Inequalities

An inequality states that one expression is greater than or less than another. Inequalities arise naturally when finding the domain and range of functions, and are closely related to quadratic functions through their graphical interpretation.

Inequality Rules

Basic Properties

Let $a$, $b$, and $c$ be real numbers. The following properties hold for inequalities:

Addition property:

$$a > b \implies a + c > b + c$$

Adding the same quantity to both sides preserves the inequality.

Multiplication by a positive number:

$$a > b,\; c > 0 \implies ac > bc$$

Multiplication by a negative number (reversal):

$$a > b,\; c < 0 \implies ac < bc$$

This is the most important rule to remember: multiplying or dividing both sides by a negative number reverses the inequality sign.

Transitivity

$$a > b \mathrm{ and } b > c \implies a > c$$

Other Properties

• If $a > b > 0$, then $a^2 > b^2$ and $\dfrac{1}{a} < \dfrac{1}{b}$.
• If $a > b$ and $c > d$, then $a + c > b + d$.
• If $a > b > 0$ and $c > d > 0$, then $ac > bd$.

Examples

• $3 > 1 \implies 3 + 5 > 1 + 5$, i.e., $8 > 6$.
• $4 > 2$ and $3 > 0 \implies 4 \times 3 > 2 \times 3$, i.e., $12 > 6$.
• $5 > 2$ and $-3 < 0 \implies 5 \times (-3) < 2 \times (-3)$, i.e., $-15 < -6$.
• $7 > 5 > 2 \implies 7 > 2$ (transitivity).
• $3 > 2 > 0 \implies 9 > 4$ and $\dfrac{1}{3} < \dfrac{1}{2}$.

Linear Inequalities

Solving Linear Inequalities

A linear inequality has the form $ax + b > c$ or $ax + b < c$, where $a \neq 0$. The solution procedure mirrors that of linear equations, with one critical exception: when multiplying or dividing by a negative number, the inequality sign must be reversed.

General method:

1. Collect like terms on each side.
2. Isolate the variable by performing the same operation on both sides.
3. Reverse the inequality sign if multiplying or dividing by a negative number.

Number Line Representation

The solution of a linear inequality in one variable is an interval, which can be represented on a number line:

• An open circle ($\circ$) indicates a strict inequality ($<$ or $>$).
• A closed circle ($\bullet$) indicates an inclusive inequality ($\leq$ or $\geq$).

Examples

• Solve $3x - 7 > 5$:

$$3x > 12 \implies x > 4$$

The solution set is $(4, \infty)$.

• Solve $2x + 3 \leq x - 4$:

$$2x - x \leq -4 - 3 \implies x \leq -7$$

The solution set is $(-\infty, -7]$.

• Solve $-2x + 6 < 10$:

$$-2x < 4 \implies x > -2$$

Note the reversal of the inequality sign when dividing by $-2$. The solution set is $(-2, \infty)$.

• Solve $5(x - 2) \geq 3x + 4$:

$$5x - 10 \geq 3x + 4 \implies 2x \geq 14 \implies x \geq 7$$

The solution set is $[7, \infty)$.

Quadratic Inequalities

A quadratic inequality has the form $ax^2 + bx + c > 0$, $ax^2 + bx + c < 0$, or their non-strict variants, where $a \neq 0$. Solving quadratic inequalities relies on understanding the graph of the corresponding quadratic function $f(x) = ax^2 + bx + c$.

Graphical Interpretation

The graph of $f(x) = ax^2 + bx + c$ is a parabola. The solution of $f(x) > 0$ corresponds to the $x$-values where the parabola lies above the $x$-axis, and $f(x) < 0$ corresponds to where the parabola lies below the $x$-axis.

The discriminant $\Delta = b^2 - 4ac$ determines the number of intersections with the $x$-axis:

Condition	Parabola and $x$-axis	$ax^2 + bx + c > 0$ (for $a > 0$)
$\Delta > 0$	Two distinct intersections at $x = \alpha, \beta$	$x < \alpha$ or $x > \beta$
$\Delta = 0$	One intersection at $x = \alpha$	$x \neq \alpha$ (all real $x$ except $\alpha$)
$\Delta < 0$	No intersection	All real $x$ (always true)

Solving Method

Using factorization:

1. Bring all terms to one side so the inequality is in the form $ax^2 + bx + c \gtrless 0$.
2. Factorize (or use the quadratic formula) to find the roots.
3. Draw a sign diagram to determine the sign of the expression in each interval.

