Problem 61 - College Algebra

Question

Describe similarities and differences between the solutions of $$ (x-2)(x+5) \geq 0 \text { and } \frac{x-2}{x+5} \geq 0 $$

Step-by-Step Solution

Answer

The solutions for both inequalities are identical for \(x\geq2\), however, the solution \(x<-5\), valid for the first inequality does not apply to the second one.

Step 1: Solving First Inequality

The given inequality is \((x-2)(x+5) \geq 0\). The product of two terms will be positive If both terms are positive or both terms are negative.\n\nTherefore, the set of solutions of the inequality are:(i) when both \(x-2 \) and \(x+5 \) are positive i.e, \(x\geq2\) and \(x\geq-5\) ; the common part is \(x\geq2\) . (ii) When both \(x-2 \) and \(x+5 \) are negative i.e, \(x<2\) and \(x<-5\) ; the common part is \(x<-5\).\n\nTherefore the solutions are \( x\geq2 \) , \(x<-5\).

Step 2: Solving Second Inequality

Solving the second inequality \(\frac{x-2}{x+5} \geq 0\). From the inequality, we can see that the solution to the inequality will be when denominator and numerator have same sign.\n\nThe set of solutions of the inequality in this case are: (i) \(x-2 \geq 0\) , \(x+5 \geq 0\) ; the common part is \(x\geq2\) . (ii) \(x-2 < 0\) , \(x+5 < 0\) ; the common part is \(x<-5\) , but this solution is extraneous because then the denominator \(x+5\) would become zero and the inequality is undefined. So The solution will be \(x\geq2\) only.

Step 3: Comparing Solutions and Observing Differences and Similarities

Comparing the solutions of both inequalities, we see that they only share the solution \(x\geq2\). The significant difference here is that the first inequality \((x-2)(x+5) \geq 0\) also includes the solution \(x<-5\), which does not exist in the solution set for the second inequality \(\frac{x-2}{x+5} \geq 0\), due to the fact that at \(x<-5\), the denominator becomes zero and the expression becomes undefined.

Key Concepts

Polynomial Inequalities

Polynomial inequalities have great importance in algebra and involve expressions where a polynomial is compared to zero using an inequality symbol, such as ">=", "<=", ">", or "<". In the provided example, \((x-2)(x+5) \geq 0\), we have a polynomial inequality involving a quadratic polynomial, \((x-2)(x+5)\). Solving polynomial inequalities usually involves finding the critical points where the polynomial equals zero, then determining the intervals where the polynomial maintains the desired inequality.

• Critical Points: These are solutions of \((x-2)(x+5) = 0\), which are \(x=2\) and \(x=-5\).
• Test Intervals: Create test intervals around the critical points: \((-\infty, -5)\), \((-5, 2)\), and \((2, \infty)\).
• Check Signs: Substitute a number from each interval into the inequality to find where the inequality holds true.

This process identifies where the expression is positive or zero, covering both negative and positive cases based on the inequality's requirement.

Rational Inequalities

Rational inequalities typically involve a ratio of two polynomials and are solved in a similar way to polynomial inequalities. An example can be seen in \(\frac{x-2}{x+5} \geq 0\). A critical aspect of solving rational inequalities is considering where the expression is undefined, generally occurring when the denominator equals zero.

• Decompose: Identify when both numerator and denominator equal zero, \(x=2\) for the numerator and \(x=-5\) for the denominator.
• Avoid Undefined Values: For intervals that include points where the denominator is zero, such as \(x=-5\), solutions are typically excluded.
• Interval Testing: After identifying critical points, select test values from intervals to determine where the inequality holds, remembering to exclude undefined points.

In rational inequalities, the solutions are contingent on the numerator and denominator having the same sign, avoiding zero in the denominator.

Solution Sets

The solution set of an inequality represents the range of values that satisfy the inequality. Determining accurate solution sets requires considering the nature and restrictions of the inequalities involved.

• Polynomial Solutions: In the polynomial inequality \((x-2)(x+5) \geq 0\), solutions come from intervals where the product is non-negative, resulting here in \(x \geq 2\) and \(x <-5\).
• Rational Solutions: The solution set for \(\frac{x-2}{x+5} \geq 0\) avoids undefined regions and ensures numerator and denominator are both positive or both negative, yielding \(x \geq 2\) exclusively after considering undefined points.
• Brackets: Use brackets [ ] to include endpoints when the inequality is non-strict ("\(\geq\)", "\(\leq\)") and parentheses ( ) for strict inequalities (">", "<").

By outlining the solutions, one can effectively visualize the ranges of values fulfilling the inequality requirements.

Comparison of Inequalities

Comparing and contrasting inequalities helps to understand their solution sets and potential overlaps or exclusions. In the given inequalities, distinct needs arise due to their polynomial and rational natures.

• Inclusive vs. Exclusive: In \((x-2)(x+5) \geq 0\), both ends of the solution interval \(x \geq 2\) are considered inclusive, but in the rational inequality \(\frac{x-2}{x+5} \geq 0\), \(x=-5\) is excluded due to undefined conditions.
• Overlap: Both inequalities share \(x \geq 2\) implying common intervals, yet differ when \(x<-5\) is rejected in the rational inequality.
• Nature of Zeroes: In the rational inequality, zero factors, such as the point where the denominator becomes zero, create points of exclusion impacting the solution set differently than in polynomial inequalities.

Understanding these comparisons aids students in grasping the nuances and particularities of different inequality types.