# How can the equation be positive in the interval $-1 < x < \frac{7-\sqrt{73}}{2}$?

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

• You seem to be focused on whether the left side is positive, but you are being asked if the left side is greater than the right. Those are two different questions. – Ross Millikan Dec 16 '16 at 20:30

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. $$\frac { x^{ 2 }-8x-7 }{ x+1 } \geq -1\\ \frac { x^{ 2 }-8x-7 }{ x+1 } +1\ge 0\\ \frac { x^{ 2 }-8x-7+x+1 }{ x+1 } \ge 0\\ \frac { x^{ 2 }-7x-6 }{ x+1 } \ge 0\\ \frac { \left( x+1 \right) \left( x^{ 2 }-7x-6 \right) }{ { \left( x+1 \right) }^{ 2 } } \ge 0\\ \left( x+1 \right) \left( x^{ 2 }-7x-6 \right) \ge 0\\ \left( x+1 \right) \left( x-\frac { 7-\sqrt { 73 } }{ 2 } \right) \left( x-\frac { 7+\sqrt { 73 } }{ 2 } \right) \ge 0\\ \\ \frac { 7-\sqrt { 73 } }{ 2 } \approx -0.772001872659\\ \frac { 7+\sqrt { 73 } }{ 2 } \approx 7.77200187266\\ -1<x\le \frac { 7-\sqrt { 73 } }{ 2 } \quad and\quad \frac { 7+\sqrt { 73 } }{ 2 } \le x<+\infty \\ \\ \\$$ • Thank you for redoing it with the right values. Could you please explain to me why the x between -1 and the negative root is an answer? I think my issue started probably after you had $\frac/x^2-7x-6}{x+1}$ and added the extra x+1's – George Dec 16 '16 at 21:04
• Note that you must restrict $x \ne -1$ from the very beginning, otherwise the original inequality is not even defined. The way it's written, when you get to the point $\left( x+1 \right) \left( x^{ 2 }-7x-6 \right) \ge 0$, it looks like $x=-1$ is a valid solution of this inequality, yet it is not a solution of the original one. – dxiv Dec 17 '16 at 2:54