Chapter 2 — Equations and Inequalities
Equations are the central language of algebra. In this chapter, you’ll learn to solve every major type — linear, quadratic, rational, radical, absolute value — and master inequalities. Every technique here is essential for all future mathematics.
Table of Contents
1 — The Rectangular Coordinate System and Graphs
2 — Linear Equations in One Variable
- 2.1 Solving Linear Equations
- 2.2 Rational Equations (Equations with Fractions)
- 2.3 Applications of Linear Equations
3 — Models and Applications
4 — Complex Numbers
- 4.1 The Imaginary Unit i
- 4.2 Arithmetic of Complex Numbers
- 4.3 Complex Conjugates and Division
- 4.4 Powers of i
5 — Quadratic Equations
- 5.1 Solving by Factoring
- 5.2 The Square Root Property
- 5.3 Completing the Square
- 5.4 The Quadratic Formula
- 5.5 The Discriminant
6 — Other Types of Equations
- 6.1 Rational Equations
- 6.2 Radical Equations
- 6.3 Equations in Quadratic Form
- 6.4 Absolute Value Equations
7 — Linear Inequalities and Absolute Value Inequalities
Glossary — Key Terms at a Glance
| Term | Meaning |
|---|---|
| Equation | A mathematical statement that two expressions are equal |
| Solution / Root | A value that makes the equation true |
| Identity | An equation true for all values of the variable |
| Contradiction | An equation with no solution |
| Conditional Equation | An equation true for only some values |
| Complex Number | A number of the form $a + bi$ where $i = \sqrt{-1}$ |
| Discriminant | $b^2 - 4ac$ — determines the nature of quadratic roots |
| Extraneous Solution | A “solution” that doesn’t satisfy the original equation |
| Inequality | A mathematical statement using $<, >, \leq, \geq$ |
| Interval Notation | A way to describe solution sets using parentheses and brackets |
1 — The Rectangular Coordinate System and Graphs
1.1 The Cartesian Coordinate System
The Cartesian coordinate system (named after René Descartes) uses two perpendicular number lines — the $x$-axis (horizontal) and $y$-axis (vertical) — intersecting at the origin $(0, 0)$.
Every point in the plane corresponds to an ordered pair $(x, y)$.
| Quadrant | Signs | Example |
|---|---|---|
| I | $(+, +)$ | $(3, 4)$ |
| II | $(-, +)$ | $(-2, 5)$ |
| III | $(-, -)$ | $(-3, -1)$ |
| IV | $(+, -)$ | $(4, -2)$ |
Points on the axes are not in any quadrant:
- On $x$-axis: $y = 0$ → $(a, 0)$
- On $y$-axis: $x = 0$ → $(0, b)$
1.2 Distance and Midpoint Formulas
Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$:
\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]This is derived from the Pythagorean Theorem: $a^2 + b^2 = c^2$.
Midpoint Formula: The midpoint of the segment connecting $(x_1, y_1)$ and $(x_2, y_2)$:
\[M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\]Find the distance and midpoint between $(3, -2)$ and $(-1, 4)$:
Distance:
\[d = \sqrt{(-1-3)^2 + (4-(-2))^2} = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}\]Midpoint:
\[M = \left(\frac{3 + (-1)}{2}, \frac{-2 + 4}{2}\right) = \left(\frac{2}{2}, \frac{2}{2}\right) = (1, 1)\]2 — Linear Equations in One Variable
2.1 Solving Linear Equations
A linear equation in one variable has the form:
\[ax + b = 0, \quad a \neq 0\]The solution is $x = -\frac{b}{a}$.
General strategy for solving linear equations:
- Simplify both sides (distribute, combine like terms)
- Collect all variable terms on one side
- Collect all constant terms on the other side
- Divide by the variable’s coefficient
Solve $3(x - 4) + 2 = 5x - 8$:
| Step | Work |
|---|---|
| Distribute | $3x - 12 + 2 = 5x - 8$ |
| Combine like terms (left) | $3x - 10 = 5x - 8$ |
| Subtract $3x$ from both sides | $-10 = 2x - 8$ |
| Add $8$ to both sides | $-2 = 2x$ |
| Divide by $2$ | $x = -1$ |
Check: $3(-1 - 4) + 2 = 3(-5) + 2 = -13$. And $5(-1) - 8 = -13$. ✓
Types of Linear Equations
| Type | # of Solutions | Example |
|---|---|---|
| Conditional | Exactly one | $2x + 3 = 7 \Rightarrow x = 2$ |
| Identity | Infinitely many (all reals) | $2(x+1) = 2x + 2$ |
| Contradiction | None (no solution) | $x + 1 = x + 3$ |
Identify: $3(x + 2) - x = 2x + 6$
\(3x + 6 - x = 2x + 6\) \(2x + 6 = 2x + 6\)
This is true for all $x$ → Identity (infinitely many solutions).
