This collection presents 20 practice problems on parametric inequalities, arranged in order of increasing difficulty and solved step by step.
The goal is to learn to identify the values of the parameter that alter the structure of the inequality: the sign of a coefficient, the direction of the inequality, the degree of the expression, the discriminant, and the concavity of the parabola.
Each exercise shows how to correctly set up the case analysis, avoiding the most common pitfalls and arriving at a complete description of the solution set as the parameter varies.
Exercise 1 — level ★☆☆☆☆
Solve:
\[ (k-3)x > 6 \]
Answer
\[ \begin{cases} x > \dfrac{6}{k-3}, & k > 3, \\[6pt] x < \dfrac{6}{k-3}, & k < 3, \\[6pt] S = \emptyset, & k = 3. \end{cases} \]
Solution
The inequality is linear in the unknown \(x\). The coefficient of \(x\) is:
\[ k - 3. \]
To isolate \(x\), we need to divide both sides by \(k-3\). However, the sign of \(k-3\) depends on the parameter \(k\). We must therefore consider three cases:
\[ k-3 > 0, \qquad k-3 < 0, \qquad k-3 = 0. \]
Case \(k > 3\)
If \(k > 3\), then \(k-3 > 0\), so we can divide by \(k-3\) without reversing the inequality:
\[ x > \frac{6}{k-3}. \]
Case \(k < 3\)
If \(k < 3\), then \(k-3 < 0\), so dividing by a negative quantity reverses the direction of the inequality:
\[ x < \frac{6}{k-3}. \]
Case \(k = 3\)
If \(k = 3\), the coefficient of \(x\) vanishes. The inequality becomes:
\[ 0 \cdot x > 6, \qquad \text{that is} \qquad 0 > 6. \]
This statement is false, so there is no real value of \(x\) satisfying the inequality. Therefore:
\[ S = \emptyset. \]
Exercise 2 — level ★☆☆☆☆
Solve:
\[ (k+1)x - 4 \leq 0 \]
Answer
\[ \begin{cases} x \leq \dfrac{4}{k+1}, & k > -1, \\[6pt] x \geq \dfrac{4}{k+1}, & k < -1, \\[6pt] S = \mathbb{R}, & k = -1. \end{cases} \]
Solution
Moving the constant term to the right-hand side:
\[ (k+1)x \leq 4. \]
Again, the coefficient of the unknown depends on the parameter: \(k+1\). To divide correctly, we need to determine whether \(k+1\) is positive, negative, or zero.
Case \(k > -1\)
If \(k > -1\), then \(k+1 > 0\). Dividing by a positive quantity does not reverse the inequality:
\[ x \leq \frac{4}{k+1}. \]
Case \(k < -1\)
If \(k < -1\), then \(k+1 < 0\). Dividing by a negative quantity reverses the inequality:
\[ x \geq \frac{4}{k+1}. \]
Case \(k = -1\)
If \(k = -1\), the coefficient of \(x\) vanishes. The original inequality becomes:
\[ 0 \cdot x - 4 \leq 0, \qquad \text{that is} \qquad -4 \leq 0. \]
This statement is true for every real value of \(x\). Therefore:
\[ S = \mathbb{R}. \]
Exercise 3 — level ★★☆☆☆
Solve:
\[ (k-2)x + 5 \geq 0 \]
Answer
\[ \begin{cases} x \geq -\dfrac{5}{k-2}, & k > 2, \\[6pt] x \leq -\dfrac{5}{k-2}, & k < 2, \\[6pt] S = \mathbb{R}, & k = 2. \end{cases} \]
Solution
Isolating the term containing the unknown:
\[ (k-2)x \geq -5. \]
The coefficient of \(x\) is \(k-2\). Since it depends on the parameter, we cannot divide without first determining its sign.
