INECUACIONES CUADRÁTICAS (MÉTODO DE PUNTOS CRÍTICOS)

Introduction to Second Degree Inequalities

Overview of the Topic

• The session is led by Licenciado Bolívar, focusing on second-degree inequalities using the critical points method. The standard form of these inequalities is ax^2 + bx + c which can be greater than, less than, or equal to zero, provided that a neq 0 .

Critical Points Method

• The critical points method arises from simplifying the consequences of order axioms. It involves determining when the expression ax^2 + bx + c is greater than or equal to zero. The first step is ensuring that coefficient a > 0 .

Steps for Solving Inequalities

1. Ensure Coefficient Positivity: Confirm that the leading coefficient a is positive.

2. Factorization: Transform the quadratic expression into two linear factors using methods like completing squares or applying the quadratic formula.

3. Finding Critical Points: Set each linear factor to zero (e.g., x+p = 0 and x+q = 0 ) to find critical points at x = -p and x = -q. Assume without loss of generality that -q > -p.

Determining Solution Sets

Sign Intervals and Solutions

• After identifying critical points, intercalate signs starting from the right (positive, negative, positive). This helps in determining solution sets based on comparisons with zero (greater than or equal to versus less than or equal to). If an inequality states "less than," it corresponds with negative intervals while "greater than" relates to positive intervals.

Example Problems

Example 1: Solving an Inequality

• For the inequality x^2 - 7 > 0:

• Step one confirms positivity of coefficient.

• Step two uses difference of squares for factorization: (x-sqrt7)(x+sqrt7) > 0.

• Step three finds critical points at x = pmsqrt7 , placing them on a number line and intercalating signs as positive-negative-positive.

• Final solution set results in open intervals from negative infinity up to minus root seven and from root seven to positive infinity.

Example 2: Another Inequality Case

• For the inequality x^2 + x ≤ 0:

• First condition checks out since leading coefficient is positive.

• Factorization reveals common term yielding factors as x(x+1) ≤ 0.

• Critical points are found at zero and minus one; sign analysis shows where solutions lie within closed intervals between these values resulting in a solution set from [-1, 0].

Handling Negative Coefficients

Example with Negative Leading Coefficient

• In cases where the leading coefficient is negative (e.g., for expression like 2 + x - 3x²), multiply through by negative one:

• This changes directionality of inequality while transforming it into a solvable format.

Quadratic Inequalities and Critical Points

Finding Critical Points of Quadratic Expressions

• The discussion begins with the identification of linear factors in place of a quadratic expression, leading to the determination of critical points on the real number line.

• The critical points are found by setting the linear factor x equal to zero, yielding solutions x = -2/3 and x = 1 .

• An analysis is conducted to determine intervals based on these critical points, concluding that the solution set includes all numbers from negative infinity to -2/3, and from 1 to positive infinity.

Solving Quadratic Inequalities

Example 1: Inequality Analysis

• The first inequality presented is y^2 - 2y + 1 > 0 . The method of completing squares is introduced for solving this inequality.

• Completing the square results in (y - 1)^2 > 0 . However, it’s noted that when y = 1 , the expression equals zero, which does not satisfy the inequality.

• Thus, the solution set excludes y = 1, resulting in all real numbers except for one.

Example 2: Another Quadratic Inequality

• The second example involves analyzing t - t^2 - 4 leq 0 . It’s observed that since the coefficient of the quadratic term is negative, an initial step involves multiplying through by -1, reversing the inequality direction.

• After rearranging and completing squares again leads to a new form. This transformation helps analyze whether there are any valid solutions for this inequality.

Conclusion on Solution Sets

• Upon further analysis, it becomes clear that while expressions can be positive or zero at certain values (like when squared terms equal zero), they cannot yield a negative result overall.