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I have the following inequation :

$$\frac{1}{x-x^2-1}< 0$$

I know that the solution set will be all $x\in R$

but how do we find the answer? if we take the root of equation, we get imaginary values.

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Hint:

Since the numerator is positive, it is sufficient to solve the inequality: $$-x^2 + x -1 < 0.$$

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If $a,b \in \Bbb{R}$, and $a,b \neq 0$, then $$ \frac{a}{b} < 0 \iff \text{ the signs of } a \text{ and } b \text{ are different}. $$ So, because $1$ is positive, we have $$ \frac{1}{x-x^2-1} < 0 \iff x - x^2 - 1 < 0\,. $$

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Note that $$x-x^2-1=-\left(x-\frac 12\right)^2-\frac 34\lt 0.$$

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The second-degree polynomial $x^2-x+1$ has a negative discriminant ($1-4=-3$), so the sign of $\frac{1}{x-x^2-1}$ is always the same for any $x\in\mathbb{R}$.

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  • $\begingroup$ what is that rule? i know thats true, but how is it so? the roots are imaginary, so how do we proceed? $\endgroup$ – Max Payne Apr 18 '15 at 15:14
  • $\begingroup$ @Tim: if a second degree polynomial has only imaginary roots (i.e. negative discriminant) it means that it never vanishes over $\mathbb{R}$, so it always has the same sign (it is a non-vanishing continuous function over $\mathbb{R}$). $\endgroup$ – Jack D'Aurizio Apr 18 '15 at 15:20
  • $\begingroup$ So denominator can have any value? it will always satisfy the inequation for all real nos? Is it true for all such (imaginary root) expressions? $\endgroup$ – Max Payne Apr 18 '15 at 15:26
  • $\begingroup$ @TimKrul: No. Yes. Yes. $\endgroup$ – Jack D'Aurizio Apr 18 '15 at 15:39
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    $\begingroup$ yea obviously not 0 as denominator! Thanks very much! $\endgroup$ – Max Payne Apr 18 '15 at 16:06

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