Case \(k > 2\)
If \(k > 2\), then \(k-2 > 0\). Dividing by \(k-2\) preserves the direction of the inequality:
\[ x \geq -\frac{5}{k-2}. \]
Case \(k < 2\)
If \(k < 2\), then \(k-2 < 0\). Dividing by a negative quantity reverses the inequality:
\[ x \leq -\frac{5}{k-2}. \]
Case \(k = 2\)
If \(k = 2\), the term containing \(x\) vanishes:
\[ 0 \cdot x + 5 \geq 0, \qquad \text{that is} \qquad 5 \geq 0. \]
This statement is always true. Consequently, every real number is a solution:
\[ S = \mathbb{R}. \]
Exercise 4 — level ★★☆☆☆
Solve:
\[ (k^2 - 1)x < 2 \]
Answer
\[ \begin{cases} x < \dfrac{2}{k^2-1}, & k < -1 \ \text{or}\ k > 1, \\[6pt] x > \dfrac{2}{k^2-1}, & -1 < k < 1, \\[6pt] S = \mathbb{R}, & k = \pm 1. \end{cases} \]
Solution
The inequality is linear in \(x\), but the coefficient of \(x\) is:
\[ k^2 - 1 = (k-1)(k+1). \]
We analyze the sign of \(k^2-1\):
\[ k^2 - 1 > 0 \quad \Longleftrightarrow \quad k < -1 \ \text{or}\ k > 1, \]
\[ k^2 - 1 < 0 \quad \Longleftrightarrow \quad -1 < k < 1, \]
\[ k^2 - 1 = 0 \quad \Longleftrightarrow \quad k = \pm 1. \]
Case \(k < -1\) or \(k > 1\)
Here \(k^2-1 > 0\). We can divide without reversing the inequality:
\[ x < \frac{2}{k^2-1}. \]
Case \(-1 < k < 1\)
Here \(k^2-1 < 0\). Dividing by a negative quantity reverses the inequality:
\[ x > \frac{2}{k^2-1}. \]
Case \(k = \pm 1\)
If \(k = \pm 1\), then \(k^2 - 1 = 0\). The inequality becomes:
\[ 0 \cdot x < 2, \qquad \text{that is} \qquad 0 < 2. \]
This statement is true for every \(x \in \mathbb{R}\). Therefore:
\[ S = \mathbb{R}. \]
Exercise 5 — level ★★☆☆☆
Solve:
\[ (k-1)x^2 - 4 > 0 \]
Answer
\[ \begin{cases} x < -\dfrac{2}{\sqrt{k-1}} \ \text{or}\ x > \dfrac{2}{\sqrt{k-1}}, & k > 1, \\[10pt] S = \emptyset, & k \leq 1. \end{cases} \]
Solution
The inequality contains the term \(x^2\), whose coefficient is \(k-1\). We must distinguish the cases in which this coefficient is positive, zero, or negative.
Case \(k > 1\)
If \(k > 1\), then \(k-1 > 0\). We can divide without reversing the inequality:
\[ x^2 > \frac{4}{k-1}. \]
The right-hand side is positive, so:
\[ |x| > \frac{2}{\sqrt{k-1}}, \]
that is:
\[ x < -\frac{2}{\sqrt{k-1}} \quad \text{or} \quad x > \frac{2}{\sqrt{k-1}}. \]
Case \(k = 1\)
If \(k = 1\), the quadratic term vanishes. The inequality becomes \(-4 > 0\), which is false. Therefore:
\[ S = \emptyset. \]
Case \(k < 1\)
If \(k < 1\), then \(k-1 < 0\). Since \(x^2 \geq 0\) for all \(x \in \mathbb{R}\), we have \((k-1)x^2 \leq 0\), and therefore:
\[ (k-1)x^2 - 4 < 0 \]
for every \(x \in \mathbb{R}\). The inequality has no solutions:
\[ S = \emptyset. \]
Exercise 6 — level ★★☆☆☆
Solve:
\[ (k+2)x^2 + 1 \geq 0 \]
Answer
\[ \begin{cases} S = \mathbb{R}, & k \geq -2, \\[8pt] -\dfrac{1}{\sqrt{-k-2}} \leq x \leq \dfrac{1}{\sqrt{-k-2}}, & k < -2. \end{cases} \]
Solution
The coefficient of the quadratic term is \(k+2\). We must consider the cases \(k+2 > 0\), \(k+2 = 0\), and \(k+2 < 0\).
Case \(k > -2\)
If \(k > -2\), then \(k+2 > 0\). Since \(x^2 \geq 0\), we have \((k+2)x^2 \geq 0\), so:
\[ (k+2)x^2 + 1 \geq 1 > 0. \]
The inequality is satisfied for every \(x \in \mathbb{R}\):
\[ S = \mathbb{R}. \]
Case \(k = -2\)
If \(k = -2\), the quadratic term vanishes and the inequality becomes \(1 \geq 0\), which is always true:
\[ S = \mathbb{R}. \]
Case \(k < -2\)
If \(k < -2\), then \(k+2 < 0\). Starting from the inequality:
\[ (k+2)x^2 \geq -1. \]
Since \(k+2 < 0\), dividing by \(k+2\) reverses the inequality:
\[ x^2 \leq \frac{-1}{k+2} = \frac{1}{-k-2}. \]
Therefore:
\[ -\frac{1}{\sqrt{-k-2}} \leq x \leq \frac{1}{\sqrt{-k-2}}. \]
Exercise 7 — level ★★★☆☆
Solve:
\[ x^2 - 2kx + k^2 - 1 > 0 \]
Answer
\[ x < k-1 \quad \text{or} \quad x > k+1. \]
Solution
We observe that the trinomial can be rewritten as a perfect square minus \(1\):
\[ x^2 - 2kx + k^2 - 1 = (x-k)^2 - 1. \]
The inequality becomes:
\[ (x-k)^2 > 1, \]
that is, \(|x-k| > 1\). Therefore:
\[ x - k < -1 \quad \text{or} \quad x - k > 1, \]
which gives:
\[ x < k-1 \quad \text{or} \quad x > k+1. \]
The solution set is:
\[ S = (-\infty,\, k-1) \cup (k+1,\, +\infty). \]
Exercise 8 — level ★★★☆☆
Solve:
\[ x^2 + (k-3)x + k < 0 \]
Answer
\[ \begin{cases} x_1 < x < x_2, & k < 1 \ \text{or}\ k > 9, \\[6pt] S = \emptyset, & 1 \leq k \leq 9, \end{cases} \]
where, for \(k < 1\) or \(k > 9\),
\[ x_1 = \frac{3-k-\sqrt{k^2-10k+9}}{2}, \qquad x_2 = \frac{3-k+\sqrt{k^2-10k+9}}{2}. \]
Solution
Consider the trinomial \(P(x) = x^2 + (k-3)x + k\). The leading coefficient is \(a = 1 > 0\), so the parabola always opens upward.
