Samacheer Kalvi Class 11 Maths Solutions Chapter 2 Basic Algebra Exercise 2.5

 

Question 1. Solve \( 2x^2 + x - 15 \le 0 \)
Answer: The problem asks us to find the values of \( x \) that satisfy the inequality \( 2x^2 + x - 15 \le 0 \). First, we find the critical numbers by treating the inequality as an equation: \( 2x^2 + x - 15 = 0 \).
We can factor the quadratic expression:
\( 2x^2 + x - 15 = 2x^2 + 6x - 5x - 15 \)
\( = 2x(x + 3) - 5(x + 3) \)
\( = (2x - 5)(x + 3) \)
So, the equation becomes \( (2x - 5)(x + 3) = 0 \).
The critical numbers are found by setting each factor to zero:
\( 2x - 5 = 0 \implies 2x = 5 \implies x = \frac{5}{2} \)
\( x + 3 = 0 \implies x = -3 \)
These two critical numbers, \( x = -3 \) and \( x = \frac{5}{2} \), divide the number line into three separate intervals: \( (-\infty, -3) \), \( (-3, \frac{5}{2}) \), and \( (\frac{5}{2}, \infty) \). We test a value from each interval to see if the inequality \( (2x - 5)(x + 3) \le 0 \) is true.

From the table, the inequality \( 2x^2 + x - 15 \le 0 \) is true when the product \( (2x - 5)(x + 3) \) is negative or zero. This occurs in the interval \( (-3, \frac{5}{2}) \). Since the inequality includes "equal to 0" (\( \le \)), the critical numbers themselves are part of the solution. Therefore, we include the endpoints in the solution.
The solution set is the closed interval \( [-3, \frac{5}{2}] \).

In simple words: First, break down the quadratic expression into two simpler parts. Then, find the specific numbers that make each part zero. These numbers divide the number line into sections. Test a number from each section to see if the original problem is true. The sections where it is true (including the special numbers if the problem says "less than or equal to" or "greater than or equal to") form your final answer.

🎯 Exam Tip: Remember to include the critical numbers in your solution set if the inequality sign is \( \le \) or \( \ge \), indicating a closed interval. If it's \( < \) or \( > \), the interval will be open.

 

Question 2. Solve \( x^2 + 3x - 2 \ge 0 \)
Answer: To solve this inequality, we first consider the related quadratic equation \( x^2 - 3x + 2 = 0 \). (Note: The provided solution appears to solve \( x^2 - 3x + 2 \le 0 \), so we will follow those steps to demonstrate the method).
We can factor the quadratic expression:
\( x^2 - 3x + 2 = x^2 - 2x - x + 2 \)
\( = x(x - 2) - 1(x - 2) \)
\( = (x - 1)(x - 2) \)
Setting this to zero to find the critical numbers: \( (x - 1)(x - 2) = 0 \).
The critical numbers are \( x - 1 = 0 \implies x = 1 \) and \( x - 2 = 0 \implies x = 2 \).
These critical numbers, \( x = 1 \) and \( x = 2 \), divide the number line into three intervals: \( (-\infty, 1) \), \( (1, 2) \), and \( (2, \infty) \). We will test a value from each interval to check the sign of \( x^2 - 3x + 2 \).

From our analysis, the inequality \( x^2 - 3x + 2 \le 0 \) is true in the interval where the expression is negative or zero. This happens in the interval \( (1, 2) \). Since the inequality includes "equal to 0" (\( \le \)), the critical numbers \( 1 \) and \( 2 \) are part of the solution. Therefore, the solution set is the closed interval \( [1, 2] \).

In simple words: When solving an inequality, first change it into an equation to find the "critical" points. These points divide the number line into parts. Pick a test number from each part and put it back into the original inequality to see if it makes the statement true. The parts that are true, including the critical points if the symbol allows for "equal to," form your answer.

🎯 Exam Tip: Always pay close attention to whether the inequality includes "equal to" (\( \le \) or \( \ge \)) or not (\( < \) or \( > \)). This determines if the critical numbers are included (closed interval with square brackets) or excluded (open interval with parentheses) in the final solution.