Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
📌 Based on National Curriculum 2023 / PECTAA 2026 Syllabus
📖 What's Inside: This exercise covers solving quadratic inequalities using the sign chart method – finding critical points, testing intervals, and writing solution sets. Perfect for Punjab Boards exam preparation.
📚 Related Resources – Unit 2: Quadratic Equations & Inequalities
Class 10 Math Unit 2 Exercise 2.5 – Quadratic Inequalities: Complete Guide
Exercise 2.5 of Unit 2 introduces students to quadratic inequalities and the sign chart method for solving them. Students learn to find critical points by solving the associated equation, divide the number line into intervals, test points, and write solution sets. These skills are essential for advanced algebra and real-world applications.
What You Will Learn
By working through Exercise 2.5, students master: solving quadratic inequalities using the sign chart method, finding critical points from the associated equation, testing intervals to determine the sign, and writing solution sets in interval notation.
Topics Covered in This Exercise
- Sign Chart Method – step-by-step approach to solving inequalities
- Critical Points – finding roots of the associated equation
- Interval Testing – determining the sign of the expression in each interval
- Solution Sets – writing answers in interval notation
- Inequality Types – \( > \), \( < \), \( \ge \), \( \le \)
Why Exercise 2.5 Is Important
Exercise 2.5 introduces a critical skill for solving inequalities. Board exam papers frequently feature questions on quadratic inequalities. This exercise builds essential skills for understanding how to find solution sets of inequalities.
Punjab Board Preparation
Students preparing for board exams under any of the 10 BISE Punjab boards should prioritise this exercise. Quadratic inequalities are frequently tested topics. Practising every part of Q1 by hand is highly recommended.
Exam Tips for Quadratic Inequalities
- Always solve the associated equation first to find critical points.
- Test one point from each interval to determine the sign.
- Use open circles for strict inequalities (\( > \), \( < \)) and closed circles for non-strict inequalities (\( \ge \), \( \le \)).
- Write the solution set in interval notation.
Common Mistakes Students Make
- Forgetting to find the critical points first.
- Testing too many points and getting confused.
- Using wrong symbols for open/closed intervals.
- Not including endpoints when the inequality is non-strict.
📖 Exercise 2.5 – Solved Problems
(i)
\( x^2 + 3x - 4 > 0 \)
Step I: Associated equation
\[
\begin{aligned}
x^2 + 3x - 4 &= 0 \\
x^2 + 4x - x - 4 &= 0 \\
x(x + 4) - 1(x + 4) &= 0 \\
(x + 4)(x - 1) &= 0
\end{aligned}
\]
Now
either
\[
\begin{aligned}
x + 4 &= 0 \\
\mathbf{x} &= \mathbf{-4}
\end{aligned}
\]
or
\[
\begin{aligned}
x - 1 &= 0 \\
\mathbf{x} &= \mathbf{1}
\end{aligned}
\]
Critical Points: \((-4,0), (1,0)\)