Since the inequality requires \(P(x) < 0\), the trinomial must have two distinct real roots (for an upward-opening parabola, the polynomial is negative only between its two roots).
We compute the discriminant:
\[ \Delta = (k-3)^2 - 4k = k^2 - 6k + 9 - 4k = k^2 - 10k + 9 = (k-1)(k-9). \]
We analyze the sign:
\[ \Delta > 0 \quad \Longleftrightarrow \quad k < 1 \ \text{or}\ k > 9, \]
\[ \Delta = 0 \quad \Longleftrightarrow \quad k = 1 \ \text{or}\ k = 9, \]
\[ \Delta < 0 \quad \Longleftrightarrow \quad 1 < k < 9. \]
Case \(k < 1\) or \(k > 9\)
The trinomial has two distinct real roots and the parabola opens upward, so it is negative between the two roots:
\[ x_1 < x < x_2. \]
Case \(k = 1\) or \(k = 9\)
The trinomial has a repeated root. Since the parabola opens upward, it is always \(\geq 0\), but the inequality requires \(P(x) < 0\). Therefore:
\[ S = \emptyset. \]
Case \(1 < k < 9\)
The discriminant is negative. The upward-opening parabola does not intersect the \(x\)-axis and the trinomial is always positive. Therefore:
\[ S = \emptyset. \]
Exercise 9 — level ★★★☆☆
Solve:
\[ (k-4)x^2 + 2x - 3 \leq 0 \]
Answer
\[ \begin{cases} S = \mathbb{R}, & k \leq \dfrac{11}{3}, \\[8pt] x \leq x_1 \ \text{or}\ x \geq x_2, & \dfrac{11}{3} < k < 4, \\[8pt] x \leq \dfrac{3}{2}, & k = 4, \\[8pt] x_1 \leq x \leq x_2, & k > 4, \end{cases} \]
where \(x_1\) and \(x_2\) denote the two ordered roots of the polynomial, with \(x_1 < x_2\).
Solution
The coefficient of the quadratic term is \(k-4\). Before analyzing the discriminant, we must check whether the inequality is actually quadratic.
Case \(k = 4\)
If \(k = 4\), the quadratic term vanishes and the inequality reduces to \(2x - 3 \leq 0\), giving:
\[ x \leq \frac{3}{2}. \]
Case \(k \neq 4\)
For \(k \neq 4\), consider \(P(x) = (k-4)x^2 + 2x - 3\). The discriminant is:
\[ \Delta = 4 - 4(k-4)(-3) = 4 + 12(k-4) = 12k - 44 = 4(3k-11). \]
We analyze the sign:
\[ \Delta > 0 \iff k > \tfrac{11}{3}, \qquad \Delta = 0 \iff k = \tfrac{11}{3}, \qquad \Delta < 0 \iff k < \tfrac{11}{3}. \]
The concavity depends on the sign of \(k-4\).
Case \(k < \dfrac{11}{3}\)
Here \(\Delta < 0\) and \(k - 4 < 0\). The parabola opens downward and does not intersect the \(x\)-axis: the polynomial is always negative. Since the inequality requires \(P(x) \leq 0\):
\[ S = \mathbb{R}. \]
Case \(k = \dfrac{11}{3}\)
Here \(\Delta = 0\) and \(k - 4 = \frac{11}{3} - 4 = -\frac{1}{3} < 0\). The parabola opens downward and is tangent to the \(x\)-axis at the repeated root: the polynomial is always \(\leq 0\). Therefore:
\[ S = \mathbb{R}. \]
Case \(\dfrac{11}{3} < k < 4\)
Here \(\Delta > 0\) and \(k - 4 < 0\). The polynomial has two distinct real roots and the parabola opens downward: it is positive between the roots and negative outside. For \(P(x) \leq 0\):
\[ x \leq x_1 \quad \text{or} \quad x \geq x_2. \]
Case \(k > 4\)
Here \(\Delta > 0\) and \(k - 4 > 0\). The polynomial has two distinct real roots and the parabola opens upward: it is negative between the roots. Since the inequality is non-strict, the roots are included:
\[ x_1 \leq x \leq x_2. \]
Exercise 10 — level ★★★★☆
Solve:
\[ (k-1)x^2 + (k+1)x + k > 0 \]
Answer
Let:
\[ k_1 = 1 - \frac{2\sqrt{3}}{3}, \qquad k_2 = 1 + \frac{2\sqrt{3}}{3}. \]
\[ \begin{cases} S = \emptyset, & k \leq k_1, \\[6pt] x_1 < x < x_2, & k_1 < k < 1, \\[6pt] x > -\dfrac{1}{2}, & k = 1, \\[8pt] x < x_1 \ \text{or}\ x > x_2, & 1 < k < k_2, \\[6pt] S = \mathbb{R} \setminus \{x_0\}, & k = k_2, \\[6pt] S = \mathbb{R}, & k > k_2. \end{cases} \]
Solution
The coefficient of the quadratic term is \(k-1\). The first value to check is \(k = 1\).
