A collection of worked problems to learn how to study the sign of a trinomial as a real parameter varies, how to analyze the discriminant as a function of \( k \), and how to handle the degenerate cases in which the coefficient of the quadratic term vanishes. You will find a complete discussion for every value of \( k \), with clear steps even in the more intricate cases, designed to help you reason methodically and never miss a single case.
Exercise 1 — level ★★★☆☆
\[ x^2 + kx + 1 > 0 \]
Answer
• \(|k| < 2\): \(S = \mathbb{R}\)
• \(|k| = 2\): \(S = \mathbb{R} \setminus \left\{-\tfrac{k}{2}\right\}\)
• \(|k| > 2\): \(S = \left(-\infty,\, x_1\right) \cup \left(x_2,\, +\infty\right)\)
Solution
Discriminant
\[ \Delta = k^2 - 4 \]
Roots (when \(\Delta \ge 0\))
\[ x_{1,2} = \frac{-k \pm \sqrt{k^2 - 4}}{2} \]
Sign analysis — upward-opening parabola
• \(\Delta < 0 \;\Leftrightarrow\; |k| < 2\): no real roots, the trinomial is always \(> 0\); \(S = \mathbb{R}\).
• \(\Delta = 0 \;\Leftrightarrow\; k = \pm 2\): double root \(x_0 = -k/2\); the parabola touches the axis but does not cross it; strict inequality; \(S = \mathbb{R} \setminus \{x_0\}\).
• \(\Delta > 0 \;\Leftrightarrow\; |k| > 2\): two distinct roots \(x_1 < x_2\); positive outside the roots; \(S = (-\infty, x_1) \cup (x_2, +\infty)\).
Exercise 2 — level ★★★☆☆
\[ x^2 + (k-2)x + k > 0 \]
Answer
• \(k \in (4-2\sqrt{3},\; 4+2\sqrt{3})\): \(S = \mathbb{R}\)
• \(k = 4 \pm 2\sqrt{3}\): \(S = \mathbb{R} \setminus \{x_0\}\)
• otherwise: \(S = (-\infty, x_1) \cup (x_2, +\infty)\)
Solution
Discriminant
\[ \Delta = (k-2)^2 - 4k = k^2 - 8k + 4 \]
Roots of \(\Delta = 0\)
\[ k = \frac{8 \pm \sqrt{48}}{2} = 4 \pm 2\sqrt{3} \]
Analysis
• \(\Delta < 0 \;\Leftrightarrow\; k \in (4-2\sqrt{3},\, 4+2\sqrt{3})\): trinomial always \(> 0\); \(S = \mathbb{R}\).
• \(\Delta = 0\): double root; strict inequality; \(S = \mathbb{R} \setminus \{x_0\}\).
• \(\Delta > 0\): upward-opening parabola with two distinct roots; \(S = (-\infty, x_1) \cup (x_2, +\infty)\).
Exercise 3 — level ★★★☆☆
\[ x^2 - 2(k+1)x + k^2 > 0 \]
Answer
• \(k < -\tfrac{1}{2}\): \(S = \mathbb{R}\)
• \(k = -\tfrac{1}{2}\): \(S = \mathbb{R} \setminus \left\{\tfrac{1}{2}\right\}\)
• \(k > -\tfrac{1}{2}\): \(S = (-\infty, x_1) \cup (x_2, +\infty)\) with \(x_{1,2} = (k+1) \mp \sqrt{2k+1}\)
Solution
Discriminant
\[ \Delta = 4(k+1)^2 - 4k^2 = 4(k^2+2k+1-k^2) = 4(2k+1) \]
Analysis
• \(\Delta < 0 \;\Leftrightarrow\; k < -\tfrac{1}{2}\): \(S = \mathbb{R}\).
• \(\Delta = 0 \;\Leftrightarrow\; k = -\tfrac{1}{2}\): double root \(x_0 = k+1 = \tfrac{1}{2}\); strict inequality; \(S = \mathbb{R} \setminus \left\{\tfrac{1}{2}\right\}\).