Sign diagram method:

1. Find the roots of $ax^2 + bx + c = 0$.
2. Mark the roots on a number line, dividing the real line into intervals.
3. Test the sign of the expression in each interval.
4. Select the intervals satisfying the inequality.

Examples

• Solve $x^2 - 5x + 6 > 0$:

Factorize: $(x-2)(x-3) > 0$.

Roots are $x = 2$ and $x = 3$. Since $a = 1 > 0$, the parabola opens upward.

Sign diagram:

$$\begin{array}{c|ccc} x & (-\infty, 2) & (2, 3) & (3, \infty) \\ \hline (x-2) & - & + & + \\ (x-3) & - & - & + \\ \hline (x-2)(x-3) & + & - & + \end{array}$$

Solution: $x < 2$ or $x > 3$, i.e., $(-\infty, 2) \cup (3, \infty)$.

• Solve $-x^2 + 4x - 3 \geq 0$:

Multiply both sides by $-1$ (reverse inequality): $x^2 - 4x + 3 \leq 0$.

Factorize: $(x-1)(x-3) \leq 0$.

Since $a = 1 > 0$, the parabola opens upward. The expression is non-positive between the roots.

Solution: $1 \leq x \leq 3$, i.e., $[1, 3]$.

• Solve $x^2 + 2x + 5 < 0$:

Discriminant: $\Delta = 4 - 20 = -16 < 0$.

Since $a = 1 > 0$ and $\Delta < 0$, the parabola is always above the $x$-axis.

Solution: $\varnothing$ (no solution).

• Solve $2x^2 - 3x - 2 > 0$:

Factorize: $(2x + 1)(x - 2) > 0$.

Roots: $x = -\dfrac{1}{2}$ and $x = 2$.

Sign diagram:

$$\begin{array}{c|ccc} x & \left(-\infty, -\tfrac{1}{2}\right) & \left(-\tfrac{1}{2}, 2\right) & (2, \infty) \\ \hline (2x+1) & - & + & + \\ (x-2) & - & - & + \\ \hline (2x+1)(x-2) & + & - & + \end{array}$$

Solution: $x < -\dfrac{1}{2}$ or $x > 2$, i.e., $\left(-\infty, -\dfrac{1}{2}\right) \cup (2, \infty)$.

Absolute Value Inequalities

The absolute value of a real number $x$, denoted $|x|$, represents its distance from zero on the number line. This geometric interpretation is the key to solving absolute value inequalities.

Fundamental Forms

$|x| < a$ (where $a > 0$):

Geometrically, $x$ is within distance $a$ from zero.

$$|x| < a \iff -a < x < a$$

$|x| > a$ (where $a > 0$):

Geometrically, $x$ is more than distance $a$ from zero.

$$|x| > a \iff x < -a \;\mathrm{ or }\; x > a$$

General Forms

$|ax + b| < c$ (where $c > 0$):

$$|ax + b| < c \iff -c < ax + b < c$$

This is equivalent to a system of two linear inequalities, which can be solved simultaneously.

$|ax + b| > c$ (where $c > 0$):

$$|ax + b| > c \iff ax + b < -c \;\mathrm{ or }\; ax + b > c$$

This gives two separate linear inequalities, each solved independently.

Special Cases

• If $c \leq 0$, then $|x| < c$ has no solution ($\varnothing$).
• If $c < 0$, then $|x| > c$ is true for all real $x$ ($\mathbb{'\{'}R{'\}'}$).
• $|x| \geq a$ and $|x| \leq a$ follow the same patterns with non-strict inequality signs.

Examples

• Solve $|x - 3| < 5$:

$$-5 < x - 3 < 5 \implies -2 < x < 8$$

Solution: $(-2, 8)$.

• Solve $|2x + 1| \geq 7$:

$$2x + 1 \leq -7 \;\mathrm{ or }\; 2x + 1 \geq 7$$$$2x \leq -8 \;\mathrm{ or }\; 2x \geq 6$$$$x \leq -4 \;\mathrm{ or }\; x \geq 3$$

Solution: $(-\infty, -4] \cup [3, \infty)$.

• Solve $|3x - 6| > 0$:

$$3x - 6 \neq 0 \implies x \neq 2$$

Solution: $\mathbb{'\{'}R{'\}'} \setminus \{2\}$.

• Solve $|x^2 - 4| < 5$:

$$-5 < x^2 - 4 < 5 \implies -1 < x^2 < 9$$

Since $x^2 \geq 0$ for all real $x$, the left inequality $-1 < x^2$ is always satisfied.