Identify: $2x + 5 = 2x + 8$
\[5 = 8 \quad \text{✗ — never true}\]→ Contradiction (no solution).
2.2 Rational Equations (Equations with Fractions)
To solve an equation with fractions, multiply every term by the LCD (Least Common Denominator) to clear all fractions, then solve the resulting equation.
Solve $\frac{x}{3} + \frac{x}{4} = 10$:
LCD = $12$. Multiply every term by $12$:
\[12 \cdot \frac{x}{3} + 12 \cdot \frac{x}{4} = 12 \cdot 10\] \[4x + 3x = 120\] \[7x = 120\] \[x = \frac{120}{7}\]2.3 Applications of Linear Equations
Formulas are equations relating multiple variables. You can solve for any variable using the same algebraic techniques.
Common formulas:
| Formula | Name |
|---|---|
| $P = 2l + 2w$ | Perimeter of a rectangle |
| $A = \frac{1}{2}bh$ | Area of a triangle |
| $I = Prt$ | Simple interest |
| $C = \frac{5}{9}(F - 32)$ | Celsius to Fahrenheit |
| $d = rt$ | Distance = rate × time |
Solve $P = 2l + 2w$ for $w$:
\[P - 2l = 2w\] \[w = \frac{P - 2l}{2}\]Solve $I = Prt$ for $r$:
\[r = \frac{I}{Pt}\]3 — Models and Applications
3.1 Setting Up Equations from Word Problems
Problem-Solving Strategy:
- Read the problem carefully — identify what is asked
- Assign a variable to the unknown
- Write an equation from the given information
- Solve the equation
- Check — does the answer make sense in the original context?
Problem: The sum of two consecutive odd integers is 76. Find them.
Let the first odd integer be $x$. The next consecutive odd integer is $x + 2$.
\[x + (x + 2) = 76\] \[2x + 2 = 76\] \[2x = 74\] \[x = 37\]The integers are 37 and 39. Check: $37 + 39 = 76$ ✓
3.2 Geometry, Mixture, and Rate Problems
Geometry Problem: A rectangle’s length is 5 cm more than twice its width. The perimeter is 46 cm. Find the dimensions.
Let width $= w$. Then length $= 2w + 5$.
\[P = 2l + 2w\] \[46 = 2(2w + 5) + 2w\] \[46 = 4w + 10 + 2w\] \[46 = 6w + 10\] \[36 = 6w\] \[w = 6 \text{ cm}, \quad l = 2(6) + 5 = 17 \text{ cm}\]Rate Problem: A train travels at $60$ mph and another at $80$ mph. They leave the same station in opposite directions. How long until they are $350$ miles apart?
\[d_1 + d_2 = 350\] \[60t + 80t = 350\] \[140t = 350\] \[t = 2.5 \text{ hours}\]4 — Complex Numbers
4.1 The Imaginary Unit $i$
The Imaginary Unit:
\[i = \sqrt{-1}, \quad i^2 = -1\]This extends the real numbers so that every polynomial equation has a solution.
Square root of a negative number:
\[\sqrt{-a} = i\sqrt{a} \quad \text{for } a > 0\]⚠️ Critical Warning: Always extract $i$ before doing any operations!
\[\sqrt{-4} \cdot \sqrt{-9} \neq \sqrt{(-4)(-9)} = \sqrt{36} = 6 \quad \text{✗ WRONG!}\]Correct: $\sqrt{-4} \cdot \sqrt{-9} = (2i)(3i) = 6i^2 = 6(-1) = -6$
4.2 Arithmetic of Complex Numbers
A complex number has the form $a + bi$ where:
- $a$ is the real part
- $b$ is the imaginary part
- Both $a$ and $b$ are real numbers
Addition/Subtraction: Combine like parts.
\[(a + bi) + (c + di) = (a+c) + (b+d)i\] \[(a + bi) - (c + di) = (a-c) + (b-d)i\]Multiplication: Use FOIL and substitute $i^2 = -1$.