• \(\Delta > 0 \;\Leftrightarrow\; k > -\tfrac{1}{2}\): two distinct roots; positive outside the roots; \(S = (-\infty, x_1) \cup (x_2, +\infty)\).
Exercise 4 — level ★★★★☆
\[ (k-1)x^2 + 2x + 1 \ge 0 \]
Answer
• \(k = 1\): \(S = \left[-\tfrac{1}{2}, +\infty\right)\)
• \(k > 2\): \(S = \mathbb{R}\)
• \(k = 2\): \(S = \mathbb{R}\)
• \(1 < k < 2\): \(S = (-\infty, x_1] \cup [x_2, +\infty)\)
• \(k < 1\): \(S = [x_1, x_2]\)
Solution
Case \(k = 1\) — linear equation
\(2x + 1 \ge 0 \;\Rightarrow\; S = \left[-\tfrac{1}{2}, +\infty\right)\).
Case \(k \neq 1\) — quadratic equation
\[ \Delta = 4 - 4(k-1) = 8 - 4k \]
• \(k > 2\): \(\Delta < 0\), \(k-1 > 0\) (concavity \(\uparrow\)); \(S = \mathbb{R}\).
• \(k = 2\): \(\Delta = 0\), double root \(x_0 = -1\), \(k-1 > 0\); non-strict inequality; \(S = \mathbb{R}\).
• \(1 < k < 2\): \(\Delta > 0\), \(k-1 > 0\) (concavity \(\uparrow\)); \(S = (-\infty, x_1] \cup [x_2, +\infty)\).
• \(k < 1\): \(\Delta > 0\), \(k-1 < 0\) (concavity \(\downarrow\)); \(S = [x_1, x_2]\).
Exercise 5 — level ★★★☆☆
\[ x^2 + kx - 4 > 0 \]
Answer
For every \(k \in \mathbb{R}\) there always exist two distinct real roots; \(S = (-\infty, x_1) \cup (x_2, +\infty)\) with \(x_{1,2} = \dfrac{-k \pm \sqrt{k^2+16}}{2}\).
Solution
Discriminant
\[ \Delta = k^2 + 16 \ge 16 > 0 \quad \forall\, k \in \mathbb{R} \]
The trinomial always has two distinct real roots. Upward-opening parabola: positive outside the roots.
Exercise 6 — level ★★★☆☆
\[ x^2 - kx + k > 0 \]
Answer
• \(k \in (0, 4)\): \(S = \mathbb{R}\)
• \(k = 0\): \(S = \mathbb{R} \setminus \{0\}\)
• \(k = 4\): \(S = \mathbb{R} \setminus \{2\}\)
• \(k < 0\) or \(k > 4\): \(S = (-\infty, x_1) \cup (x_2, +\infty)\)
Solution
Discriminant
\[ \Delta = k^2 - 4k = k(k-4) \]
Analysis
• \(\Delta < 0 \;\Leftrightarrow\; k \in (0,4)\): \(S = \mathbb{R}\).
• \(\Delta = 0 \;\Leftrightarrow\; k=0\) (root \(x_0=0\)) or \(k=4\) (root \(x_0=2\)): strict inequality; \(S = \mathbb{R} \setminus \{x_0\}\).
• \(\Delta > 0 \;\Leftrightarrow\; k < 0\) or \(k > 4\): \(S = (-\infty, x_1) \cup (x_2, +\infty)\).
Exercise 7 — level ★★★★☆
\[ x^2 + (k+1)x + k < 0 \]
Answer
• \(k = 1\): \(S = \emptyset\)
• \(k < 1\): \(S = (-1,\; -k)\)
• \(k > 1\): \(S = (-k,\; -1)\)
Solution
Factorization
\[ x^2 + (k+1)x + k = (x+1)(x+k) \]
Discriminant (check)
\[ \Delta = (k+1)^2 - 4k = k^2 - 2k + 1 = (k-1)^2 \ge 0 \quad \forall\, k \]
Sign analysis
Roots: \(x = -1\) and \(x = -k\).