From $x^2 < 9$: $-3 < x < 3$.

Solution: $(-3, 3)$.

• Solve $|x + 2| \leq -1$:

Since $|x + 2| \geq 0$ for all real $x$, it can never be $\leq -1$.

Solution: $\varnothing$.

Systems of Inequalities

A system of inequalities requires finding the set of values that satisfy all inequalities simultaneously. The solution set of the system is the intersection of the solution sets of the individual inequalities.

Method for Systems of Linear Inequalities

1. Solve each inequality separately.
2. Find the intersection of all solution sets.
3. Represent the combined solution on a number line or using interval notation.

Systems Involving Quadratic and Absolute Value Inequalities

The same principle applies: solve each inequality independently, then take the intersection of all solution sets.

Examples

• Find all $x$ satisfying $x^2 - 4x + 3 < 0$ and $2x - 1 > 3$:

From $x^2 - 4x + 3 < 0$: $(x-1)(x-3) < 0 \implies 1 < x < 3$.

From $2x - 1 > 3$: $2x > 4 \implies x > 2$.

Intersection: $2 < x < 3$, i.e., $(2, 3)$.

• Find all $x$ satisfying $|x - 1| \leq 3$ and $x^2 - 9 \leq 0$:

From $|x - 1| \leq 3$: $-3 \leq x - 1 \leq 3 \implies -2 \leq x \leq 4$.

From $x^2 - 9 \leq 0$: $(x-3)(x+3) \leq 0 \implies -3 \leq x \leq 3$.

Intersection: $-2 \leq x \leq 3$, i.e., $[-2, 3]$.

• Find all $x$ satisfying $x^2 - 2x - 8 > 0$ and $|x + 1| < 6$:

From $x^2 - 2x - 8 > 0$: $(x-4)(x+2) > 0 \implies x < -2$ or $x > 4$.

From $|x + 1| < 6$: $-6 < x + 1 < 6 \implies -7 < x < 5$.

Intersection: $-7 < x < -2$, i.e., $(-7, -2)$.

(Note: the second branch $x > 4$ from the quadratic has no overlap with $x < 5$ beyond $(4, 5)$, but $x > 4$ and $x < 5$ gives $4 < x < 5$. The full intersection is $(-7, -2) \cup (4, 5)$.)

• Find all $x$ satisfying $x^2 + 1 > 0$ and $x - 3 < 0$:

From $x^2 + 1 > 0$: always true (discriminant $\Delta = -4 < 0$, and $a = 1 > 0$).

From $x - 3 < 0$: $x < 3$.

Intersection: $(-\infty, 3)$.

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Wrap-up Questions

1. Question: Solve the inequality $\dfrac{2x - 1}{3} \leq \dfrac{x + 2}{4} + 1$.

Details

Answer

Multiply through by $12$ (the LCM of $3$ and $4$, which is positive so the inequality sign is preserved):

$$4(2x - 1) \leq 3(x + 2) + 12$$$$8x - 4 \leq 3x + 6 + 12$$$$5x \leq 22 \implies x \leq \frac{22}{5}$$

Solution: $\left(-\infty, \dfrac{22}{5}\right]$.

2. Question: Solve $x^2 - 6x + 9 \geq 0$.

Details

Answer

Factorize: $(x - 3)^2 \geq 0$.

Since $(x - 3)^2 \geq 0$ for all real $x$ (a square is always non-negative), the solution is all real numbers.

Solution: $\mathbb{'\{'}R{'\}'}$.

3. Question: Find the range of $x$ for which $x^2 - 3x - 10 < 0$ and $2x + 1 > 0$ both hold.

Details

Answer

From $x^2 - 3x - 10 < 0$: $(x - 5)(x + 2) < 0 \implies -2 < x < 5$.

From $2x + 1 > 0$: $x > -\dfrac{1}{2}$.

Intersection: $-\dfrac{1}{2} < x < 5$, i.e., $\left(-\dfrac{1}{2}, 5\right)$.

4. Question: Solve $|3x - 5| < 7$.

Answer

$$-7 < 3x - 5 < 7$$$$-2 < 3x < 12$$$$-\frac{2}{3} < x < 4$$

Solution: $\left(-\dfrac{2}{3}, 4\right)$.

5. Question: Solve $|2x + 3| \geq x^2 + 2$.

Details

Answer

This inequality combines absolute value and quadratic expressions. Consider two cases.