\[(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i\]Addition: $(3 + 4i) + (2 - 7i) = 5 - 3i$
Subtraction: $(5 + 2i) - (1 + 6i) = 4 - 4i$
Multiplication:
\((3 + 2i)(4 - 5i)\) \(= 12 - 15i + 8i - 10i^2\) \(= 12 - 7i - 10(-1)\) \(= 12 - 7i + 10 = 22 - 7i\)
4.3 Complex Conjugates and Division
Complex Conjugate: The conjugate of $a + bi$ is $a - bi$.
\[\overline{a + bi} = a - bi\]Key Property:
\[(a + bi)(a - bi) = a^2 + b^2 \quad \text{(always a real number!)}\]Division: Multiply numerator and denominator by the conjugate of the denominator.
\[\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(a+bi)(c-di)}{c^2 + d^2}\]Divide: $\frac{3 + 2i}{4 - i}$
\[= \frac{(3+2i)(4+i)}{(4-i)(4+i)} = \frac{12 + 3i + 8i + 2i^2}{16 + 1} = \frac{12 + 11i - 2}{17} = \frac{10 + 11i}{17} = \frac{10}{17} + \frac{11}{17}i\]4.4 Powers of $i$
The powers of $i$ cycle with period 4:
| Power | Value |
|---|---|
| $i^1$ | $i$ |
| $i^2$ | $-1$ |
| $i^3$ | $-i$ |
| $i^4$ | $1$ |
| $i^5$ | $i$ (cycle repeats) |
To evaluate $i^n$: Divide $n$ by $4$ and use the remainder.
\[i^n = i^{n \bmod 4}\]5 — Quadratic Equations
A quadratic equation has the standard form $ax^2 + bx + c = 0$ where $a \neq 0$. There are four methods to solve them.
5.1 Solving by Factoring
Zero Product Property: If $AB = 0$, then $A = 0$ or $B = 0$.
Steps:
- Write the equation in standard form ($= 0$)
- Factor the left side
- Set each factor equal to zero
- Solve each equation
Solve $x^2 - 5x + 6 = 0$:
\[= (x - 2)(x - 3) = 0\] \[x - 2 = 0 \Rightarrow x = 2 \quad \text{or} \quad x - 3 = 0 \Rightarrow x = 3\]Solutions: $x = 2$ and $x = 3$
Solve $3x^2 + 14x - 5 = 0$:
Using AC method: $a \cdot c = 3(-5) = -15$. Numbers: $15$ and $-1$ (multiply to $-15$, add to $14$).
\[3x^2 + 15x - x - 5 = 0\] \[3x(x + 5) - 1(x + 5) = 0\] \[(3x - 1)(x + 5) = 0\] \[x = \frac{1}{3} \quad \text{or} \quad x = -5\]5.2 The Square Root Property
Square Root Property: If $x^2 = k$, then:
\[x = \pm\sqrt{k}\]More generally, if $(ax + b)^2 = k$, then $ax + b = \pm\sqrt{k}$.
Solve $(2x - 3)^2 = 25$:
\[2x - 3 = \pm 5\] \[2x - 3 = 5 \Rightarrow 2x = 8 \Rightarrow x = 4\] \[2x - 3 = -5 \Rightarrow 2x = -2 \Rightarrow x = -1\]Solutions: $x = 4$ and $x = -1$
5.3 Completing the Square
Completing the Square transforms $ax^2 + bx + c = 0$ into the form $(x + h)^2 = k$.
Steps (when $a = 1$):
- Move the constant to the right side: $x^2 + bx = -c$
- Take half of $b$, square it: $\left(\frac{b}{2}\right)^2$
- Add this value to both sides
- Factor the left side as a perfect square
- Take the square root of both sides
Solve $x^2 + 6x + 2 = 0$ by completing the square:
| Step | Work |
|---|---|
| Move constant | $x^2 + 6x = -2$ |
| Half of $6 = 3$; square it: $9$ | Add $9$ to both sides |
| $x^2 + 6x + 9 = -2 + 9$ | |
| Factor left side | $(x + 3)^2 = 7$ |
| Square root both sides | $x + 3 = \pm\sqrt{7}$ |
| Solve | $x = -3 \pm \sqrt{7}$ |
When $a \neq 1$: Solve $2x^2 - 8x + 3 = 0$:
| Step | Work |
|---|---|
| Divide by $a = 2$ | $x^2 - 4x + \frac{3}{2} = 0$ |
| Move constant | $x^2 - 4x = -\frac{3}{2}$ |
| Half of $-4 = -2$; square: $4$ | $x^2 - 4x + 4 = -\frac{3}{2} + 4$ |
| Factor | $(x - 2)^2 = \frac{5}{2}$ |
| Solve | $x = 2 \pm \sqrt{\frac{5}{2}} = 2 \pm \frac{\sqrt{10}}{2}$ |
5.4 The Quadratic Formula
For $ax^2 + bx + c = 0$ (with $a \neq 0$), the solutions are:
\[\boxed{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\]This formula always works — it’s derived by completing the square on the general quadratic.
