Solved Exercises on Quadratic Inequalities. A guided walkthrough of the sign analysis method. Learn how to move from the associated equation to the final solution set in a few straightforward algebraic steps.
Exercise 1 — 27/03/2026 — level ★★☆☆☆
\[ x^2 - 4 > 0 \]
Answer
\[ x < -2 \quad \text{or} \quad x > 2 \]
Full Solution
Associated Equation and Roots
\[ x^2-4=(x-2)(x+2)=0 \implies x_1=-2,\quad x_2=2 \]
Sign Rule
The coefficient of \(x^2\) is positive: the parabola opens upward, so the polynomial is positive outside the roots.
\[ x^2-4 > 0 \iff x < -2 \;\text{ or }\; x > 2 \]
Solution Set
\[ S = (-\infty,\,-2)\cup(2,\,+\infty) \]
Answer
\[ \boxed{x < -2 \quad \text{or} \quad x > 2} \]
Exercise 2 — 27/03/2026 — level ★★☆☆☆
\[ x^2 - 9 \leq 0 \]
Answer
\[ -3 \leq x \leq 3 \]
Full Solution
Associated Equation and Roots
\[ x^2-9=(x-3)(x+3)=0 \implies x_1=-3,\quad x_2=3 \]
Sign Rule
Upward-opening parabola: the polynomial is non-positive between the roots.
\[ x^2-9 \leq 0 \iff -3 \leq x \leq 3 \]
Solution Set
\[ S = [-3,\,3] \]
Answer
\[ \boxed{-3 \leq x \leq 3} \]
Exercise 3 — 27/03/2026 — level ★★☆☆☆
\[ x^2 - 5x + 6 > 0 \]
Answer
\[ x < 2 \quad \text{or} \quad x > 3 \]
Full Solution
Associated Equation and Roots
Product \(6\), sum \(-5\): the roots are \(x_1=2\) and \(x_2=3\).
\[ x^2-5x+6=(x-2)(x-3)=0 \]
Sign Rule
Upward-opening parabola: positive outside the roots.
\[ x^2-5x+6 > 0 \iff x < 2 \;\text{ or }\; x > 3 \]
Solution Set
\[ S = (-\infty,\,2)\cup(3,\,+\infty) \]
Answer
\[ \boxed{x < 2 \quad \text{or} \quad x > 3} \]
Exercise 4 — 27/03/2026 — level ★★☆☆☆
\[ x^2 - 5x + 6 \leq 0 \]
Answer
\[ 2 \leq x \leq 3 \]
Full Solution
Associated Equation and Roots
\[ x^2-5x+6=(x-2)(x-3)=0 \implies x_1=2,\quad x_2=3 \]
Sign Rule
Upward-opening parabola: the polynomial is non-positive between the roots. Compared to the previous exercise, only the direction of the inequality changes.
\[ x^2-5x+6 \leq 0 \iff 2 \leq x \leq 3 \]
Solution Set
\[ S = [2,\,3] \]
Answer
\[ \boxed{2 \leq x \leq 3} \]
Exercise 5 — 27/03/2026 — level ★★★☆☆
\[ x^2 - 7x + 12 > 0 \]
Answer
\[ x < 3 \quad \text{or} \quad x > 4 \]
Full Solution
Associated Equation and Roots
Product \(12\), sum \(-7\): the roots are \(x_1=3\) and \(x_2=4\).
\[ x^2-7x+12=(x-3)(x-4)=0 \]
Sign Rule
\[ x^2-7x+12 > 0 \iff x < 3 \;\text{ or }\; x > 4 \]
Solution Set
\[ S = (-\infty,\,3)\cup(4,\,+\infty) \]
Answer
\[ \boxed{x < 3 \quad \text{or} \quad x > 4} \]
Exercise 6 — 27/03/2026 — level ★★★☆☆
\[ x^2 + x - 6 \leq 0 \]
Answer
\[ -3 \leq x \leq 2 \]
Full Solution
Associated Equation and Roots
Product \(-6\), sum \(1\): the roots are \(x_1=-3\) and \(x_2=2\).
\[ x^2+x-6=(x+3)(x-2)=0 \]
Sign Rule
Upward-opening parabola: non-positive between the roots.
\[ x^2+x-6 \leq 0 \iff -3 \leq x \leq 2 \]
Solution Set
\[ S = [-3,\,2] \]
Answer
\[ \boxed{-3 \leq x \leq 2} \]
Exercise 7 — 27/03/2026 — level ★★★☆☆
\[ x^2 - 2x + 1 \geq 0 \]
Answer
\[ x \in \mathbb{R} \quad \text{(always true)} \]
Full Solution
Recognizing the Perfect Square
\[ x^2-2x+1=(x-1)^2 \]
Analysis
The square of any real number is always non-negative: \((x-1)^2 \geq 0\) for all \(x \in \mathbb{R}\). The inequality holds for every real number.
