac{8 \pm 4}{4} = \frac{12}{4} = 3,\quad \frac{4}{4} = 1\n]", "So, the equation is effectively:", "[\nx^2 - 4x + 12 = 0\n]", "(since sum = 4, product = 12, and ( b = -4 ), but this contradicts ( b = -8 ) in numerator)", "Clarification: The original equation ( x = \frac{-(-8) \pm \sqrt{16}}{4} ) inherently sets ( b = 8 ), ( \sqrt{16} = 4 ), so ( -b = -8 ), indicating a derivative from dividing through by 2, suggesting the full equation was scaled or simplified.", "Nonetheless, solving the stated expression cleanly yields:", "[\nx = 3,\quad x = 1\n]", "## Practical Applications of This Solution", "Quadratic equations model motion, profit maximization, geometry, and physics problems. This specific solution set shows that when a quadratic factors nicely—such as into ( (x - 1)(x - 3) = 0 )—the roots are 1 and 3, representing critical points like break-even levels or spatial positions.", "Understanding how to manipulate and solve equations of this form equips learners to tackle more complex problems in science, engineering, and economics.", "## Summary", "- The expression ( x = \frac{-(-8) \pm \sqrt{16}}{4} ) simplifies to ( x = \frac{8 \pm 4}{4} ), giving ( x = 3 ) and ( x = 1 ).\n- It reflects a quadratic solution method applied to a simplified or derived equation.\n- The discriminant confirms a real and distinct root story.\n- Mastery of such expressions strengthens algebraic fluency and real-world problem solving.", "## Key Takeaways", "- Always simplify expressions carefully, noting signs and operations.\n- Use the quadratic formula as a bridge from linear and general forms.\n- Verify that derived equations match given roots through substitution.\n- Practice solving quadratics through multiple pathways—factoring, completing the square, and formula—growth your versatility.", "---", "Keywords: quadratic equation, solving quadratic, quadratic formula, algebra solution, (x = \frac{-(-8) \pm \sqrt{16}}{4}), step-by-step quadratic, find roots, math practice, equation solving, mathematics tutorial."]