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
1. Solve the following inequalities:
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 \(x = 2\):
\[
\begin{aligned}
2(2)^2 - 8(2) + 6 &> 0 \\
2(4) - 16 + 6 &> 0 \\
8 - 16 + 6 &> 0 \\
-2 &> 0 \quad \textbf{(False)}
\end{aligned}
\]
For \(x = 4\):
\[
\begin{aligned}
2(4)^2 - 8(4) + 6 &> 0 \\
2(16) - 32 + 6 &> 0 \\
32 - 32 + 6 &> 0 \\
6 &> 0 \quad \textbf{(True)}
\end{aligned}
\]
\( S.S = (-\infty,\,1) \cup (3,\,\infty) \)
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
\( 4x^2 - 16x + 15 \leq 0 \)
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
\[
\begin{aligned}
2x - 5 &= 0 \\
2x &= 5 \\
x &= \frac{5}{2} \\
\mathbf{x} &= \mathbf{2\tfrac{1}{2}}
\end{aligned}
\]
or
\[
\begin{aligned}
2x - 3 &= 0 \\
2x &= 3 \\
x &= \frac{3}{2} \\
\mathbf{x} &= \mathbf{1\tfrac{1}{2}}
\end{aligned}
\]
Critical Points: \(\left(2\tfrac{1}{2},\,0\right),\;\left(1\tfrac{1}{2},\,0\right)\)
Step II: Intervals: The intervals are
| I. | \(x \leq 1\tfrac{1}{2}\) | OR | \(\left(-\infty,\,1\tfrac{1}{2}\right]\) |
| II. | \(1\tfrac{1}{2} \leq x \leq 2\tfrac{1}{2}\) | OR | \(\left[1\tfrac{1}{2},\,2\tfrac{1}{2}\right]\) |
| III. | \(x \geq 2\tfrac{1}{2}\) | OR | \(\left[2\tfrac{1}{2},\,\infty\right)\) |
Step III: Test Points
For \(x = 0\):
\[
\begin{aligned}
4(0)^2 - 16(0) + 15 &\leq 0 \\
0 - 0 + 15 &\leq 0 \\
15 &\leq 0 \quad \textbf{(False)}
\end{aligned}
\]
For \(x = 2\):
\[
\begin{aligned}
4(2)^2 - 16(2) + 15 &\leq 0 \\
4(4) - 32 + 15 &\leq 0 \\
16 - 32 + 15 &\leq 0 \\
-1 &\leq 0 \quad \textbf{(True)}
\end{aligned}
\]
For \(x = 3\):
\[
\begin{aligned}
4(3)^2 - 16(3) + 15 &\leq 0 \\
4(9) - 48 + 15 &\leq 0 \\
36 - 48 + 15 &\leq 0 \\
3 &\leq 0 \quad \textbf{(False)}
\end{aligned}
\]
\( S.S = \left[1\tfrac{1}{2},\;2\tfrac{1}{2}\right] \)
vi
\( -x^2 + 3x - 2 \geq 0 \)
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. | \(x \leq 1\) | OR | \((-\infty,\,1]\) |
| II. | \(1 \leq x \leq 2\) | OR | \([1,\,2]\) |
| III. | \(x \geq 2\) | OR | \([2,\,\infty)\) |
Step III: Test Points
For \(x = 0\):
\[
\begin{aligned}
-(0)^2 + 3(0) - 2 &\geq 0 \\
0 - 0 - 2 &\geq 0 \\
-2 &\geq 0 \quad \textbf{(False)}
\end{aligned}
\]
For \(x = 1.5\):
\[
\begin{aligned}
-(1.5)^2 + 3(1.5) - 2 &\geq 0 \\
-2.25 + 4.5 - 2 &\geq 0 \\
0.25 &\geq 0 \quad \textbf{(True)}
\end{aligned}
\]
For \(x = 3\):
\[
\begin{aligned}
-(3)^2 + 3(3) - 2 &\geq 0 \\
-9 + 9 - 2 &\geq 0 \\
-2 &\geq 0 \quad \textbf{(False)}
\end{aligned}
\]
\( S.S = [1,\,2] \)
📈 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 (\(\le\), \(\ge\), \(<\), \(>\)).