Case 1: $2x + 3 \geq 0$ (i.e., $x \geq -\dfrac{3}{2}$), so $|2x + 3| = 2x + 3$:

$$2x + 3 \geq x^2 + 2 \implies x^2 - 2x - 1 \leq 0$$

Roots: $x = \dfrac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$.

So $1 - \sqrt{2} \leq x \leq 1 + \sqrt{2}$.

Combined with $x \geq -\dfrac{3}{2}$: $\max\!\left(1 - \sqrt{2},\; -\dfrac{3}{2}\right) \leq x \leq 1 + \sqrt{2}$.

Since $1 - \sqrt{2} \approx -0.414 > -\dfrac{3}{2} = -1.5$, the constraint is $1 - \sqrt{2} \leq x \leq 1 + \sqrt{2}$.

Case 2: $2x + 3 < 0$ (i.e., $x < -\dfrac{3}{2}$), so $|2x + 3| = -(2x + 3)$:

$$-2x - 3 \geq x^2 + 2 \implies x^2 + 2x + 5 \leq 0$$

Discriminant: $\Delta = 4 - 20 = -16 < 0$. Since $a = 1 > 0$, the expression is always positive. No solution in this case.

Solution: $[1 - \sqrt{2},\; 1 + \sqrt{2}]$.

6. Question: For what values of $k$ does the quadratic equation $x^2 + 2kx + k + 6 = 0$ have two distinct real roots?

Details

Answer

For two distinct real roots, the discriminant must satisfy $\Delta > 0$:

$$\Delta = (2k)^2 - 4(1)(k + 6) = 4k^2 - 4k - 24 > 0$$$$k^2 - k - 6 > 0$$

Factorize: $(k - 3)(k + 2) > 0$.

Since $a = 1 > 0$, the parabola opens upward. The expression is positive outside the roots.

Solution: $k < -2$ or $k > 3$, i.e., $(-\infty, -2) \cup (3, \infty)$.

7. Question: Solve the system of inequalities $x^2 - 5x + 4 \leq 0$, $|x - 2| \leq 3$, and $x > 0$.

Details

Answer

From $x^2 - 5x + 4 \leq 0$: $(x-1)(x-4) \leq 0 \implies 1 \leq x \leq 4$.

From $|x - 2| \leq 3$: $-3 \leq x - 2 \leq 3 \implies -1 \leq x \leq 5$.

From $x > 0$: $x \in (0, \infty)$.

Intersection of all three:

• From the first: $[1, 4]$.
• From the second: $[-1, 5]$.
• From the third: $(0, \infty)$.

Combined: $[1, 4]$.

Solution: $[1, 4]$.

8. Question: A ball is thrown upward from a height of $2$ m with an initial velocity of $20$ m/s. The height $h$ (in metres) after $t$ seconds is given by $h(t) = -5t^2 + 20t + 2$. During what time interval is the ball at a height greater than $17$ m?

Details

Answer

We need $h(t) > 17$:

$$-5t^2 + 20t + 2 > 17$$$$-5t^2 + 20t - 15 > 0$$

Divide by $-5$ (reverse inequality):

$$t^2 - 4t + 3 < 0$$

Factorize: $(t - 1)(t - 3) < 0 \implies 1 < t < 3$.

The ball is above $17$ m during the interval $(1, 3)$ seconds.

9. Question: Solve $\dfrac{x^2 - 4}{x - 1} \geq 0$.

Details

Answer

First note that $x \neq 1$ (the denominator cannot be zero).

Factorize the numerator: $\dfrac{(x-2)(x+2)}{x-1} \geq 0$.

Critical points: $x = -2$, $x = 1$, $x = 2$.

Sign diagram:

$$\begin{array}{c|cccc} x & (-\infty, -2) & (-2, 1) & (1, 2) & (2, \infty) \\ \hline (x-2) & - & - & - & + \\ (x+2) & - & + & + & + \\ (x-1) & - & - & + & + \\ \hline \dfrac{(x-2)(x+2)}{x-1} & - & + & - & + \end{array}$$

The expression is $\geq 0$ when $-2 \leq x < 1$ or $x \geq 2$.

(Note: $x = -2$ is included because the numerator is zero there; $x = 1$ is excluded; $x = 2$ is included.)

Solution: $[-2, 1) \cup [2, \infty)$.

10. Question: Solve $|x^2 - 3x + 1| < 3$.

Answer

$$-3 < x^2 - 3x + 1 < 3$$

Split into two inequalities:

Left inequality: $x^2 - 3x + 1 > -3 \implies x^2 - 3x + 4 > 0$.

