yx^2 + y = 2x^2 - 3x + 1 \quad \Rightarrow \quad (y - 2)x^2 + 3x + (y - 1) = 0

["Solving the Quadratic Equation: Analyzing yx² + y = 2x² - 3x + 1", "In algebra, solving quadratic equations correctly is essential for understanding curves and relationships in equations. Today, we dive into the equation:", "[ yx^2 + y = 2x^2 - 3x + 1 \quad \Rightarrow \quad (y - 2)x^2 + 3x + (y - 1) = 0 ]", "This transformation helps identify the nature of the quadratic in ( x ), making it easier to solve and interpret.", "---", "### Step 1: Rearranging the Equation", "Start by moving all terms to one side:\n[\nyx^2 + y - 2x^2 + 3x - 1 = 0\n]\nGroup like terms:\n[\n(y - 2)x^2 + 3x + (y - 1) = 0\n]", "This is a standard quadratic equation in the form:\n[\nAx^2 + Bx + C = 0\n]\nwhere\n- ( A = y - 2 )\n- ( B = 3 )\n- ( C = y - 1 )", "---", "### Step 2: Understanding the Roots", "For this quadratic to have real solutions in ( x ), the discriminant ( D = B^2 - 4AC ) must be non-negative:\n[\nD = 3^2 - 4(y - 2)(y - 1) \geq 0\n]", "Calculate:\n[\n9 - 4(y - 2)(y - 1) \geq 0\n]", "Expand the product:\n[\n(y - 2)(y - 1) = y^2 - 3y + 2\n]", "Substitute back:\n[\n9 - 4(y^2 - 3y + 2) \geq 0\n]\n[\n9 - 4y^2 + 12y - 8 \geq 0\n]\n[\n-4y^2 + 12y + 1 \geq 0\n]", "Multiply both sides by –1 (reversing inequality):\n[\n4y^2 - 12y - 1 \leq 0\n]", "---", "### Step 3: Solve the Inequality for ( y )", "Solve the quadratic inequality ( 4y^2 - 12y - 1 \leq 0 ):\nUse the quadratic formula to find roots:\n[\ny = \frac{12 \pm \sqrt{(-12)^2 - 4(4)(-1)}}{2 \cdot 4} = \frac{12 \pm \sqrt{144 + 16}}{8} = \frac{12 \pm \sqrt{160}}{8}\n]\nSimplify:\n[\n\sqrt{160} = \sqrt{16 \cdot 10} = 4\sqrt{10}\n]\n[\ny = \frac{12 \pm 4\sqrt{10}}{8} = \frac{3 \pm \sqrt{10}}{2}\n]", "Thus, the inequality holds for:\n[\n\frac{3 - \sqrt{10}}{2} \leq y \leq \frac{3 + \sqrt{10}}{2}\n]", "---", "### Step 4: Analyzing When the Equation Has Real Solutions in ( x )", "- For ( y ) outside this range, no real ( x ) satisfies the equation.\n- When ( y ) is inside the interval, two real solutions exist.\n- At ( y = \frac{3 \pm \sqrt{10}}{2} ), the discriminant is zero, resulting in exactly one real solution (a repeated root).", "---", "### Step 5: Solving for ( x ) When Real Solutions Exist", "Using the quadratic formula:\n[\nx = \frac{-3 \pm \sqrt{D}}{2(y - 2)}, \quad \ ext{where } D = 9 - 4(y - 2)(y - 1)\n]", "Since ( D = 4y^2 - 12y - 1 ), substituting gives:\n[\nx = \frac{-3 \pm \sqrt{4y^2 - 12y - 1}}{2(y - 2)}\n]", "---", "### Conclusion", "The equation ( yx^2 + y = 2x^2 - 3x + 1 ) translates neatly into a quadratic in ( x ) with coefficients dependent on ( y ). Analyzing its discriminant reveals the values of ( y ) for which real solutions exist, offering insight into the behavior of the equation across the ( y )-plane. Recognizing when real ( x ) solutions occur enhances problem-solving precision in algebra and mathematical modeling.", "---", "### SEO Keywords:\nyx² + y = 2x² - 3x + 1, quadratic equation solved, discriminant analysis, real solutions quadratic, solving for x in quadratic form, algebra solutions, discriminant inequality, algebraic manipulation, quadratic coefficients, y in quadratic equation", "---", "For further reading, explore how changing ( y ) affects the graph of the equation or how to graph this family of parabolas using the derived constraints."]