• \(k = 1\): double root \(x = -1\); \((x+1)^2 < 0\) is impossible; \(S = \emptyset\).
• \(k < 1\): \(-k > -1\), roots ordered \(-1 < -k\); product \(< 0\) between the roots; \(S = (-1,\, -k)\).
• \(k > 1\): \(-k < -1\), roots ordered \(-k < -1\); product \(< 0\) between the roots; \(S = (-k,\, -1)\).
Exercise 8 — level ★★★★☆
\[ x^2 - 3kx + 2k > 0 \]
Answer
• \(k \in \left(0,\, \tfrac{8}{9}\right)\): \(S = \mathbb{R}\)
• \(k = 0\) or \(k = \tfrac{8}{9}\): \(S = \mathbb{R} \setminus \{x_0\}\)
• \(k < 0\) or \(k > \tfrac{8}{9}\): \(S = (-\infty, x_1) \cup (x_2, +\infty)\)
Solution
Discriminant
\[ \Delta = 9k^2 - 8k = k(9k-8) \]
Analysis
• \(\Delta < 0 \;\Leftrightarrow\; k \in \left(0,\, \tfrac{8}{9}\right)\): \(S = \mathbb{R}\).
• \(\Delta = 0 \;\Leftrightarrow\; k = 0\) or \(k = \tfrac{8}{9}\): strict inequality; \(S = \mathbb{R} \setminus \{x_0\}\).
• \(\Delta > 0 \;\Leftrightarrow\; k < 0\) or \(k > \tfrac{8}{9}\): \(S = (-\infty, x_1) \cup (x_2, +\infty)\).
Exercise 9 — level ★★★☆☆
\[ x^2 - kx + 2 > 0 \]
Answer
• \(|k| < 2\sqrt{2}\): \(S = \mathbb{R}\)
• \(|k| = 2\sqrt{2}\): \(S = \mathbb{R} \setminus \left\{\tfrac{k}{2}\right\}\)
• \(|k| > 2\sqrt{2}\): \(S = (-\infty, x_1) \cup (x_2, +\infty)\)
Solution
Discriminant
\[ \Delta = k^2 - 8 \]
Analysis
• \(\Delta < 0 \;\Leftrightarrow\; |k| < 2\sqrt{2}\): \(S = \mathbb{R}\).
• \(\Delta = 0 \;\Leftrightarrow\; k = \pm 2\sqrt{2}\): double root \(x_0 = k/2\); strict inequality; \(S = \mathbb{R} \setminus \{x_0\}\).
• \(\Delta > 0 \;\Leftrightarrow\; |k| > 2\sqrt{2}\): \(S = (-\infty, x_1) \cup (x_2, +\infty)\).
Exercise 10 — level ★★★☆☆
\[ x^2 + (k-3)x + 2k \le 0 \]
Answer
• \(k \in (7-2\sqrt{10},\; 7+2\sqrt{10})\): \(S = \emptyset\)
• \(k = 7 \pm 2\sqrt{10}\): \(S = \{x_0\}\)
• otherwise: \(S = [x_1, x_2]\)
Solution
Discriminant
\[ \Delta = (k-3)^2 - 8k = k^2 - 14k + 9 \]
Roots of \(\Delta = 0\)
\[ k = \frac{14 \pm \sqrt{196-36}}{2} = 7 \pm 2\sqrt{10} \]
Analysis
• \(\Delta < 0 \;\Leftrightarrow\; k \in (7-2\sqrt{10},\, 7+2\sqrt{10})\): upward-opening parabola, always \(> 0\); \(S = \emptyset\).
• \(\Delta = 0\): single root \(x_0\); non-strict inequality; \(S = \{x_0\}\).