Step II: Intervals: The intervals are
| I. | \(x < -4\) | OR | \((-\infty, -4)\) |
| II. | \(-4 < x < 1\) | OR | \((-4, 1)\) |
| III. | \(x > 1\) | OR | \((1, \infty)\) |
Step III: Test Points
For \(x = -5\):
\[
\begin{aligned}
(-5)^2 + 3(-5) - 4 &> 0 \\
25 - 15 - 4 &> 0 \\
6 &> 0 \quad \textbf{(True)}
\end{aligned}
\]
For \(x = 0\):
\[
\begin{aligned}
(0)^2 + 3(0) - 4 &> 0 \\
0 + 0 - 4 &> 0 \\
-4 &> 0 \quad \textbf{(False)}
\end{aligned}
\]
For \(x = 2\):
\[
\begin{aligned}
(2)^2 + 3(2) - 4 &> 0 \\
4 + 6 - 4 &> 0 \\
6 &> 0 \quad \textbf{(True)}
\end{aligned}
\]
\( S.S = (-\infty, -4) \cup (1, \infty) \)
(ii)
\( 2x^2 - 8x + 6 > 0 \)
Step I: Associated equation
\[
\begin{aligned}
2x^2 - 8x + 6 &= 0 \\
2(x^2 - 4x + 3) &= 0 \\
\Rightarrow x^2 - 4x + 3 &= 0 \\
x^2 - 3x - x + 3 &= 0 \\
x(x - 3) - 1(x - 3) &= 0 \\
(x - 3)(x - 1) &= 0
\end{aligned}
\]
Now
either
\[
\begin{aligned}
x - 3 &= 0 \\
\mathbf{x} &= \mathbf{3}
\end{aligned}
\]
or
\[
\begin{aligned}
x - 1 &= 0 \\
\mathbf{x} &= \mathbf{1}
\end{aligned}
\]
Critical Points: \((3,0), (1,0)\)
Step II: Intervals: The intervals are
| I. | \(x < 1\) | OR | \((-\infty, 1)\) |
| II. | \(1 < x < 3\) | OR | \((1, 3)\) |
| III. | \(x > 3\) | OR | \((3, \infty)\) |
Step III: Test Points
For \(x = 0\):
\[
\begin{aligned}
2(0)^2 - 8(0) + 6 &> 0 \\
0 - 0 + 6 &> 0 \\
6 &> 0 \quad \textbf{(True)}
\end{aligned}
\]
For \(x = 2\):
\[
\begin{aligned}
2(2)^2 - 8(2) + 6 &> 0 \\
8 - 16 + 6 &> 0 \\
-2 &> 0 \quad \textbf{(False)}
\end{aligned}
\]
For \(x = 4\):
\[
\begin{aligned}
2(4)^2 - 8(4) + 6 &> 0 \\
32 - 32 + 6 &> 0 \\
6 &> 0 \quad \textbf{(True)}
\end{aligned}
\]
\( S.S = (-\infty, 1) \cup (3, \infty) \)
(iii)
\( x^2 + x - 6 < 0 \)
Step I: Associated equation
\[
\begin{aligned}
x^2 + x - 6 &= 0 \\
x^2 + 3x - 2x - 6 &= 0 \\
x(x + 3) - 2(x + 3) &= 0 \\
(x + 3)(x - 2) &= 0
\end{aligned}
\]
Now
either
\[
\begin{aligned}
x + 3 &= 0 \\
\mathbf{x} &= \mathbf{-3}
\end{aligned}
\]
or
\[
\begin{aligned}
x - 2 &= 0 \\
\mathbf{x} &= \mathbf{2}
\end{aligned}
\]
Critical Points: \((-3,0), (2,0)\)
Step II: Intervals: The intervals are
| I. | \(x < -3\) | OR | \((-\infty, -3)\) |
| II. | \(-3 < x < 2\) | OR | \((-3, 2)\) |
| III. | \(x > 2\) | OR | \((2, \infty)\) |
Step III: Test Points
For \(x = -4\):
\[
\begin{aligned}
(-4)^2 + (-4) - 6 &< 0 \\
16 - 4 - 6 &< 0 \\
6 &< 0 \quad \textbf{(False)}
\end{aligned}
\]
For \(x = 0\):
\[
\begin{aligned}
(0)^2 + 0 - 6 &< 0 \\
0 + 0 - 6 &< 0 \\
-6 &< 0 \quad \textbf{(True)}
\end{aligned}
\]
For \(x = 3\):
\[
\begin{aligned}
(3)^2 + 3 - 6 &< 0 \\
9 + 3 - 6 &< 0 \\
6 &< 0 \quad \textbf{(False)}
\end{aligned}
\]
\( S.S = (-3, 2) \)
(iv)
\( x^2 - 6x + 9 < 0 \)
Step I: Associated equation
\[
\begin{aligned}
x^2 - 6x + 9 &= 0 \\
x^2 - 3x - 3x + 9 &= 0 \\
x(x - 3) - 3(x - 3) &= 0 \\
(x - 3)(x - 3) &= 0 \\
(x - 3)^2 &= 0 \\
\Rightarrow x - 3 &= 0 \\
x &= 3
\end{aligned}
\]
Critical Points: \((3,0)\)
Step II: Intervals: The intervals are