Solve $2x^2 + 5x - 3 = 0$:
$a = 2, \; b = 5, \; c = -3$
\[x = \frac{-5 \pm \sqrt{25 - 4(2)(-3)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}\] \[x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3\]5.5 The Discriminant
The discriminant is $\Delta = b^2 - 4ac$ (the expression under the square root in the quadratic formula). It determines the nature of the solutions:
| Discriminant | # of Solutions | Type | ||
|---|---|---|---|---|
| $\Delta > 0$ | 2 distinct real solutions | May be rational (if $\Delta$ is a perfect square) or irrational | ||
| $\Delta = 0$ | 1 repeated real solution | $x = \frac{-b}{2a}$ (a double root) | ||
| $\Delta < 0$ | 2 complex conjugate solutions | $x = \frac{-b \pm i\sqrt{ | \Delta | }}{2a}$ |
Determine the nature of solutions without solving:
| Equation | $\Delta = b^2 - 4ac$ | Result |
|---|---|---|
| $x^2 - 4x + 4 = 0$ | $16 - 16 = 0$ | 1 repeated real solution |
| $x^2 + 3x + 1 = 0$ | $9 - 4 = 5 > 0$ | 2 distinct real (irrational) solutions |
| $x^2 + x + 1 = 0$ | $1 - 4 = -3 < 0$ | 2 complex conjugate solutions |
| $2x^2 - 7x + 3 = 0$ | $49 - 24 = 25 > 0$ | 2 distinct real (rational) solutions |
Which method to use?
| Situation | Best Method |
|---|---|
| Equation is easily factorable | Factoring |
| No $x$ term (e.g., $x^2 = k$ or $(x+a)^2 = k$) | Square Root Property |
| Leading coefficient is 1, $b$ is even | Completing the Square |
| General case / messy coefficients | Quadratic Formula |
6 — Other Types of Equations
6.1 Rational Equations
A rational equation contains at least one rational expression (fraction with a variable in the denominator).
Strategy:
- Find the LCD
- Multiply every term by the LCD to clear fractions
- Solve the resulting polynomial equation
- Check for extraneous solutions — any value that makes a denominator zero must be rejected
Solve $\frac{2}{x} + \frac{3}{x-2} = \frac{7}{x(x-2)}$:
LCD = $x(x-2)$. Restrictions: $x \neq 0, \; x \neq 2$
Multiply through by $x(x-2)$:
\[2(x-2) + 3x = 7\] \[2x - 4 + 3x = 7\] \[5x = 11\] \[x = \frac{11}{5}\]Check: $x = \frac{11}{5} \neq 0$ and $\neq 2$, so the solution is valid. ✓
⚠️ Extraneous Solutions in Rational Equations:
Solve $\frac{x}{x-3} + \frac{2}{x-3} = \frac{3x}{x^2-9}$:
Note: $x^2 - 9 = (x-3)(x+3)$. LCD = $(x-3)(x+3)$.
Restrictions: $x \neq 3, \; x \neq -3$
\[x(x+3) + 2(x+3) = 3x\] \[x^2 + 3x + 2x + 6 = 3x\] \[x^2 + 5x + 6 = 3x\] \[x^2 + 2x + 6 = 0 \quad \text{Hmm, let me redo...}\]Actually: $(x+2)(x+3) = 3x$
\[x^2 + 5x + 6 = 3x\] \[x^2 + 2x + 6 = 0\]$\Delta = 4 - 24 = -20 < 0$ → No real solutions!
