Comprehensive Guide to Equations

1. Linear Equations

A linear equation is an equation where each term is either a constant or the product of a constant and a single variable raised to the power of 1.

General form: ax + b = 0, where a ≠ 0

Example 1: Solve 2x + 5 = 13

Subtract 5 from both sides: 2x + 5 - 5 = 13 - 5

Simplify: 2x = 8

Divide both sides by 2: x = 4

Solution: x = 4

Methods to Solve Linear Equations:

Isolation Method: Isolate the variable by performing inverse operations.
Substitution Method: If there are multiple equations, solve one for a variable and substitute.
Elimination Method: Add or subtract equations to eliminate a variable.
Graphical Method: Plot the equation and find the x-intercept.

Example 2: System of Linear Equations

Solve the system:

3x + 2y = 7

x - y = 1

From the second equation: x = y + 1

Substitute into the first equation: 3(y + 1) + 2y = 7

Expand: 3y + 3 + 2y = 7

Simplify: 5y + 3 = 7

Subtract 3 from both sides: 5y = 4

Divide by 5: y = 4/5

Substitute back: x = 4/5 + 1 = 9/5

Solution: x = 9/5, y = 4/5

2. Quadratic Equations

A quadratic equation is a polynomial equation of the second degree.

Standard form: ax² + bx + c = 0, where a ≠ 0

Methods to Solve Quadratic Equations:

Example 1: Factoring Method

Solve x² - 5x + 6 = 0

Find factors of 6 that add up to -5: -2 and -3

Rewrite as: (x - 2)(x - 3) = 0

Apply zero-product property: x - 2 = 0 or x - 3 = 0

Solve: x = 2 or x = 3

Solutions: x = 2 or x = 3

Example 2: Quadratic Formula

Solve 2x² - 7x + 3 = 0

Using the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a)

Identify a = 2, b = -7, c = 3

Calculate discriminant: b² - 4ac = (-7)² - 4(2)(3) = 49 - 24 = 25

Apply formula: x = (7 ± √25) / (2×2) = (7 ± 5) / 4

Calculate solutions: x = (7 + 5) / 4 = 12/4 = 3 or x = (7 - 5) / 4 = 2/4 = 0.5

Solutions: x = 3 or x = 0.5

Example 3: Completing the Square

Solve x² - 6x + 8 = 0

Rearrange: x² - 6x = -8

Take half the coefficient of x and square it: (-6/2)² = (-3)² = 9

Add 9 to both sides: x² - 6x + 9 = -8 + 9

Left side is now a perfect square: (x - 3)² = 1

Take square root of both sides: x - 3 = ±1

Solve: x = 3 ± 1

Solutions: x = 4 or x = 2

Nature of Roots Based on Discriminant:

If b² - 4ac > 0: Two distinct real roots
If b² - 4ac = 0: One real root (repeated)
If b² - 4ac < 0: Two complex conjugate roots

3. Polynomial Equations

Polynomial equations have the form: a₀xⁿ + a₁xⁿ⁻¹ + ... + aₙ₋₁x + aₙ = 0

Methods to Solve Polynomial Equations:

Factoring: When possible, factor the polynomial and set each factor to zero.
Synthetic Division: Find roots systematically using synthetic division.
Rational Root Theorem: Identify potential rational roots.
Descartes' Rule of Signs: Determine possible number of positive and negative roots.
Numerical Methods: For higher degree polynomials (Newton-Raphson, etc.)

Example: Solve x³ - 6x² + 11x - 6 = 0

Try a simple root (like x = 1): 1³ - 6×1² + 11×1 - 6 = 1 - 6 + 11 - 6 = 0

x = 1 is a root, so (x - 1) is a factor

Use synthetic division to find the quotient:

1	-6	11	-6
	1	-5	6
1	-5	6	0

The quotient is x² - 5x + 6

Factor this quadratic: x² - 5x + 6 = (x - 2)(x - 3)

Therefore, x³ - 6x² + 11x - 6 = (x - 1)(x - 2)(x - 3)

Solutions: x = 1, x = 2, and x = 3

4. Rational Equations

Rational equations contain fractions with variables in the denominators.

