2r - (3 + \sqrt{5})r + 2 + 3 + \sqrt{5} = 0
["Understanding the Equation: 2r - (3 + \sqrt{5})r + 2 + 3 + \sqrt{5} = 0", "In the realm of algebra, solving equations involving radicals and rational coefficients is both challenging and enlightening. This article explores the quadratic equation:", "[\n2r - (3 + \sqrt{5})r + 2 + 3 + \sqrt{5} = 0\n]", "We will simplify, analyze, and solve this equation step-by-step, highlighting key algebraic techniques and the significance of irrational numbers—particularly (\sqrt{5})—in quadratic expressions.", "---", "### 1. Simplify the Equation", "Start by simplifying all terms in the equation:", "[\n2r - (3 + \sqrt{5})r + (2 + 3 + \sqrt{5}) = 0\n]", "Simplify the coefficients and constants:", "- Coefficient of ( r ):\n ( 2 - (3 + \sqrt{5}) = 2 - 3 - \sqrt{5} = -1 - \sqrt{5} )", "- Constant term:\n ( 2 + 3 + \sqrt{5} = 5 + \sqrt{5} )", "So, the equation becomes:", "[\n(-1 - \sqrt{5})r + 5 + \sqrt{5} = 0\n]", "---", "### 2. Isolate the Variable ( r )", "Move the constant term to the right side:", "[\n(-1 - \sqrt{5})r = - (5 + \sqrt{5})\n]", "To solve for ( r ), divide both sides by ( (-1 - \sqrt{5}) ):", "[\nr = \frac{-(5 + \sqrt{5})}{-1 - \sqrt{5}} = \frac{5 + \sqrt{5}}{1 + \sqrt{5}}\n]", "Note: The negative signs cancel.", "---", "### 3. Rationalize the Denominator", "The expression ( \frac{5 + \sqrt{5}}{1 + \sqrt{5}} ) features an irrational denominator. Rationalize it by multiplying numerator and denominator by the conjugate ( 1 - \sqrt{5} ):", "[\nr = \frac{5 + \sqrt{5}}{1 + \sqrt{5}} \cdot \frac{1 - \sqrt{5}}{1 - \sqrt{5}} = \frac{(5 + \sqrt{5})(1 - \sqrt{5})}{(1 + \sqrt{5})(1 - \sqrt{5})}\n]", "Compute numerator:\n[\n(5 + \sqrt{5})(1 - \sqrt{5}) = 5(1) - 5\sqrt{5} + \sqrt{5}(1) - \sqrt{5}\cdot\sqrt{5} = 5 - 5\sqrt{5} + \sqrt{5} - 5 = (5 - 5) + (-5\sqrt{5} + \sqrt{5}) = -4\sqrt{5}\n]", "Compute denominator:\n[\n(1 + \sqrt{5})(1 - \sqrt{5}) = 1^2 - (\sqrt{5})^2 = 1 - 5 = -4\n]", "Thus:", "[\nr = \frac{-4\sqrt{5}}{-4} = \sqrt{5}\n]", "---", "### 4. Final Solution", "The solution to the equation\n[\n2r - (3 + \sqrt{5})r + 2 + 3 + \sqrt{5} = 0\n]\nis", "[\n\boxed{r = \sqrt{5}}\n]", "---", "### 5. Why This Root Matters: The Role of ( \sqrt{5} )", "The number ( \sqrt{5} ) is an irrational number, representing the positive solution to the quadratic equation ( x^2 - 5 = 0 ). Here, its presence within coefficients leads to an exact, non-radical solution — a striking example of how irrational constants can yield elegant algebraic results.", "This equation exemplifies the utility of combining like terms, simplifying radical expressions, and rationalizing denominators — fundamental skills in solving equations with irrational coefficients.", "---", "### 6. Verifying the Solution", "Substitute ( r = \sqrt{5} ) into the original equation:", "Left-hand side:", "[\n2\sqrt{5} - (3 + \sqrt{5})\sqrt{5} + 5 + \sqrt{5}\n= 2\sqrt{5} - (3\sqrt{5} + 5) + 5 + \sqrt{5}\n]", "Simplify:", "[\n2\sqrt{5} - 3\sqrt{5} - 5 + 5 + \sqrt{5} = (2 - 3 + 1)\sqrt{5} + (-5 + 5) = 0 + 0 = 0\n]", "The equation holds true.", "---", "### Conclusion", "The equation ( 2r - (3 + \sqrt{5})r + 2 + 3 + \sqrt{5} = 0 ) may appear complex due to the irrational term ( \sqrt{5} ), but through careful simplification and rationalization, we find the elegant solution ( r = \sqrt{5} ). This problem illustrates the power of algebraic manipulation and underscores the importance of handling irrational numbers with precision — a cornerstone of higher-level mathematics.", "For students and enthusiasts alike, mastering such equations builds a strong foundation for tackling advanced algebraic theory and real-world applications involving non-integer solutions.", "---", "Keywords:\nquadratic equation, irrational numbers, solving equations with radicals, simplifying expressions, rationalizing denominators, algebraic solutions, √5, algebra homework help, equation solution steps, 2r - (3 + √5)r + 2 + 3 + √5 = 0, exact solutions, math tutorial"]