| I. | \(x < 3\) | OR | \((-\infty, 3)\) |
| II. | \(x > 3\) | OR | \((3, \infty)\) |
Step III: Test Points
For \(x = 2\):
\[
\begin{aligned}
(2)^2 - 6(2) + 9 &< 0 \\
4 - 12 + 9 &< 0 \\
1 &< 0 \quad \textbf{(False)}
\end{aligned}
\]
For \(x = 4\):
\[
\begin{aligned}
(4)^2 - 6(4) + 9 &< 0 \\
16 - 24 + 9 &< 0 \\
1 &< 0 \quad \textbf{(False)}
\end{aligned}
\]
\( S.S = \{\} \quad \textbf{OR} \quad \phi \)
(v)
\( 4x^2 - 16x + 15 \le 0 \)
Step I: Associated equation
\[
\begin{aligned}
4x^2 - 16x + 15 &= 0 \\
4x^2 - 10x - 6x + 15 &= 0 \\
2x(2x - 5) - 3(2x - 5) &= 0 \\
(2x - 5)(2x - 3) &= 0
\end{aligned}
\]
Now
either
\[
\begin{aligned}
2x - 5 &= 0 \\
2x &= 5 \\
x &= \frac{5}{2} \\
\mathbf{x} &= \mathbf{2\frac{1}{2}}
\end{aligned}
\]
or
\[
\begin{aligned}
2x - 3 &= 0 \\
2x &= 3 \\
x &= \frac{3}{2} \\
\mathbf{x} &= \mathbf{1\frac{1}{2}}
\end{aligned}
\]
Critical Points: \((2\frac{1}{2},0), (1\frac{1}{2},0)\)
Step II: Intervals: The intervals are
| I. | \(x \le 1\frac{1}{2}\) | OR | \((-\infty, 1\frac{1}{2}]\) |
| II. | \(1\frac{1}{2} \le x \le 2\frac{1}{2}\) | OR | \([1\frac{1}{2}, 2\frac{1}{2}]\) |
| III. | \(x \ge 2\frac{1}{2}\) | OR | \([2\frac{1}{2}, \infty)\) |
Step III: Test Points
For \(x = 0\):
\[
\begin{aligned}
4(0)^2 - 16(0) + 15 &\le 0 \\
0 - 0 + 15 &\le 0 \\
15 &\le 0 \quad \textbf{(False)}
\end{aligned}
\]
For \(x = 2\):
\[
\begin{aligned}
4(2)^2 - 16(2) + 15 &\le 0 \\
16 - 32 + 15 &\le 0 \\
-1 &\le 0 \quad \textbf{(True)}
\end{aligned}
\]
For \(x = 3\):
\[
\begin{aligned}
4(3)^2 - 16(3) + 15 &\le 0 \\
36 - 48 + 15 &\le 0 \\
3 &\le 0 \quad \textbf{(False)}
\end{aligned}
\]
\( S.S = [1\frac{1}{2}, 2\frac{1}{2}] \)
(vi)
\( -x^2 + 3x - 2 \ge 0 \)
Step I: Associated equation
\[
\begin{aligned}
-x^2 + 3x - 2 &= 0 \\
-1(x^2 - 3x + 2) &= 0 \\
\Rightarrow x^2 - 3x + 2 &= 0 \\
x^2 - 2x - x + 2 &= 0 \\
x(x - 2) - 1(x - 2) &= 0 \\
(x - 2)(x - 1) &= 0
\end{aligned}
\]
Now
either
\[
\begin{aligned}
x - 2 &= 0 \\
\mathbf{x} &= \mathbf{2}
\end{aligned}
\]
or
\[
\begin{aligned}
x - 1 &= 0 \\
\mathbf{x} &= \mathbf{1}
\end{aligned}
\]
Critical Points: \((2,0), (1,0)\)
Step II: Intervals: The intervals are
| I. | \(x \le 1\) | OR | \((-\infty, 1]\) |
| II. | \(1 \le x \le 2\) | OR | \([1, 2]\) |
| III. | \(x \ge 2\) | OR | \([2, \infty)\) |
Step III: Test Points
For \(x = 0\):
\[
\begin{aligned}
-(0)^2 + 3(0) - 2 &\ge 0 \\
0 - 0 - 2 &\ge 0 \\
-2 &\ge 0 \quad \textbf{(False)}
\end{aligned}
\]
For \(x = 1.5\):
\[
\begin{aligned}
-(1.5)^2 + 3(1.5) - 2 &\ge 0 \\
-2.25 + 4.5 - 2 &\ge 0 \\
0.25 &\ge 0 \quad \textbf{(True)}
\end{aligned}
\]
For \(x = 3\):
\[
\begin{aligned}
-(3)^2 + 3(3) - 2 &\ge 0 \\
-9 + 9 - 2 &\ge 0 \\
-2 &\ge 0 \quad \textbf{(False)}
\end{aligned}
\]
\( S.S = [1, 2] \)
📝 Multiple Choice Questions (Unit 2 Review)
1. The critical points of \(x^2 + 3x - 4 > 0\) are:
✅ Correct Answer: (A) \(-4, 1\)
Solving \(x^2 + 3x - 4 = 0\) gives \((x+4)(x-1)=0\), so \(x=-4\) and \(x=1\).
