If \( x^{2}-7 x+1=0 \), what is \( x+\frac{1}{x}-7 ? \)

To solve for \( x + \frac{1}{x} - 7 \) given the equation \( x^2 - 7x + 1 = 0 \), we first need to find the values of \( x \). Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -7 \), and \( c = 1 \): \[ x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \] \[ x = \frac{7 \pm \sqrt{49 - 4}}{2} \] \[ x = \frac{7 \pm \sqrt{45}}{2} \] \[ x = \frac{7 \pm 3\sqrt{5}}{2} \] Now we have two possible values for \( x \): \[ x_1 = \frac{7 + 3\sqrt{5}}{2}, \quad x_2 = \frac{7 - 3\sqrt{5}}{2} \] Next, we need to calculate \( x + \frac{1}{x} \). We can find \( \frac{1}{x} \) using the identity \( \frac{1}{x} = \frac{a}{b} \) where \( x = \frac{b}{a} \). To find \( x + \frac{1}{x} \), we can use the relationship derived from the original quadratic equation. We know: \[ x^2 - 7x + 1 = 0 \implies x^2 = 7x - 1 \] Dividing both sides by \( x \) (assuming \( x \neq 0 \)) gives: \[ x + \frac{1}{x} = 7 - \frac{1}{x} \] Now, we can express \( \frac{1}{x} \) in terms of \( x \): \[ \frac{1}{x} = \frac{1}{\frac{7 \pm 3\sqrt{5}}{2}} = \frac{2}{7 \pm 3\sqrt{5}} \] However, we can also find \( x + \frac{1}{x} \) directly from the quadratic equation. We can use the identity: \[ x + \frac{1}{x} = \frac{x^2 + 1}{x} \] Substituting \( x^2 = 7x - 1 \): \[ x + \frac{1}{x} = \frac{(7x - 1) + 1}{x} = \frac{7x}{x} = 7 \] Thus, we have: \[ x + \frac{1}{x} = 7 \] Finally, we can compute \( x + \frac{1}{x} - 7 \): \[ x + \frac{1}{x} - 7 = 7 - 7 = 0 \] Therefore, the final answer is: \[ \boxed{0} \]