• \(\Delta > 0\): upward-opening parabola, negative between the roots; \(S = [x_1, x_2]\).
Exercise 11 — level ★★★★☆
\[ (k+2)x^2 - x + 1 > 0 \]
Answer
• \(k = -2\): \(S = (-\infty, 1)\)
• \(k > -\tfrac{7}{4}\) (with \(k \neq -2\)): \(S = \mathbb{R}\)
• \(k = -\tfrac{7}{4}\): \(S = \mathbb{R} \setminus \{2\}\)
• \(-2 < k < -\tfrac{7}{4}\): \(S = (-\infty, x_1) \cup (x_2, +\infty)\)
• \(k < -2\): \(S = (x_1, x_2)\)
Solution
Case \(k = -2\) — linear equation
\(-x + 1 > 0 \;\Rightarrow\; x < 1\); \(S = (-\infty, 1)\).
Case \(k \neq -2\) — quadratic equation
\[ \Delta = 1 - 4(k+2) = -4k - 7 \]
Note: \(-2 < -\tfrac{7}{4}\) (that is, \(-2 < -1.75\)), so on the real line the ordering is \(k < -2\), then \(-2 < k < -\tfrac{7}{4}\), then \(k \ge -\tfrac{7}{4}\).
• \(k > -\tfrac{7}{4}\): \(\Delta < 0\), \(k+2 > 0\) (concavity \(\uparrow\)); \(S = \mathbb{R}\).
• \(k = -\tfrac{7}{4}\): \(\Delta = 0\), \(k+2 = \tfrac{1}{4} > 0\); double root \(x_0 = \tfrac{1}{2(k+2)} = 2\); strict inequality; \(S = \mathbb{R} \setminus \{2\}\).
• \(-2 < k < -\tfrac{7}{4}\): \(\Delta > 0\), \(k+2 > 0\) (concavity \(\uparrow\)); positive outside the roots; \(S = (-\infty, x_1) \cup (x_2, +\infty)\).
• \(k < -2\): \(\Delta > 0\), \(k+2 < 0\) (concavity \(\downarrow\)); positive between the roots; \(S = (x_1, x_2)\).
Exercise 12 — level ★★★★☆
\[ x^2 - 2kx + k^2 - 1 \ge 0 \]
Answer
For every \(k\), \(S = (-\infty,\; k-1] \cup [k+1,\; +\infty)\).
Solution
Discriminant
\[ \Delta = 4k^2 - 4(k^2-1) = 4 > 0 \quad \forall\, k \]
Roots
\[ x_{1,2} = \frac{2k \pm 2}{2} = k \pm 1 \]
Upward-opening parabola with roots \(k-1 < k+1\); non-strict inequality; \(S = (-\infty,\, k-1] \cup [k+1,\, +\infty)\).
Exercise 13 — level ★★★☆☆
\[ x^2 + kx + k > 0 \]
Answer
• \(k \in (0, 4)\): \(S = \mathbb{R}\)
• \(k = 0\): \(S = \mathbb{R} \setminus \{0\}\)
• \(k = 4\): \(S = \mathbb{R} \setminus \{-2\}\)
• \(k < 0\) or \(k > 4\): \(S = (-\infty, x_1) \cup (x_2, +\infty)\)
Solution
Discriminant
\[ \Delta = k^2 - 4k = k(k-4) \]
Analysis
• \(\Delta < 0 \;\Leftrightarrow\; k \in (0,4)\): \(S = \mathbb{R}\).
• \(\Delta = 0 \;\Leftrightarrow\; k=0\) (root \(x_0=0\)) or \(k=4\) (root \(x_0=-2\)): strict inequality; \(S = \mathbb{R} \setminus \{x_0\}\).
• \(\Delta > 0 \;\Leftrightarrow\; k < 0\) or \(k > 4\): \(S = (-\infty, x_1) \cup (x_2, +\infty)\).