Case \(k = 1\)
If \(k = 1\), the inequality becomes \(2x + 1 > 0\), giving:
\[ x > -\frac{1}{2}. \]
Case \(k \neq 1\)
For \(k \neq 1\), we compute the discriminant:
\[ \Delta = (k+1)^2 - 4(k-1)k = k^2 + 2k + 1 - 4k^2 + 4k = -3k^2 + 6k + 1. \]
Setting \(\Delta = 0\): \(3k^2 - 6k - 1 = 0\), with solutions:
\[ k = \frac{6 \pm \sqrt{36+12}}{6} = \frac{6 \pm 4\sqrt{3}}{6} = 1 \pm \frac{2\sqrt{3}}{3}. \]
Since the discriminant (as a function of \(k\)) is a downward-opening parabola:
\[ \Delta > 0 \iff k_1 < k < k_2, \quad \Delta = 0 \iff k = k_1 \ \text{or}\ k = k_2, \quad \Delta < 0 \iff k < k_1 \ \text{or}\ k > k_2. \]
Case \(k < k_1\)
\(\Delta < 0\) and \(k - 1 < 0\): downward-opening parabola, polynomial always negative. Hence \(S = \emptyset\).
Case \(k = k_1\)
\(\Delta = 0\) and \(k - 1 < 0\): downward-opening parabola, polynomial always \(\leq 0\). The inequality is strict, so \(S = \emptyset\).
Case \(k_1 < k < 1\)
\(\Delta > 0\) and \(k - 1 < 0\): downward-opening parabola, polynomial positive between the roots. Therefore:
\[ x_1 < x < x_2. \]
Case \(1 < k < k_2\)
\(\Delta > 0\) and \(k - 1 > 0\): upward-opening parabola, polynomial positive outside the roots. Therefore:
\[ x < x_1 \quad \text{or} \quad x > x_2. \]
Case \(k = k_2\)
\(\Delta = 0\) and \(k - 1 > 0\): upward-opening parabola, polynomial always \(\geq 0\), but vanishes at the repeated root \(x_0 = -\dfrac{k+1}{2(k-1)}\). Since the inequality is strict:
\[ S = \mathbb{R} \setminus \{x_0\}. \]
Case \(k > k_2\)
\(\Delta < 0\) and \(k - 1 > 0\): upward-opening parabola, polynomial always positive. Therefore \(S = \mathbb{R}\).
Exercise 11 — level ★★★★☆
Solve:
\[ (k^2 - 4)x^2 - 1 < 0 \]
Answer
\[ \begin{cases} -\dfrac{1}{\sqrt{k^2-4}} < x < \dfrac{1}{\sqrt{k^2-4}}, & |k| > 2, \\[10pt] S = \mathbb{R}, & |k| \leq 2. \end{cases} \]
Solution
The coefficient of \(x^2\) is \(k^2 - 4 = (k-2)(k+2)\). We analyze its sign:
\[ k^2 - 4 > 0 \iff |k| > 2, \quad k^2 - 4 = 0 \iff k = \pm 2, \quad k^2 - 4 < 0 \iff |k| < 2. \]
Case \(|k| > 2\)
Since \(k^2 - 4 > 0\), we divide by it without reversing the inequality:
\[ x^2 < \frac{1}{k^2 - 4}. \]
Therefore:
\[ -\frac{1}{\sqrt{k^2-4}} < x < \frac{1}{\sqrt{k^2-4}}. \]
Case \(|k| < 2\)
Since \(k^2 - 4 < 0\) and \(x^2 \geq 0\), we have \((k^2-4)x^2 \leq 0\), so:
\[ (k^2-4)x^2 - 1 < 0 \]
for every \(x \in \mathbb{R}\). Hence \(S = \mathbb{R}\).
Case \(k = \pm 2\)
If \(k = \pm 2\), the inequality becomes \(-1 < 0\), which is always true. Therefore \(S = \mathbb{R}\).
Exercise 12 — level ★★★★☆
Solve:
\[ (k-2)x^2 + (k+1)x + 1 \geq 0 \]
Answer
\[ \begin{cases} x_1 \leq x \leq x_2, & k < 2, \\[6pt] x \geq -\dfrac{1}{3}, & k = 2, \\[8pt] x \leq x_1 \ \text{or}\ x \geq x_2, & k > 2, \end{cases} \]
where \(x_1\) and \(x_2\) are the ordered roots of the polynomial, with \(x_1 < x_2\).
Solution
The coefficient of the quadratic term is \(k-2\).
