Solve for all values of x: (1)/(x-1)+4=(1)/(x^2)-x Answer Attemptiout of 2 Additional Solution (C) No Solution x=square

To solve the equation \(\frac{1}{x-1} + 4 = \frac{1}{x^2 - x}\), we first simplify the equation by finding a common denominator.<br /><br />The common denominator for the fractions on the left-hand side and the right-hand side is \(x^2 - x\). We can rewrite the equation as:<br /><br />\[<br />\frac{x^2 - x}{x^2 - x} + \frac{4(x^2 - x)}{x^2 - x} = \frac{1}{x^2 - x}<br />\]<br /><br />Simplifying the left-hand side, we get:<br /><br />\[<br />\frac{x^2 - x + 4(x^2 - x)}{x^2 - x} = \frac{1}{x^2 - x}<br />\]<br /><br />Combining like terms in the numerator, we have:<br /><br />\[<br />\frac{5x^2 - 5x}{x^2 - x} = \frac{1}{x^2 - x}<br />\]<br /><br />Since the denominators are the same, we can equate the numerators:<br /><br />\[<br />5x^2 - 5x = 1<br />\]<br /><br />Rearranging the equation, we get:<br /><br />\[<br />5x^2 - 5x - 1 = 0<br />\]<br /><br />This is a quadratic equation in the form \(ax^2 + bx + c = 0\), where \(a = 5\), \(b = -5\), and \(c = -1\). We can solve this equation using the quadratic formula:<br /><br />\[<br />x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}<br />\]<br /><br />Substituting the values of \(a\), \(b\), and \(c\), we get:<br /><br />\[<br />x = \frac{5 \pm \sqrt{25 + 20}}{10}<br />\]<br /><br />Simplifying further, we have:<br /><br />\[<br />x = \frac{5 \pm \sqrt{45}}{10}<br />\]<br /><br />\[<br />x = \frac{5 \pm 3\sqrt{5}}{10}<br />\]<br /><br />Therefore, the solutions to the equation are:<br /><br />\[<br />x = \frac{5 + 3\sqrt{5}}{10} \quad \text{and} \quad x = \frac{5 - 3\sqrt{5}}{10}<br />\]<br /><br />So, the final answer is:<br /><br />\[<br />x = \frac{5 + 3\sqrt{5}}{10} \quad \text{or} \quad x = \frac{5 - 3\sqrt{5}}{10}<br />\]