I've got the inequality: $$\frac{x^2-8x-7}{x+1}\geq -1$$
For the inequality to be larger than zero we need either both numerator and denominator to be positive OR both numerator and denominator to be negative.
I draw the number line with all the zeros in question, i.e, -1 (root of the bottom equation) and $\frac{7\pm\sqrt{73}}{2}$ (roots of the quadratic). This way I have the intervals of interest. I "move" the 1 to the left hand side and then simplifying I obtain $x^2-7x-6$ in the numerator and $x-1$ in the denominator. Let's call the quadratic 'A' and the denominator equation 'B'.
For $x \leq-1$, I get that A is positive and B is negative. For $-1\leq x \leq \frac{7-\sqrt{73}}{2}$ I get A is negative and B is positive. For $\frac{7-\sqrt{73}}{2} \leq x \leq \frac{7+\sqrt{73}}{2}$ I get that A is negative and B is positive. And finally for $x\geq\frac{7+\sqrt{73}}{2}$ I get that both A and B are positive.
Thus, the only interval that satisfies my inequality is the last one, because both numerator and denominator are positive. But in the official result, the interval between $-1$ and $\frac{7-\sqrt{73}}{2}$ is also included in the result. But if you replace X with any value from that interval you'll get a negative result, which would render the whole equation negative. What am I missing here?
Thank you
It should be as follows :$$\frac { x^{ 2 }-8x-7 }{ x+1 } \geq x+1\\ \frac { x^{ 2 }-8x-7 }{ x+1 } -x-1\ge 0\\ \frac { x^{ 2 }-8x-7-{ x }^{ 2 }-2x-1 }{ x+1 } \ge 0\\ \frac { -10x-8 }{ x+1 } \ge 0\\ \frac { 5x+4 }{ x+1 } \le 0\\ \frac { \left( x+1 \right) \left( 5x+4 \right) }{ { \left( x+1 \right) }^{ 2 } } \le 0\\ \left( x+1 \right) \left( 5x+4 \right) \le 0\\ -1<x\le -\frac { 4 }{ 5 } \\ \\ $$
EDIT version. $$\frac { x^{ 2 }-8x-7 }{ x+1 } \geq -1\\ \frac { x^{ 2 }-8x-7 }{ x+1 } +1\ge 0\\ \frac { x^{ 2 }-8x-7+x+1 }{ x+1 } \ge 0\\ \frac { x^{ 2 }-7x-6 }{ x+1 } \ge 0\\ \frac { \left( x+1 \right) \left( x^{ 2 }-7x-6 \right) }{ { \left( x+1 \right) }^{ 2 } } \ge 0\\ \left( x+1 \right) \left( x^{ 2 }-7x-6 \right) \ge 0\\ \left( x+1 \right) \left( x-\frac { 7-\sqrt { 73 } }{ 2 } \right) \left( x-\frac { 7+\sqrt { 73 } }{ 2 } \right) \ge 0\\ \\ \frac { 7-\sqrt { 73 } }{ 2 } \approx -0.772001872659\\ \frac { 7+\sqrt { 73 } }{ 2 } \approx 7.77200187266\\ -1<x\le \frac { 7-\sqrt { 73 } }{ 2 } \quad and\quad \frac { 7+\sqrt { 73 } }{ 2 } \le x<+\infty \\ \\ \\ $$