Generalized Solutions for Quadratic Inequalities

Abstract

INTRODUCTION

Today, Mathematics is typically not the most preferred subject among students in the Philippines and perhaps in other countries. Based on the study of Gafoor and Kurukkan (2015), one of the reasons why students lose their interest in the subject is complex methods or ways in solving Mathematical problems. The main thrust of this study is to present an easier way of solving certain mathematical problems involving inequalities. Inequalities are one of the topics in Algebra, which is a fundamental subject in basic education. In one way or another, it may contribute to the development of mathematics by solving the generalized solutions for quadratic inequalities and providing an easier method.

METHODS

The researcher used an inductive reasoning method to find a pattern to make a conjecture in certain cases. Then deductive reasoning was also utilized in proving the conjecture using sign chart method to assure the generalization of the claim that is true for all. The results are the baseline of new methods of finding the solution set of quadratic inequality. Finally, the researcher used this method and other existing methods to solve some examples involving quadratic inequalities and compare the results.

RESULTS

Result 1. The generalized solution for quadratic inequality (x-r_1 )(x-r_2 )>0 where r_1<r_2 and element of real numbers is {x∈(-∞, r_1 )∪(r_2,+∞)} however if (x-r_1 )(x-r_2 ) ≥0 the solution set is {x∈(-∞, r_1 ]∪[ r_2,+∞)}. Result 2. The generalized solution for quadratic inequality(x-r_1 )(x-r_2 )<0 where r_1<r_2 and element of real numbers is {x∈(r_1,r_2 )} however if (x-r_1 )(x-r_2 )≤0 the solution set is {x∈[〖r〗_1,r_2 ]}.

DISCUSSIONS

Since the solution set for quadratic inequalities was generalized, the method of solving for quadratic inequalities will be shortened and easier, for instance, Step 1. Determine the district roots of quadratic inequality. Step 2. Determine the intervals (-∞,r_1), (r_1, r_2), and, (r_2, +∞). Step 3. If (x-r_1) (x-r_2)>0 based on result 1, the solutions set is {x∈ (-∞, r_1) ∪ (r_2, +∞)}. Also, if (x-r_1) (x-r_2) ≥0 the solution set is {x∈ (-∞, r_1] ∪ [ r_2, +∞)}. Otherwise if (x-r_1) (x-r_2) <0 based on result 2, the solution set is {x∈ (r_1, r_2)}. Also, if (x-r_1) (x-r_2) ≤0 the solution set is {x∈ [r_1, r_2]}.