Discriminant: $\Delta = 9 - 16 = -7 < 0$. Since $a = 1 > 0$, this is always true.

Right inequality: $x^2 - 3x + 1 < 3 \implies x^2 - 3x - 2 < 0$.

Roots: $x = \dfrac{3 \pm \sqrt{9 + 8}}{2} = \dfrac{3 \pm \sqrt{17}}{2}$.

Since $a = 1 > 0$, the expression is negative between the roots:

$$\frac{3 - \sqrt{17}}{2} < x < \frac{3 + \sqrt{17}}{2}$$

Since the left inequality imposes no restriction, the solution is:

Solution: $\left(\dfrac{3 - \sqrt{17}}{2},\; \dfrac{3 + \sqrt{17}}{2}\right)$.

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tip

Diagnostic Test Ready to test your understanding of Inequalities? The diagnostic test contains the hardest questions within the DSE specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Inequalities with other DSE mathematics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.

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Rational Inequalities

Method

To solve $\dfrac{f(x)}{g(x)} > 0$ (or $\geq, <, \leq$):

1. Find the zeros of $f(x)$ and $g(x)$ (the critical points).
2. Note that $g(x) \neq 0$ -- these points are always excluded.
3. Construct a sign diagram across all intervals defined by the critical points.
4. Select intervals satisfying the inequality.
5. Include critical points from the numerator (where $f(x) = 0$) only for $\geq$ or $\leq$.

Worked Example

Solve $\dfrac{x^2 - 3x + 2}{x + 1} \leq 0$.

Solution

Factor: $\dfrac{(x - 1)(x + 2)}{x + 1} \leq 0$.

Critical points: $x = -2$, $x = -1$, $x = 1$. Note $x \neq -1$.

Interval	$(-\infty, -2)$	$(-2, -1)$	$(-1, 1)$	$(1, \infty)$
Sign	$-$	$+$	$-$	$+$

The expression is $\leq 0$ when $x \leq -2$ or $-1 < x \leq 1$.

Solution: $(-\infty, -2] \cup (-1, 1]$.

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DSE Exam Technique

Showing Working

For inequality problems in DSE Paper 1:

1. When solving quadratic inequalities, always find the roots and sketch the parabola or draw a sign chart.
2. When multiplying or dividing by a negative, explicitly state that the inequality sign reverses.
3. For rational inequalities, clearly identify points where the denominator is zero and exclude them.
4. For system of inequalities, draw each solution on a number line and identify the intersection.

Significant Figures

Exact answers are preferred. If an approximate numerical answer is required, use 3 significant figures.

Common DSE Question Types

1. Quadratic inequalities with parameters (find the range of a parameter).
2. Absolute value inequalities (split into cases).
3. Rational inequalities (sign diagram method).
4. Systems of inequalities (intersection of solution sets).
5. Inequalities involving the discriminant (condition for real roots).

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Additional Worked Examples

Worked Example: Inequality with quadratic and absolute value

Solve $|x^2 - 4x + 3| < 3x - 5$.

Solution

The RHS must be positive: $3x - 5 > 0 \implies x > \dfrac{5}{3}$.

Case 1: $x^2 - 4x + 3 \geq 0$, i.e., $(x-1)(x-3) \geq 0 \implies x \leq 1$ or $x \geq 3$.

Combined with $x > \dfrac{5}{3}$: $x \geq 3$.

The inequality becomes $x^2 - 4x + 3 < 3x - 5 \implies x^2 - 7x + 8 < 0$.

Roots: $x = \dfrac{7 \pm \sqrt{49 - 32}}{2} = \dfrac{7 \pm \sqrt{17}}{2}$.

$\dfrac{7 - \sqrt{17}}{2} \approx 1.44$ and $\dfrac{7 + \sqrt{17}}{2} \approx 5.56$.

Intersection with $x \geq 3$: $3 \leq x < \dfrac{7 + \sqrt{17}}{2}$.

Case 2: $x^2 - 4x + 3 < 0$, i.e., $1 < x < 3$.

Combined with $x > \dfrac{5}{3}$: $\dfrac{5}{3} < x < 3$.

The inequality becomes $-(x^2 - 4x + 3) < 3x - 5 \implies -x^2 + 4x - 3 < 3x - 5 \implies -x^2 + x + 2 < 0 \implies x^2 - x - 2 > 0$.