6.2 Radical Equations
A radical equation has a variable inside a radical. To solve:
- Isolate the radical on one side
- Raise both sides to the appropriate power to eliminate the radical
- Solve the resulting equation
- Check all solutions — squaring can introduce extraneous solutions
Solve $\sqrt{2x + 3} = x$:
Square both sides:
\[2x + 3 = x^2\] \[x^2 - 2x - 3 = 0\] \[(x - 3)(x + 1) = 0\] \[x = 3 \quad \text{or} \quad x = -1\]Check $x = 3$: $\sqrt{6 + 3} = \sqrt{9} = 3$ ✓
Check $x = -1$: $\sqrt{-2 + 3} = \sqrt{1} = 1 \neq -1$ ✗ (extraneous!)
Solution: $x = 3$ only.
Solve $\sqrt{x + 4} - \sqrt{x - 1} = 1$ (two radicals):
Isolate one radical: $\sqrt{x+4} = 1 + \sqrt{x-1}$
Square both sides:
\[x + 4 = 1 + 2\sqrt{x-1} + (x-1)\] \[x + 4 = x + 2\sqrt{x-1}\] \[4 = 2\sqrt{x-1}\] \[2 = \sqrt{x-1}\] \[4 = x - 1\] \[x = 5\]Check: $\sqrt{9} - \sqrt{4} = 3 - 2 = 1$ ✓
6.3 Equations in Quadratic Form
An equation is in quadratic form if it can be written as:
\[a[f(x)]^2 + b[f(x)] + c = 0\]Strategy: Substitute $u = f(x)$, solve the quadratic in $u$, then substitute back.
Solve $x^4 - 5x^2 + 4 = 0$:
Let $u = x^2$:
\[u^2 - 5u + 4 = 0\] \[(u - 1)(u - 4) = 0\] \[u = 1 \quad \text{or} \quad u = 4\]Back-substitute:
\[x^2 = 1 \Rightarrow x = \pm 1 \quad \text{or} \quad x^2 = 4 \Rightarrow x = \pm 2\]Solutions: $x = -2, -1, 1, 2$
Solve $x^{2/3} - 5x^{1/3} + 6 = 0$:
Let $u = x^{1/3}$:
\[u^2 - 5u + 6 = 0\] \[(u - 2)(u - 3) = 0\] \[u = 2 \Rightarrow x^{1/3} = 2 \Rightarrow x = 8\] \[u = 3 \Rightarrow x^{1/3} = 3 \Rightarrow x = 27\]6.4 Absolute Value Equations
| For $ | A | = B$ (where $B \geq 0$): |
If $B < 0$: no solution (absolute value is never negative).
| For $ | A | = | B | $: |
| **Solve $ | 2x - 5 | = 7$:** |
Solutions: $x = 6$ and $x = -1$
| **Solve $ | 3x + 1 | = -4$:** |
No solution — absolute value cannot equal a negative number.
7 — Linear Inequalities and Absolute Value Inequalities
7.1 Interval Notation
Interval notation describes sets of real numbers:
| Inequality | Interval | Type | Number Line |
|---|---|---|---|
| $a \leq x \leq b$ | $[a, b]$ | Closed | Solid dots at both ends |
| $a < x < b$ | $(a, b)$ | Open | Open circles at both ends |
| $a \leq x < b$ | $[a, b)$ | Half-open | Solid at $a$, open at $b$ |
| $x \geq a$ | $[a, \infty)$ | Closed-infinite | Solid at $a$, arrow right |
| $x < a$ | $(-\infty, a)$ | Open-infinite | Arrow left, open at $a$ |
| All reals | $(-\infty, \infty)$ | Entire line | Both arrows |
Key: Use $[$ or $]$ for $\leq / \geq$ (included). Use $($ or $)$ for $< / >$ (excluded). Always use $($ with $\pm\infty$.
7.2 Solving Linear Inequalities
Solve linear inequalities exactly like equations with one critical difference:
When you multiply or divide by a NEGATIVE number, REVERSE the inequality sign!
\[-2x > 6 \implies x < -3 \quad \text{(sign flipped!)}\]Solve $3x - 7 \leq 5x + 1$:
\[3x - 5x \leq 1 + 7\] \[-2x \leq 8\] \[x \geq -4 \quad \text{(divided by } {-2}\text{, flipped sign)}\]Solution: $[-4, \infty)$
7.3 Compound Inequalities
“And” compound inequality ($a < x < b$): Solve each part simultaneously. The solution is the intersection of the individual solutions.
“Or” compound inequality ($x < a$ or $x > b$): The solution is the union of the individual solutions.