Case \(k = 2\)
If \(k = 2\), the inequality becomes \(3x + 1 \geq 0\), giving:
\[ x \geq -\frac{1}{3}. \]
Case \(k \neq 2\)
We compute the discriminant:
\[ \Delta = (k+1)^2 - 4(k-2) = k^2 + 2k + 1 - 4k + 8 = k^2 - 2k + 9 = (k-1)^2 + 8. \]
Since \((k-1)^2 + 8 > 0\) for all \(k \in \mathbb{R}\), the polynomial always has two distinct real roots when \(k \neq 2\). It only remains to determine the concavity, i.e., the sign of \(k-2\).
Case \(k < 2\)
Downward-opening parabola, polynomial positive between the roots. For \(P(x) \geq 0\):
\[ x_1 \leq x \leq x_2. \]
Case \(k > 2\)
Upward-opening parabola, polynomial positive outside the roots. For \(P(x) \geq 0\):
\[ x \leq x_1 \quad \text{or} \quad x \geq x_2. \]
Exercise 13 — level ★★★★☆
Solve:
\[ (k+1)x^2 - 2(k-1)x + k \leq 0 \]
Answer
\[ \begin{cases} x \leq \dfrac{1}{4}, & k = -1, \\[8pt] x_1 \leq x \leq x_2, & -1 < k < \dfrac{1}{3}, \\[8pt] x = x_0, & k = \dfrac{1}{3}, \\[8pt] S = \emptyset, & k > \dfrac{1}{3}, \\[8pt] x \leq x_1 \ \text{or}\ x \geq x_2, & k < -1. \end{cases} \]
When two distinct real roots exist, \(x_1\) and \(x_2\) denote the ordered roots, with \(x_1 < x_2\). When \(k = \dfrac{1}{3}\), \(x_0\) is the repeated root.
Solution
The coefficient of the quadratic term is \(k+1\). The first value to consider is \(k = -1\).
Case \(k = -1\)
Substituting \(k = -1\), the inequality becomes \(4x - 1 \leq 0\), giving:
\[ x \leq \frac{1}{4}. \]
Case \(k \neq -1\)
We compute the discriminant:
\[ \Delta = 4(k-1)^2 - 4k(k+1) = 4\bigl[(k-1)^2 - k(k+1)\bigr] = 4(1-3k). \]
We analyze the sign:
\[ \Delta > 0 \iff k < \tfrac{1}{3}, \quad \Delta = 0 \iff k = \tfrac{1}{3}, \quad \Delta < 0 \iff k > \tfrac{1}{3}. \]
The concavity depends on the sign of \(k+1\).
Case \(k < -1\)
\(\Delta > 0\) and \(k+1 < 0\): downward-opening parabola, polynomial negative outside the roots. For \(P(x) \leq 0\):
\[ x \leq x_1 \quad \text{or} \quad x \geq x_2. \]
Case \(-1 < k < \dfrac{1}{3}\)
\(\Delta > 0\) and \(k+1 > 0\): upward-opening parabola, polynomial negative between the roots. For \(P(x) \leq 0\):
\[ x_1 \leq x \leq x_2. \]
Case \(k = \dfrac{1}{3}\)
\(\Delta = 0\) and \(k+1 > 0\): upward-opening parabola, polynomial always \(\geq 0\), vanishing only at the repeated root. Therefore:
\[ x = x_0. \]
Case \(k > \dfrac{1}{3}\)
\(\Delta < 0\) and \(k+1 > 0\): upward-opening parabola, polynomial always positive. Therefore \(S = \emptyset\).
Exercise 14 — level ★★★★☆
Solve:
\[ x^2 - 2(k+1)x + k^2 + k < 0 \]
Answer
\[ \begin{cases} k+1-\sqrt{k+1} < x < k+1+\sqrt{k+1}, & k > -1, \\[8pt] S = \emptyset, & k \leq -1. \end{cases} \]
Solution
The leading coefficient is \(a = 1 > 0\): the parabola always opens upward. We compute the discriminant:
\[ \Delta = 4(k+1)^2 - 4(k^2+k) = 4\bigl[(k+1)^2 - (k^2+k)\bigr] = 4(k+1). \]
We analyze the sign:
\[ \Delta > 0 \iff k > -1, \quad \Delta = 0 \iff k = -1, \quad \Delta < 0 \iff k < -1. \]
Case \(k > -1\)
Two distinct real roots. Since \(\sqrt{4(k+1)} = 2\sqrt{k+1}\):
\[ x_{1,2} = \frac{2(k+1) \pm 2\sqrt{k+1}}{2} = k+1 \mp \sqrt{k+1}. \]
The parabola opens upward, so the polynomial is negative between the two roots:
\[ k+1-\sqrt{k+1} < x < k+1+\sqrt{k+1}. \]
Case \(k = -1\)
\(\Delta = 0\): upward-opening parabola, polynomial always \(\geq 0\). The inequality requires \(P(x) < 0\), so \(S = \emptyset\).
Case \(k < -1\)
\(\Delta < 0\): polynomial always positive. Therefore \(S = \emptyset\).