$(x - 2)(x + 1) > 0 \implies x < -1$ or $x > 2$.

Intersection with $\dfrac{5}{3} < x < 3$: $2 < x < 3$.

Combined solution: $2 < x < \dfrac{7 + \sqrt{17}}{2}$.

Worked Example: Quadratic inequality with parameter

Find the range of $m$ such that $mx^2 + (m - 1)x + m > 0$ for all real $x$.

Solution

Case 1: $m = 0$. The inequality becomes $-x > 0 \implies x < 0$, which is not true for all real $x$. Reject.

Case 2: $m \neq 0$. For $mx^2 + (m-1)x + m > 0$ for all real $x$, we need $m > 0$ and $\Delta < 0$:

$\Delta = (m - 1)^2 - 4m^2 = m^2 - 2m + 1 - 4m^2 = -3m^2 - 2m + 1 < 0$

$3m^2 + 2m - 1 > 0 \implies (3m - 1)(m + 1) > 0 \implies m < -1 \;\text{or}\; m > \dfrac{1}{3}$

Combined with $m > 0$: $m > \dfrac{1}{3}$.

Worked Example: System with three inequalities

Solve the system: $x^2 - 2x - 15 \leq 0$, $|x - 1| \leq 4$, $x > 0$.

Solution

From $x^2 - 2x - 15 \leq 0$: $(x - 5)(x + 3) \leq 0 \implies -3 \leq x \leq 5$.

From $|x - 1| \leq 4$: $-4 \leq x - 1 \leq 4 \implies -3 \leq x \leq 5$.

From $x > 0$: $x \in (0, \infty)$.

Intersection: $(0, 5]$.

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DSE Exam-Style Questions

DSE Practice 1. Solve $\dfrac{2x - 1}{x + 3} \geq 1$.

Solution

$\frac{2x - 1}{x + 3} - 1 \geq 0 \implies \frac{2x - 1 - x - 3}{x + 3} \geq 0 \implies \frac{x - 4}{x + 3} \geq 0$

Critical points: $x = -3$ (excluded) and $x = 4$ (included).

Interval	$(-\infty, -3)$	$(-3, 4)$	$(4, \infty)$
Sign	$+$	$-$	$+$

Solution: $(-\infty, -3) \cup [4, \infty)$.

DSE Practice 2. Find the range of $k$ for which $kx^2 - 2kx + 3 > 0$ for all real $x$.

Solution

Case $k = 0$: $3 > 0$ for all real $x$. So $k = 0$ works.

Case $k \neq 0$: Need $k > 0$ and $\Delta < 0$:

$\Delta = 4k^2 - 12k = 4k(k - 3) < 0 \implies 0 < k < 3$

Combined with $k = 0$: the answer is $0 \leq k < 3$.

DSE Practice 3. Solve $|2x - 3| > |x + 1|$.

Solution

Square both sides (both sides non-negative):

$(2x - 3)^2 > (x + 1)^2 \implies 4x^2 - 12x + 9 > x^2 + 2x + 1$

$3x^2 - 14x + 8 > 0 \implies (3x - 2)(x - 4) > 0$

Solution: $x < \dfrac{2}{3}$ or $x > 4$.

DSE Practice 4. Find all real values of $x$ satisfying $x^2 - 2|x| - 8 < 0$.

Solution

Let $t = |x| \geq 0$: $t^2 - 2t - 8 < 0 \implies (t - 4)(t + 2) < 0 \implies -2 < t < 4$.

Since $t \geq 0$: $0 \leq t < 4$, i.e., $|x| < 4 \implies -4 < x < 4$.

DSE Practice 5. Given that $x^2 + 2(k + 1)x + 9 > 0$ for all real $x$, find the range of $k$.

Solution

$\Delta < 0$: $4(k + 1)^2 - 36 < 0 \implies (k + 1)^2 < 9 \implies -3 < k + 1 < 3 \implies -4 < k < 2$.

DSE Practice 6. Solve the inequality $\dfrac{x^2 - x - 6}{x^2 - 4} \leq 0$.

Solution

$\frac{x^2 - x - 6}{x^2 - 4} = \frac{(x - 3)(x + 2)}{(x - 2)(x + 2)} = \frac{x - 3}{x - 2}$

for $x \neq -2$.

$\dfrac{x - 3}{x - 2} \leq 0$: $2 < x \leq 3$.

But $x \neq -2$, which is not in $[2, 3]$ anyway.

Solution: $(2, 3]$.