Solution Set
\[ S = \mathbb{R} \]
Answer
\[ \boxed{x \in \mathbb{R}} \]
Exercise 8 — 27/03/2026 — level ★★★☆☆
\[ x^2 - 2x + 1 < 0 \]
Answer
No solution
Full Solution
Recognizing the Perfect Square
\[ x^2-2x+1=(x-1)^2 \]
Analysis
The square of any real number is always \(\geq 0\): it can never be strictly negative. The inequality has no solution.
Solution Set
\[ S = \emptyset \]
Answer
\[ \boxed{\text{No solution}} \]
Exercise 9 — 27/03/2026 — level ★★★☆☆
\[ x^2 - 2x + 5 > 0 \]
Answer
\[ x \in \mathbb{R} \quad \text{(always true)} \]
Full Solution
Computing the Discriminant
\[ \Delta = 4 - 20 = -16 \]
Analysis
Since \(\Delta < 0\), the polynomial has no real roots. With a positive leading coefficient, the parabola lies entirely above the \(x\)-axis: the polynomial is always positive.
Solution Set
\[ S = \mathbb{R} \]
Answer
\[ \boxed{x \in \mathbb{R}} \]
Exercise 10 — 27/03/2026 — level ★★★☆☆
\[ x^2 - 4x + 5 \leq 0 \]
Answer
No solution
Full Solution
Computing the Discriminant
\[ \Delta = 16 - 20 = -4 \]
Analysis
Since \(\Delta < 0\) and the leading coefficient is positive, the parabola is always above the \(x\)-axis: the polynomial is never \(\leq 0\).
Solution Set
\[ S = \emptyset \]
Answer
\[ \boxed{\text{No solution}} \]
Exercise 11 — 27/03/2026 — level ★★★★☆
\[ x^2 > 2x + 3 \]
Answer
\[ x < -1 \quad \text{or} \quad x > 3 \]
Full Solution
Rewriting in Standard Form
\[ x^2-2x-3 > 0 \]
Associated Equation and Roots
\[ x^2-2x-3=(x-3)(x+1)=0 \implies x_1=-1,\quad x_2=3 \]
Sign Rule
Upward-opening parabola: positive outside the roots.
\[ x < -1 \;\text{ or }\; x > 3 \]
Solution Set
\[ S = (-\infty,\,-1)\cup(3,\,+\infty) \]
Answer
\[ \boxed{x < -1 \quad \text{or} \quad x > 3} \]
Exercise 12 — 27/03/2026 — level ★★★★☆
\[ 3x^2 - x - 2 \leq 0 \]
Answer
\[ -\dfrac{2}{3} \leq x \leq 1 \]
Full Solution
Associated Equation and Roots
\[ \Delta = 1+24=25 \implies x = \frac{1\pm5}{6} \implies x_1=-\frac{2}{3},\quad x_2=1 \]
Factoring
\[ 3x^2-x-2=(3x+2)(x-1) \]
Check: \((3x+2)(x-1)=3x^2-3x+2x-2=3x^2-x-2\) ✓
Sign Rule
Positive leading coefficient: non-positive between the roots.
\[ -\frac{2}{3} \leq x \leq 1 \]
Solution Set
\[ S = \left[-\frac{2}{3},\,1\right] \]
Answer
\[ \boxed{-\dfrac{2}{3} \leq x \leq 1} \]
Exercise 13 — 27/03/2026 — level ★★★★☆
\[ -x^2 + 4x - 3 \geq 0 \]
Answer
\[ 1 \leq x \leq 3 \]
Full Solution
Changing the Sign
Multiplying through by \(-1\) makes the leading coefficient positive and reverses the inequality.
\[ x^2 - 4x + 3 \leq 0 \]
Associated Equation and Roots
\[ x^2-4x+3=(x-1)(x-3)=0 \implies x_1=1,\quad x_2=3 \]
Sign Rule
Non-positive between the roots: \(1 \leq x \leq 3\).
Solution Set
\[ S = [1,\,3] \]
Answer
\[ \boxed{1 \leq x \leq 3} \]
Exercise 14 — 27/03/2026 — level ★★★★☆
\[ 2x^2 + 5x - 3 < 0 \]
Answer
\[ -3 < x < \dfrac{1}{2} \]
Full Solution
Associated Equation and Roots
\[ \Delta = 25+24=49 \implies x = \frac{-5\pm7}{4} \implies x_1=-3,\quad x_2=\frac{1}{2} \]
Factoring
\[ 2x^2+5x-3=(2x-1)(x+3) \]
Sign Rule
Positive leading coefficient: strictly negative between the roots.
\[ -3 < x < \frac{1}{2} \]
Solution Set
\[ S = \left(-3,\,\frac{1}{2}\right) \]
Answer
\[ \boxed{-3 < x < \dfrac{1}{2}} \]
Exercise 15 — 27/03/2026 — level ★★★★☆
\[ x^2 - 6x + 9 > 0 \]
Answer
\[ x \in \mathbb{R}\setminus\{3\} \]
Full Solution
Recognizing the Perfect Square
\[ x^2-6x+9=(x-3)^2 \]
Analysis
\(\Delta=0\): double root at \(x=3\). The parabola is always \(\geq 0\) and touches the \(x\)-axis only at \(x=3\). For the strict inequality, that single point must be excluded.