Example: Solve 1/x + 1/(x-1) = 1

Find the LCD: x(x-1)

Multiply all terms by the LCD: x(x-1) × [1/x + 1/(x-1) = 1]

Distribute: (x-1) + x = x(x-1)

Simplify: 2x - 1 = x² - x

Rearrange in standard form: x² - 3x + 1 = 0

Use the quadratic formula: x = (3 ± √(9-4)) / 2 = (3 ± √5) / 2

Solutions: x = (3 + √5)/2 or x = (3 - √5)/2

Important: Always check for extraneous solutions that make denominators zero.

5. Radical Equations

Radical equations contain variables under radical signs.

Example: Solve √(2x + 3) - x = 1

Isolate the radical: √(2x + 3) = x + 1

Square both sides: 2x + 3 = (x + 1)²

Expand: 2x + 3 = x² + 2x + 1

Rearrange: 0 = x² + 0x - 2

Factorize: 0 = (x + √2)(x - √2)

Solve: x = -√2 or x = √2

Check solutions: For x = √2, √(2√2 + 3) - √2 = √(2√2 + 3) - √2 ≈ 3 - 1.414 ≈ 1.586 ≠ 1

Check x = -√2: √(2(-√2) + 3) - (-√2) = √(3 - 2√2) + √2

Since 3 - 2√2 > 0 (as √2 ≈ 1.414), we have √(3 - 2√2) + √2 ≈ 1 + 1.414 ≈ 2.414 ≠ 1

The equation has no solutions (both potential solutions are extraneous).

Important: Always verify solutions by substituting back into the original equation. Squaring both sides can introduce extraneous solutions.

6. Exponential Equations

Exponential equations have the variable in the exponent.

Example 1: Solve 2ˣ = 8

Rewrite 8 as a power of 2: 2ˣ = 2³

Since the bases are the same, the exponents must be equal: x = 3

Solution: x = 3

Example 2: Solve 3ˣ = 5

Take the natural log of both sides: ln(3ˣ) = ln(5)

Use properties of logarithms: x·ln(3) = ln(5)

Solve for x: x = ln(5)/ln(3) ≈ 1.465

Solution: x ≈ 1.465

7. Logarithmic Equations

Logarithmic equations contain logarithms with variables.

Example: Solve log₂(x + 3) - log₂(x - 1) = 2

Use logarithm property to combine: log₂((x + 3)/(x - 1)) = 2

Convert to exponential form: (x + 3)/(x - 1) = 2²

Simplify: (x + 3)/(x - 1) = 4

Cross multiply: x + 3 = 4(x - 1)

Expand: x + 3 = 4x - 4

Rearrange: -3x = -7

Solve: x = 7/3

Check domain constraints: x - 1 > 0 implies x > 1, so x = 7/3 ≈ 2.33 is valid

Solution: x = 7/3

Important: Always check domain constraints for logarithmic expressions.

8. Trigonometric Equations

Trigonometric equations involve trigonometric functions.

Example 1: Solve sin(x) = 0.5

Identify: sin(x) = 0.5 corresponds to sin(x) = sin(π/6)

Find solutions in [0, 2π): x = π/6 or x = 5π/6

General solution: x = π/6 + 2nπ or x = 5π/6 + 2nπ, where n is an integer

Solutions: x = π/6 + 2nπ or x = 5π/6 + 2nπ, where n is an integer

Example 2: Solve 2cos²(x) - 1 = 0

Rearrange: 2cos²(x) = 1

Solve: cos²(x) = 1/2

Take square root: cos(x) = ±1/√2 = ±√2/2

Find solutions in [0, 2π): For cos(x) = √2/2: x = π/4 or x = 7π/4 For cos(x) = -√2/2: x = 3π/4 or x = 5π/4

General solution: x = π/4 + 2nπ or x = 7π/4 + 2nπ or x = 3π/4 + 2nπ or x = 5π/4 + 2nπ, where n is an integer

Solutions: x = π/4 + nπ, where n is an integer

9. Differential Equations

Differential equations involve derivatives of functions.

Example: Solve dy/dx = 2x with y(0) = 1

Integrate both sides: ∫ dy = ∫ 2x dx

Apply the antiderivative: y = x² + C

Use initial condition y(0) = 1 to find C: 1 = 0² + C, so C = 1

The solution is y = x² + 1

Solution: y = x² + 1