Solving \(x^2 + 3x - 4 = 0\) gives \((x+4)(x-1)=0\), so \(x=-4\) and \(x=1\).
2. The solution of \(x^2 + x - 6 < 0\) is:
✅ Correct Answer: (A) \((-3, 2)\)
Critical points are \(-3\) and \(2\). Testing shows the expression is negative between \(-3\) and \(2\).
Critical points are \(-3\) and \(2\). Testing shows the expression is negative between \(-3\) and \(2\).
3. The solution of \(x^2 - 6x + 9 < 0\) is:
✅ Correct Answer: (C) \(\emptyset\)
\(x^2 - 6x + 9 = (x-3)^2\) is always \(\ge 0\), never less than 0.
\(x^2 - 6x + 9 = (x-3)^2\) is always \(\ge 0\), never less than 0.
4. For \(4x^2 - 16x + 15 \le 0\), the critical points are:
✅ Correct Answer: (A) 1.5, 2.5
Solving \(4x^2 - 16x + 15 = 0\) gives \(x = \frac{3}{2}\) and \(x = \frac{5}{2}\).
Solving \(4x^2 - 16x + 15 = 0\) gives \(x = \frac{3}{2}\) and \(x = \frac{5}{2}\).
5. The solution of \(-x^2 + 3x - 2 \ge 0\) is:
✅ Correct Answer: (A) \([1, 2]\)
Multiplying by \(-1\) gives \(x^2 - 3x + 2 \le 0\), so \((x-1)(x-2) \le 0\), giving \([1, 2]\).
Multiplying by \(-1\) gives \(x^2 - 3x + 2 \le 0\), so \((x-1)(x-2) \le 0\), giving \([1, 2]\).
📈 Key Concepts – Quadratic Inequalities
- Sign Chart Method: Find critical points by solving the associated equation.
- Intervals: Divide the number line into intervals using critical points.
- Test Points: Choose a test value from each interval to determine the sign.
- Solution Set: Include or exclude endpoints based on the inequality sign (\(\le\), \(\ge\), \(<\), \(>\)).
- Empty Set: When no values satisfy the inequality (e.g., \(x^2 + 1 < 0\)).
❓ Frequently Asked Questions
What is taught in Exercise 2.5 of Class 10 Math Unit 2?
Exercise 2.5 covers solving quadratic inequalities using the sign chart method. Students learn to find critical points by solving the associated equation, divide the number line into intervals, test points, and write solution sets.
How many questions are there in Unit 2 Exercise 2.5?
Exercise 2.5 has 6 questions covering various quadratic inequalities including \( > \), \( < \), \( \le \), and \( \ge \) with different critical points and solution sets.
What is the sign chart method for quadratic inequalities?
The sign chart method involves: 1) Solving the associated equation to find critical points, 2) Dividing the number line into intervals using critical points, 3) Testing a point from each interval to determine the sign, and 4) Writing the solution set based on the inequality sign.
Is this solution according to the PECTAA 2026 syllabus?
Yes, these solutions are prepared according to the PECTAA 2026 / National Curriculum 2023 syllabus for Class 10 Mathematics.
Is this Exercise 2.5 solution valid for all Punjab Boards?
Yes, the content follows the unified Punjab textbook and is applicable to students of all 10 BISE Punjab boards.
Are solved PDF notes available for Exercise 2.5?
Yes, a complete solved PDF for Exercise 2.5 is embedded on this page and available to download for free.
Can I download the Unit 2 Exercise 2.5 solution as a PDF?
Yes, use the Download PDF button on this page to save the complete solved Exercise 2.5 notes to your device.
Is Exercise 2.5 important for Class 10 board exams?
Yes, quadratic inequalities are frequently tested in board exams. Exercise 2.5 builds essential skills in solving inequalities using the sign chart method.
Who prepared these Class 10 Math Unit 2 notes?
These notes were prepared by Muhammad Tayyab, Subject Specialist Mathematics at Govt Christian High School Daska, for Hira Science Academy.