Exercise 15 — level ★★★★★
Solve:
\[ (k-3)x^2 + (2k-1)x + k > 0 \]
Answer
\[ \begin{cases} S = \emptyset, & k < -\dfrac{1}{8}, \\[8pt] S = \emptyset, & k = -\dfrac{1}{8}, \\[8pt] x_1 < x < x_2, & -\dfrac{1}{8} < k < 3, \\[8pt] x > -\dfrac{3}{5}, & k = 3, \\[8pt] x < x_1 \ \text{or}\ x > x_2, & k > 3. \end{cases} \]
When two distinct real roots exist, \(x_1\) and \(x_2\) are the ordered roots, with \(x_1 < x_2\).
Solution
The coefficient of the quadratic term is \(k-3\). The first case to treat is \(k = 3\).
Case \(k = 3\)
If \(k = 3\), the inequality becomes \(5x + 3 > 0\), giving:
\[ x > -\frac{3}{5}. \]
Case \(k \neq 3\)
We compute the discriminant:
\[ \Delta = (2k-1)^2 - 4(k-3)k = 4k^2 - 4k + 1 - 4k^2 + 12k = 8k + 1. \]
We analyze the sign:
\[ \Delta > 0 \iff k > -\tfrac{1}{8}, \quad \Delta = 0 \iff k = -\tfrac{1}{8}, \quad \Delta < 0 \iff k < -\tfrac{1}{8}. \]
Case \(k < -\dfrac{1}{8}\)
\(\Delta < 0\) and \(k - 3 < 0\): downward-opening parabola, polynomial always negative. \(S = \emptyset\).
Case \(k = -\dfrac{1}{8}\)
\(\Delta = 0\) and \(k - 3 < 0\): downward-opening parabola, polynomial always \(\leq 0\). The inequality is strict, so \(S = \emptyset\).
Case \(-\dfrac{1}{8} < k < 3\)
\(\Delta > 0\) and \(k - 3 < 0\): downward-opening parabola, polynomial positive between the roots:
\[ x_1 < x < x_2. \]
Case \(k > 3\)
\(\Delta > 0\) and \(k - 3 > 0\): upward-opening parabola, polynomial positive outside the roots:
\[ x < x_1 \quad \text{or} \quad x > x_2. \]
Exercise 16 — level ★★★★★
Solve:
\[ (k^2-1)x^2 + 2kx + 1 \geq 0 \]
Answer
\[ \begin{cases} x \leq x_1 \ \text{or}\ x \geq x_2, & |k| > 1, \\[8pt] x \geq -\dfrac{1}{2}, & k = 1, \\[8pt] x \leq \dfrac{1}{2}, & k = -1, \\[8pt] x_1 \leq x \leq x_2, & |k| < 1. \end{cases} \]
In the quadratic cases, \(x_1\) and \(x_2\) are the two ordered roots, with \(x_1 < x_2\).
Solution
The coefficient of the quadratic term is \(k^2 - 1 = (k-1)(k+1)\). The degenerate cases occur for \(k = \pm 1\).
Case \(k = 1\)
The inequality becomes \(2x + 1 \geq 0\), giving \(x \geq -\dfrac{1}{2}\).
Case \(k = -1\)
The inequality becomes \(-2x + 1 \geq 0\), giving \(x \leq \dfrac{1}{2}\).
Case \(k \neq \pm 1\)
We compute the discriminant:
\[ \Delta = (2k)^2 - 4(k^2-1) = 4k^2 - 4k^2 + 4 = 4. \]
The discriminant is always positive: the polynomial always has two distinct real roots. It only remains to determine the concavity:
\[ k^2 - 1 > 0 \iff |k| > 1, \qquad k^2 - 1 < 0 \iff |k| < 1. \]
Case \(|k| > 1\)
Upward-opening parabola, \(P(x) \geq 0\) outside the roots:
\[ x \leq x_1 \quad \text{or} \quad x \geq x_2. \]
Case \(|k| < 1\)
Downward-opening parabola, polynomial positive between the roots. For \(P(x) \geq 0\):
\[ x_1 \leq x \leq x_2. \]
Exercise 17 — level ★★★★★
Solve:
\[ (k-2)x^2 - 4x + k < 0 \]
Answer
\[ \begin{cases} S = \mathbb{R}, & k < 1-\sqrt{5}, \\[6pt] S = \mathbb{R} \setminus \{x_0\}, & k = 1-\sqrt{5}, \\[6pt] x < x_1 \ \text{or}\ x > x_2, & 1-\sqrt{5} < k < 2, \\[8pt] x > \dfrac{1}{2}, & k = 2, \\[8pt] x_1 < x < x_2, & 2 < k < 1+\sqrt{5}, \\[6pt] S = \emptyset, & k \geq 1+\sqrt{5}. \end{cases} \]
When two distinct real roots exist, \(x_1\) and \(x_2\) are the ordered roots, with \(x_1 < x_2\). When \(k = 1-\sqrt{5}\), \(x_0\) is the repeated root.
Solution
The coefficient of the quadratic term is \(k-2\).