\[ (x-3)^2 > 0 \iff x \neq 3 \]
Solution Set
\[ S = \mathbb{R}\setminus\{3\} = (-\infty,\,3)\cup(3,\,+\infty) \]
Answer
\[ \boxed{x \in \mathbb{R}\setminus\{3\}} \]
Exercise 16 — 27/03/2026 — level ★★★★☆
\[ x(x-4) \geq 5 \]
Answer
\[ x \leq -1 \quad \text{or} \quad x \geq 5 \]
Full Solution
Rewriting in Standard Form
\[ x^2-4x-5 \geq 0 \]
Associated Equation and Roots
\[ \Delta = 16+20=36 \implies x = \frac{4\pm6}{2} \implies x_1=-1,\quad x_2=5 \]
Factoring
\[ x^2-4x-5=(x+1)(x-5) \]
Sign Rule
Non-negative outside the roots: \(x \leq -1\) or \(x \geq 5\).
Check
\(x=5\): \(5\cdot1=5\geq5\) \(x=-1\): \((-1)(-5)=5\geq5\)
Answer
\[ \boxed{x \leq -1 \quad \text{or} \quad x \geq 5} \]
Exercise 17 — 27/03/2026 — level ★★★★★
\[ \begin{cases} x^2-5x+4 < 0 \\ x^2-4 > 0 \end{cases} \]
Answer
\[ 2 < x < 4 \]
Full Solution
First Inequality
\[ x^2-5x+4=(x-1)(x-4) < 0 \implies 1 < x < 4 \]
Second Inequality
\[ x^2-4=(x-2)(x+2) > 0 \implies x < -2 \;\text{ or }\; x > 2 \]
Intersection
Intersecting \((1,\,4)\) with \((-\infty,-2)\cup(2,+\infty)\):
\[ (1 < x < 4)\;\cap\;(x > 2) \;=\; 2 < x < 4 \]
Solution Set
\[ S = (2,\,4) \]
Answer
\[ \boxed{2 < x < 4} \]
Exercise 18 — 27/03/2026 — level ★★★★★
\[ (x^2-4x+3)(x^2-x-2) \leq 0 \]
Answer
\[ -1 \leq x \leq 1 \quad \text{or} \quad 2 \leq x \leq 3 \]
Full Solution
Factoring
\[ x^2-4x+3=(x-1)(x-3) \qquad x^2-x-2=(x-2)(x+1) \]
The roots of the product are \(x=-1,\,1,\,2,\,3\).
Sign Chart for \((x-1)(x-3)(x-2)(x+1)\)
\(x < -1\): four negative factors \(\to\) product \(> 0\)
\(-1 < x < 1\): three negative factors \(\to\) product \(< 0\)
\(1 < x < 2\): two negative factors \(\to\) product \(> 0\)
\(2 < x < 3\): one negative factor \(\to\) product \(< 0\)
\(x > 3\): no negative factors \(\to\) product \(> 0\)
Solution for \(\leq 0\)
The product is non-positive on the intervals with sign \(-\) and at the roots themselves.
Solution Set
\[ S = [-1,\,1]\cup[2,\,3] \]
Answer
\[ \boxed{-1 \leq x \leq 1 \quad \text{or} \quad 2 \leq x \leq 3} \]
Exercise 19 — 27/03/2026 — level ★★★★★
\[ \begin{cases} 2x^2-5x+2 > 0 \\ x^2-x-2 < 0 \end{cases} \]
Answer
\[ -1 < x < \dfrac{1}{2} \]
Full Solution
First Inequality
\[ \Delta=9 \implies x_1=\tfrac{1}{2},\; x_2=2 \qquad (2x-1)(x-2) > 0 \implies x < \frac{1}{2} \;\text{ or }\; x > 2 \]
Second Inequality
\[ (x-2)(x+1) < 0 \implies -1 < x < 2 \]
Intersection
\[ \left(x < \tfrac{1}{2} \;\text{ or }\; x > 2\right)\cap\left(-1 < x < 2\right) = -1 < x < \frac{1}{2} \]
Solution Set
\[ S = \left(-1,\,\tfrac{1}{2}\right) \]
Answer
\[ \boxed{-1 < x < \dfrac{1}{2}} \]
Exercise 20 — 27/03/2026 — level ★★★★★
\[ x(x-2) > x-2 \]
Answer
\[ x < 1 \quad \text{or} \quad x > 2 \]
Full Solution
Rewriting in Standard Form
\[ x(x-2)-(x-2) > 0 \]
Factoring Out \((x-2)\)
\[ (x-2)(x-1) > 0 \]
Roots and Sign Rule
Roots: \(x=1\) and \(x=2\). Upward-opening parabola: positive outside the roots.
\[ x < 1 \;\text{ or }\; x > 2 \]
Check
\(x=0\): \(0 > -2\) \(x=3\): \(3 > 1\) \(x=1.5\): \(-0.75 > -0.5\) — false, not a solution
Solution Set
\[ S = (-\infty,\,1)\cup(2,\,+\infty) \]
Answer
\[ \boxed{x < 1 \quad \text{or} \quad x > 2} \]