Exercise 14 — level ★★★★☆
\[ (k-1)x^2 + (k+1)x + 1 < 0 \]
Answer
• \(k = 1\): \(S = \left(-\infty,\, -\tfrac{1}{2}\right)\)
• \(k > 1\): \(S = (x_1, x_2)\)
• \(k < 1\): \(S = (-\infty, x_1) \cup (x_2, +\infty)\)
Solution
Case \(k = 1\) — linear equation
\(2x + 1 < 0 \;\Rightarrow\; x < -\tfrac{1}{2}\); \(S = \left(-\infty, -\tfrac{1}{2}\right)\).
Case \(k \neq 1\) — quadratic equation
\[ \Delta = (k+1)^2 - 4(k-1) = k^2 - 2k + 5 = (k-1)^2 + 4 \ge 4 > 0 \quad \forall\, k \]
The discriminant is always positive: the trinomial always has two distinct real roots.
• \(k > 1\): \(k-1 > 0\) (concavity \(\uparrow\)); negative between the roots; \(S = (x_1, x_2)\).
• \(k < 1\): \(k-1 < 0\) (concavity \(\downarrow\)); negative outside the roots; \(S = (-\infty, x_1) \cup (x_2, +\infty)\).
Exercise 15 — level ★★★☆☆
\[ x^2 - k^2 x + 1 > 0 \]
Answer
• \(|k| < \sqrt{2}\): \(S = \mathbb{R}\)
• \(|k| = \sqrt{2}\): \(S = \mathbb{R} \setminus \{1\}\)
• \(|k| > \sqrt{2}\): \(S = (-\infty, x_1) \cup (x_2, +\infty)\)
Solution
Discriminant
\[ \Delta = k^4 - 4 \]
Analysis
• \(\Delta < 0 \;\Leftrightarrow\; k^4 < 4 \;\Leftrightarrow\; k^2 < 2 \;\Leftrightarrow\; |k| < \sqrt{2}\): \(S = \mathbb{R}\).
• \(\Delta = 0 \;\Leftrightarrow\; |k| = \sqrt{2}\): \(k^2 = 2\), double root \(x_0 = k^2/2 = 1\); strict inequality; \(S = \mathbb{R} \setminus \{1\}\).
• \(\Delta > 0 \;\Leftrightarrow\; |k| > \sqrt{2}\): \(S = (-\infty, x_1) \cup (x_2, +\infty)\).
Exercise 16 — level ★★★★☆
\[ x^2 + 2kx + k^2 - 4 \ge 0 \]
Answer
For every \(k\), \(S = (-\infty,\; -k-2] \cup [-k+2,\; +\infty)\).
Solution
Discriminant
\[ \Delta = 4k^2 - 4(k^2-4) = 16 > 0 \quad \forall\, k \]
Roots
\[ x_{1,2} = \frac{-2k \pm 4}{2} = -k \pm 2 \]
Upward-opening parabola with roots \(-k-2 < -k+2\); non-strict inequality; \(S = (-\infty,\, -k-2] \cup [-k+2,\, +\infty)\).
Exercise 17 — level ★★★★☆
\[ x^2 - (k+1)x + k > 0 \]
Answer
• \(k = 1\): \(S = \mathbb{R} \setminus \{1\}\)
• \(k < 1\): \(S = (-\infty, k) \cup (1, +\infty)\)
• \(k > 1\): \(S = (-\infty, 1) \cup (k, +\infty)\)
Solution
Factorization
\[ x^2 - (k+1)x + k = (x-1)(x-k) \]
Discriminant (check)
\[ \Delta = (k+1)^2 - 4k = k^2 - 2k + 1 = (k-1)^2 \ge 0 \quad \forall\, k \]
Sign analysis
• \(k = 1\): double root \(x = 1\); \((x-1)^2 > 0\) for every \(x \neq 1\); \(S = \mathbb{R} \setminus \{1\}\).
• \(k < 1\): roots ordered \(k < 1\); product \(> 0\) outside the roots; \(S = (-\infty, k) \cup (1, +\infty)\).