Case \(k = 2\)
If \(k = 2\), the inequality becomes \(-4x + 2 < 0\), giving:
\[ x > \frac{1}{2}. \]
Case \(k \neq 2\)
We compute the discriminant:
\[ \Delta = 16 - 4(k-2)k = 16 - 4k^2 + 8k = -4(k^2 - 2k - 4). \]
The roots of \(k^2 - 2k - 4 = 0\) are \(k = 1 \pm \sqrt{5}\). Since \(\Delta = -4(k^2 - 2k - 4)\):
\[ \Delta > 0 \iff 1-\sqrt{5} < k < 1+\sqrt{5}, \]
\[ \Delta = 0 \iff k = 1-\sqrt{5} \ \text{or}\ k = 1+\sqrt{5}, \]
\[ \Delta < 0 \iff k < 1-\sqrt{5} \ \text{or}\ k > 1+\sqrt{5}. \]
Case \(k < 1-\sqrt{5}\)
\(\Delta < 0\) and \(k - 2 < 0\): downward-opening parabola, polynomial always negative. The inequality \(P(x) < 0\) is satisfied for every \(x\):
\[ S = \mathbb{R}. \]
Case \(k = 1-\sqrt{5}\)
\(\Delta = 0\) and \(k - 2 < 0\): downward-opening parabola, polynomial always \(\leq 0\), vanishing at the repeated root \(x_0\). Since the inequality is strict:
\[ S = \mathbb{R} \setminus \{x_0\}. \]
Case \(1-\sqrt{5} < k < 2\)
\(\Delta > 0\) and \(k - 2 < 0\): downward-opening parabola, polynomial negative outside the roots:
\[ x < x_1 \quad \text{or} \quad x > x_2. \]
Case \(2 < k < 1+\sqrt{5}\)
\(\Delta > 0\) and \(k - 2 > 0\): upward-opening parabola, polynomial negative between the roots:
\[ x_1 < x < x_2. \]
Case \(k = 1+\sqrt{5}\)
\(\Delta = 0\) and \(k - 2 > 0\): upward-opening parabola, polynomial always \(\geq 0\). The inequality \(P(x) < 0\) has no solutions:
\[ S = \emptyset. \]
Case \(k > 1+\sqrt{5}\)
\(\Delta < 0\) and \(k - 2 > 0\): upward-opening parabola, polynomial always positive. Therefore \(S = \emptyset\).
Exercise 18 — level ★★★★★
Solve:
\[ (k+1)x^2 + (k-1)x - 2 > 0 \]
Answer
\[ \begin{cases} x_1 < x < x_2, & k < -3, \\[6pt] S = \emptyset, & k = -3, \\[6pt] x_1 < x < x_2, & -3 < k < -1, \\[6pt] x < -1, & k = -1, \\[6pt] x < x_1 \ \text{or}\ x > x_2, & k > -1. \end{cases} \]
When two distinct real roots exist, \(x_1\) and \(x_2\) denote the ordered roots of the polynomial, with \(x_1 < x_2\).
Solution
The coefficient of the quadratic term is \(k+1\).
Case \(k = -1\)
If \(k = -1\), the inequality becomes \(-2x - 2 > 0\), giving \(x < -1\).
Case \(k \neq -1\)
We compute the discriminant:
\[ \Delta = (k-1)^2 + 8(k+1) = k^2 - 2k + 1 + 8k + 8 = k^2 + 6k + 9 = (k+3)^2. \]
The discriminant is always \(\geq 0\):
\[ \Delta = 0 \iff k = -3, \qquad \Delta > 0 \iff k \neq -3. \]
The concavity depends on the sign of \(k+1\).
Case \(k < -3\)
\(\Delta > 0\) and \(k+1 < 0\): downward-opening parabola, polynomial positive between the roots:
\[ x_1 < x < x_2. \]
Case \(k = -3\)
\(\Delta = 0\) and \(k+1 < 0\): downward-opening parabola, polynomial always \(\leq 0\). The inequality is strict, so \(S = \emptyset\).
Case \(-3 < k < -1\)
\(\Delta > 0\) and \(k+1 < 0\): downward-opening parabola, polynomial positive between the roots:
\[ x_1 < x < x_2. \]
Case \(k > -1\)
\(\Delta > 0\) and \(k+1 > 0\): upward-opening parabola, polynomial positive outside the roots:
\[ x < x_1 \quad \text{or} \quad x > x_2. \]
Exercise 19 — level ★★★★★
Solve:
\[ (k-1)x^2 + (k-1)x + k > 0 \]
Answer
\[ \begin{cases} S = \emptyset, & k \leq -\dfrac{1}{3}, \\[8pt] x_1 < x < x_2, & -\dfrac{1}{3} < k < 1, \\[8pt] S = \mathbb{R}, & k \geq 1. \end{cases} \]
When \(-\dfrac{1}{3} < k < 1\), \(x_1\) and \(x_2\) are the two ordered roots, with \(x_1 < x_2\).
Solution
The coefficient of the quadratic term is \(k-1\). The first case to consider is \(k = 1\).
Case \(k = 1\)
If \(k = 1\), the inequality becomes \(1 > 0\), which is always true. Therefore \(S = \mathbb{R}\).
