We consider the numerator placing it greater than or equal to 0

The equation associated with this inequality is We trace the axis of ordinate and abscissas and we highlight the points (0;0) and (-1;0). We find the vertex of the parabola and we trace it having a concavity upwards passing thorough the two points specified above. The positive part of the graph (parabola branches in the half-plane positive ordinate) is the image of the solution of the inequality so, the values of less than or equal to -1 and the values of greater than or equal to 0. We make the table of signs bringing back in ascending order from left to right the solution of the equation associated and represent the set of solution of the inequality of the numerator and under the denominator using at intervals of values of that make positive inequality

and in the intervals that make the inequality negative.

Determines the sign of the quotient in each interval that has been created, so the solutions of the inequality are: (spurious equation). To find this solution collect for common factor . For the annulment law, the equation is verified for and Solutions correspond to the values of that have a positive image in the function We consider the denominator placing it greater than 0 . The equation associated is (full equation). To find the solutions in case the is greater than

or equal to 0 we apply the formula

For the annulment law, the equation is verified for and . Solutions correspond to that have a pasitive image the values of in the function We trace the axis of ordinate and abscissas and we highlight the points and (1;0). We find the vertex of the parabola and we trace it having a concavity upwards passing thorough the two points specified above. The positive part of the graph (parabola branches in the half-plane positive ordinate) is the image of the solution of the inequality so, the values of less than or equal to -1 and the values of greater than or equal to 0.

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# Inequality - Claudia Ferraretto

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