which value (s) of m will (x+m)^2-2(x+m)-11=x-1 have Two different negative roots?

To find the value of \( n \) such that the equation \((x+n)^2 - 2(x+n) - 11 = x - 1\) has two different negative roots, we first need to simplify and rearrange the equation.<br /><br />Start by expanding the left side:<br /><br />\[<br />(x+n)^2 - 2(x+n) - 11 = (x^2 + 2nx + n^2) - 2x - 2n - 11<br />\]<br /><br />Simplify this expression:<br /><br />\[<br />x^2 + 2nx + n^2 - 2x - 2n - 11<br />\]<br /><br />Set it equal to the right side of the original equation:<br /><br />\[<br />x^2 + 2nx + n^2 - 2x - 2n - 11 = x - 1<br />\]<br /><br />Rearrange all terms to one side:<br /><br />\[<br />x^2 + (2n-3)x + (n^2 - 2n - 10) = 0<br />\]<br /><br />This is a quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = 2n - 3 \), and \( c = n^2 - 2n - 10 \).<br /><br />For the quadratic equation to have two different negative roots, the following conditions must be satisfied:<br /><br />1. The discriminant must be positive: \( b^2 - 4ac > 0 \).<br />2. Both roots must be negative.<br /><br />First, calculate the discriminant:<br /><br />\[<br />b^2 - 4ac = (2n - 3)^2 - 4(1)(n^2 - 2n - 10)<br />\]<br /><br />Simplify:<br /><br />\[<br />= (4n^2 - 12n + 9) - (4n^2 - 8n - 40)<br />\]<br />\[<br />= 4n^2 - 12n + 9 - 4n^2 + 8n + 40<br />\]<br />\[<br />= -4n + 49<br />\]<br /><br />Set the discriminant greater than zero:<br /><br />\[<br />-4n + 49 > 0<br />\]<br /><br />Solve for \( n \):<br /><br />\[<br />-4n > -49<br />\]<br />\[<br />n < \frac{49}{4}<br />\]<br /><br />Next, ensure both roots are negative. For a quadratic \( ax^2 + bx + c = 0 \) with \( a > 0 \), both roots are negative if:<br /><br />- \( b > 0 \)<br />- \( c > 0 \)<br /><br />Check these conditions:<br /><br />1. \( b = 2n - 3 > 0 \)<br /><br /> \[<br /> 2n > 3 \quad \Rightarrow \quad n > \frac{3}{2}<br /> \]<br /><br />2. \( c = n^2 - 2n - 10 > 0 \)<br /><br /> Solve \( n^2 - 2n - 10 > 0 \) using the quadratic formula or factorization:<br /><br /> The roots of \( n^2 - 2n - 10 = 0 \) are found using the quadratic formula:<br /><br /> \[<br /> n = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1}<br /> \]<br /> \[<br /> n = \frac{2 \pm \sqrt{4 + 40}}{2}<br /> \]<br /> \[<br /> n = \frac{2 \pm \sqrt{44}}{2}<br /> \]<br /> \[<br /> n = \frac{2 \pm 2\sqrt{11}}{2}<br /> \]<br /> \[<br /> n = 1 \pm \sqrt{11}<br /> \]<br /><br /> The critical points are \( n = 1 + \sqrt{11} \) and \( n = 1 - \sqrt{11} \).<br /><br /> Since \( \sqrt{11} \approx 3.32 \), we have approximately \( n = 4.32 \) and \( n = -2.32 \).<br /><br /> The inequality \( n^2 - 2n - 10 > 0 \) holds for \( n < 1 - \sqrt{11} \) or \( n > 1 + \sqrt{11} \).<br /><br />Combine all conditions:<br /><br />- \( \frac{3}{2} < n < \frac{49}{4} \)<br />- \( n > 1 + \sqrt{11} \approx 4.32 \)<br /><br />The only interval satisfying all conditions is \( n > 1 + \sqrt{11} \).<br /><br />Thus, the integer value of \( n \) that satisfies all conditions is \( n = 5 \).