Case \(k \neq 1\)
We compute the discriminant:
\[ \Delta = (k-1)^2 - 4(k-1)k = (k-1)\bigl[(k-1) - 4k\bigr] = (k-1)(-3k-1). \]
The critical values are \(k = 1\) and \(k = -\dfrac{1}{3}\). We obtain:
\[ \Delta > 0 \iff -\tfrac{1}{3} < k < 1, \quad \Delta = 0 \iff k = -\tfrac{1}{3} \ \text{or}\ k = 1, \quad \Delta < 0 \iff k < -\tfrac{1}{3} \ \text{or}\ k > 1. \]
Case \(k < -\dfrac{1}{3}\)
\(\Delta < 0\) and \(k - 1 < 0\): downward-opening parabola, polynomial always negative. \(S = \emptyset\).
Case \(k = -\dfrac{1}{3}\)
\(\Delta = 0\) and \(k - 1 < 0\): downward-opening parabola, polynomial always \(\leq 0\). The inequality is strict, so \(S = \emptyset\).
Case \(-\dfrac{1}{3} < k < 1\)
\(\Delta > 0\) and \(k - 1 < 0\): downward-opening parabola, polynomial positive between the roots:
\[ x_1 < x < x_2. \]
Case \(k > 1\)
\(\Delta < 0\) and \(k - 1 > 0\): upward-opening parabola, polynomial always positive. Therefore \(S = \mathbb{R}\).
Exercise 20 — level ★★★★★
Solve:
\[ (k^2-9)x^2 + 6x + 1 \leq 0 \]
Answer
\[ \begin{cases} S = \emptyset, & k < -3\sqrt{2}, \\[6pt] x = x_0, & k = -3\sqrt{2}, \\[6pt] x_1 \leq x \leq x_2, & -3\sqrt{2} < k < -3, \\[8pt] x \leq -\dfrac{1}{6}, & k = -3, \\[8pt] x \leq x_1 \ \text{or}\ x \geq x_2, & -3 < k < 3, \\[8pt] x \leq -\dfrac{1}{6}, & k = 3, \\[8pt] x_1 \leq x \leq x_2, & 3 < k < 3\sqrt{2}, \\[6pt] x = x_0, & k = 3\sqrt{2}, \\[6pt] S = \emptyset, & k > 3\sqrt{2}. \end{cases} \]
When two distinct real roots exist, \(x_1\) and \(x_2\) are the ordered roots, with \(x_1 < x_2\). When \(k = \pm 3\sqrt{2}\), \(x_0\) is the repeated root.
Solution
The coefficient of the quadratic term is \(k^2 - 9 = (k-3)(k+3)\). The degenerate cases occur for \(k = \pm 3\).
Case \(k = -3\) or \(k = 3\)
If \(k = \pm 3\), the quadratic term vanishes and the inequality becomes \(6x + 1 \leq 0\), giving:
\[ x \leq -\frac{1}{6}. \]
Case \(k \neq \pm 3\)
We compute the discriminant:
\[ \Delta = 36 - 4(k^2-9) = 36 - 4k^2 + 36 = 72 - 4k^2 = 4(18 - k^2). \]
Since \(\sqrt{18} = 3\sqrt{2}\):
\[ \Delta > 0 \iff -3\sqrt{2} < k < 3\sqrt{2}, \quad \Delta = 0 \iff k = \pm 3\sqrt{2}, \quad \Delta < 0 \iff |k| > 3\sqrt{2}. \]
The concavity depends on the sign of \(k^2 - 9\):
\[ k^2 - 9 > 0 \iff |k| > 3, \qquad k^2 - 9 < 0 \iff |k| < 3. \]
Case \(k < -3\sqrt{2}\)
\(\Delta < 0\) and \(k^2 - 9 > 0\): upward-opening parabola, polynomial always positive. \(S = \emptyset\).
Case \(k = -3\sqrt{2}\)
\(\Delta = 0\) and \(k^2 - 9 > 0\): upward-opening parabola, polynomial always \(\geq 0\). Since the inequality is non-strict, the only solution is the repeated root:
\[ x = x_0. \]
Case \(-3\sqrt{2} < k < -3\)
\(\Delta > 0\) and \(k^2 - 9 > 0\): upward-opening parabola, polynomial negative between the roots:
\[ x_1 \leq x \leq x_2. \]
Case \(-3 < k < 3\)
\(k^2 - 9 < 0\) (and certainly \(\Delta > 0\)): downward-opening parabola, polynomial negative outside the roots. For \(P(x) \leq 0\):
\[ x \leq x_1 \quad \text{or} \quad x \geq x_2. \]
Case \(3 < k < 3\sqrt{2}\)
\(\Delta > 0\) and \(k^2 - 9 > 0\): upward-opening parabola, polynomial negative between the roots:
\[ x_1 \leq x \leq x_2. \]
Case \(k = 3\sqrt{2}\)
\(\Delta = 0\) and \(k^2 - 9 > 0\): upward-opening parabola, polynomial vanishes only at the repeated root:
\[ x = x_0. \]
Case \(k > 3\sqrt{2}\)
\(\Delta < 0\) and \(k^2 - 9 > 0\): upward-opening parabola, polynomial always positive. Therefore \(S = \emptyset\).