• \(k > 1\): roots ordered \(1 < k\); product \(> 0\) outside the roots; \(S = (-\infty, 1) \cup (k, +\infty)\).
Exercise 18 — level ★★★★★
\[ (k-2)x^2 + 2x + 1 \ge 0 \]
Answer
• \(k = 2\): \(S = \left[-\tfrac{1}{2}, +\infty\right)\)
• \(k > 3\): \(S = \mathbb{R}\)
• \(k = 3\): \(S = \mathbb{R}\)
• \(2 < k < 3\): \(S = (-\infty, x_1] \cup [x_2, +\infty)\)
• \(k < 2\): \(S = [x_1, x_2]\)
Solution
Case \(k = 2\) — linear equation
\(2x + 1 \ge 0 \;\Rightarrow\; S = \left[-\tfrac{1}{2}, +\infty\right)\).
Case \(k \neq 2\) — quadratic equation
\[ \Delta = 4 - 4(k-2) = 12 - 4k \]
• \(k > 3\): \(\Delta < 0\), \(k-2 > 0\) (concavity \(\uparrow\)); \(S = \mathbb{R}\).
• \(k = 3\): \(\Delta = 0\), double root \(x_0 = -1\), \(k-2 = 1 > 0\); non-strict inequality; \(S = \mathbb{R}\).
• \(2 < k < 3\): \(\Delta > 0\), \(k-2 > 0\) (concavity \(\uparrow\)); \(S = (-\infty, x_1] \cup [x_2, +\infty)\).
• \(k < 2\): \(\Delta > 0\), \(k-2 < 0\) (concavity \(\downarrow\)); \(S = [x_1, x_2]\).
Exercise 19 — level ★★★★★
\[ x^2 + kx + k^2 - 1 < 0 \]
Answer
• \(|k| < \tfrac{2\sqrt{3}}{3}\): \(S = (x_1, x_2)\) with \(x_{1,2} = \dfrac{-k \pm \sqrt{4-3k^2}}{2}\)
• \(|k| \ge \tfrac{2\sqrt{3}}{3}\): \(S = \emptyset\)
Solution
Discriminant
\[ \Delta = k^2 - 4(k^2-1) = -3k^2 + 4 \]
Analysis
• \(\Delta > 0 \;\Leftrightarrow\; 3k^2 < 4 \;\Leftrightarrow\; |k| < \tfrac{2}{\sqrt{3}} = \tfrac{2\sqrt{3}}{3}\): upward-opening parabola, negative between the roots; \(S = (x_1, x_2)\).
• \(\Delta = 0 \;\Leftrightarrow\; |k| = \tfrac{2\sqrt{3}}{3}\): double root, trinomial \(\ge 0\); strict inequality; \(S = \emptyset\).
• \(\Delta < 0 \;\Leftrightarrow\; |k| > \tfrac{2\sqrt{3}}{3}\): no real roots, trinomial always \(> 0\); \(S = \emptyset\).
Exercise 20 — level ★★★★★
\[ (x-k)(x-1) > 0 \]
Answer
• \(k < 1\): \(S = (-\infty, k) \cup (1, +\infty)\)
• \(k = 1\): \(S = \mathbb{R} \setminus \{1\}\)
• \(k > 1\): \(S = (-\infty, 1) \cup (k, +\infty)\)
Solution
Roots
\[ x = k \qquad x = 1 \]
Sign analysis
• \(k < 1\): roots ordered \(k < 1\); product \(> 0\) outside the roots; \(S = (-\infty, k) \cup (1, +\infty)\).
• \(k = 1\): double root \((x-1)^2 > 0\) for every \(x \neq 1\); \(S = \mathbb{R} \setminus \{1\}\).
• \(k > 1\): roots ordered \(1 < k\); product \(> 0\) outside the roots; \(S = (-\infty, 1) \cup (k, +\infty)\).