Solve $-3 < 2x + 1 \leq 7$:
Subtract $1$ from all parts:
\[-4 < 2x \leq 6\]Divide by $2$:
\[-2 < x \leq 3\]Solution: $(-2, 3]$
Solve $3x + 2 < -4$ or $2x - 1 > 5$:
\[3x < -6 \Rightarrow x < -2\] \[2x > 6 \Rightarrow x > 3\]Solution: $(-\infty, -2) \cup (3, \infty)$
7.4 Absolute Value Inequalities
| Inequality | Equivalent | Solution Type |
|---|---|---|
| $\lvert A \rvert < B$ | $-B < A < B$ | “And” — intersection (interval between) |
| $\lvert A \rvert > B$ | $A < -B$ or $A > B$ | “Or” — union (outside the interval) |
| $\lvert A \rvert \leq B$ | $-B \leq A \leq B$ | Closed version of above |
| $\lvert A \rvert \geq B$ | $A \leq -B$ or $A \geq B$ | Closed version of above |
Memory Aid: “Less than” → “and” (between). “Greater than” → “or” (outside).
| **Solve $ | 2x - 3 | < 7$:** |
Solution: $(-2, 5)$
| **Solve $ | 4x + 1 | \geq 9$:** |
Solution: $\left(-\infty, -\frac{5}{2}\right] \cup [2, \infty)$
Special cases:
- $\lvert x \rvert < 0$: No solution (absolute value is never negative)
- $\lvert x \rvert > 0$: All reals except $x = 0$ → $(-\infty, 0) \cup (0, \infty)$
- $\lvert x \rvert \geq 0$: All real numbers (always true)
- $\lvert x \rvert \leq 0$: Only $x = 0$
Key Takeaways
- The distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ comes from the Pythagorean theorem
- Linear equations always have one solution (conditional), infinite solutions (identity), or no solution (contradiction)
- Complex numbers extend the reals with $i = \sqrt{-1}$; powers of $i$ cycle every 4
- There are four methods for solving quadratics: factoring, square root property, completing the square, quadratic formula
- The discriminant $\Delta = b^2 - 4ac$ reveals the nature of solutions without solving
- Always check for extraneous solutions when solving rational and radical equations
- When solving inequalities: flip the sign when multiplying/dividing by a negative
- Absolute value inequalities: less than → “and” (between); greater than → “or” (outside)
Practice Questions
Q1. Solve: $5(x - 3) + 2 = 3(x + 1) - 4$
Answer: $5x - 15 + 2 = 3x + 3 - 4 \Rightarrow 5x - 13 = 3x - 1 \Rightarrow 2x = 12 \Rightarrow x = 6$
Q2. Write $(3 - 2i)(4 + 5i)$ in standard form.
Answer: $12 + 15i - 8i - 10i^2 = 12 + 7i + 10 = 22 + 7i$
Q3. Solve $x^2 - 6x + 13 = 0$ using the quadratic formula.
Answer: $\Delta = 36 - 52 = -16$. $x = \frac{6 \pm \sqrt{-16}}{2} = \frac{6 \pm 4i}{2} = 3 \pm 2i$
Q4. Solve $\sqrt{3x + 1} = x - 1$.
Answer: Square: $3x + 1 = x^2 - 2x + 1 \Rightarrow x^2 - 5x = 0 \Rightarrow x(x-5) = 0$. $x = 0$: $\sqrt{1} = -1$ ✗. $x = 5$: $\sqrt{16} = 4 = 5 - 1$ ✓. Solution: $x = 5$
| Q5. Solve $ | 3x - 2 | > 10$ and express in interval notation. |
Answer: $3x - 2 < -10$ or $3x - 2 > 10$. So $x < -\frac{8}{3}$ or $x > 4$. Solution: $\left(-\infty, -\frac{8}{3}\right) \cup (4, \infty)$
Q6. Find the distance between $(-3, 7)$ and $(5, -1)$.
Answer: $d = \sqrt{(5-(-3))^2 + (-1-7)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2}$
Q7. Solve: $x^4 - 13x^2 + 36 = 0$
Answer: Let $u = x^2$: $u^2 - 13u + 36 = 0 \Rightarrow (u-4)(u-9) = 0$. $x^2 = 4 \Rightarrow x = \pm 2$. $x^2 = 9 \Rightarrow x = \pm 3$. Solutions: $x = -3, -2, 2, 3$