Linear and Quadratic Equations
Equations are mathematical statements that assert the equality of two expressions. Solving equations means finding the value(s) of the unknown variable(s) that make the equation true. Linear equations involve variables raised only to the power of 1, while quadratic equations involve variables raised to the power of 2. Both types are extensively tested in Post-UTME.
Linear Equations in One Variable
A linear equation has the form ax + b = 0, where a ≠ 0.
Solution Method
Use inverse operations to isolate the variable.
Example 1: Solve 3x + 7 = 22 → 3x = 15 → x = 5
Example 2: Solve 4(2x − 3) = 3(x + 5) + 2 8x − 12 = 3x + 15 + 2 → 8x − 12 = 3x + 17 → 5x = 29 → x = 29/5 = 5.8
Equations with Fractions
Clear fractions by multiplying through by the LCD.
Example: (x/2) + (x/3) = 10 → LCD = 6 → 3x + 2x = 60 → 5x = 60 → x = 12
Example: (x+1)/4 − (x−2)/3 = 1 → LCD = 12 → 3(x+1) − 4(x−2) = 12 → 3x+3−4x+8 = 12 → −x+11 = 12 → x = −1
Quadratic Equations
A quadratic equation has the general form ax² + bx + c = 0 (a ≠ 0). It has at most two solutions.
Method 1: Factorization
Example: x² − 5x + 6 = 0 → factors of 6 that add to 5: 2 and 3 (x − 2)(x − 3) = 0 → x = 2 or x = 3
Example: x² + 2x − 15 = 0 → factors of −15 that add to 2: 5 and −3 (x + 5)(x − 3) = 0 → x = −5 or x = 3
Method 2: The Quadratic Formula
For ax² + bx + c = 0, the solutions are: x = (−b ± √(b² − 4ac)) / 2a
The expression b² − 4ac is called the discriminant (Δ):
| Δ Value | Nature of Roots |
|---|---|
| Δ > 0 (perfect square) | Two distinct rational roots |
| Δ > 0 (not perfect square) | Two distinct irrational roots |
| Δ = 0 | Two equal roots |
| Δ < 0 | No real roots |
Example: Solve 2x² − 5x − 3 = 0 a=2, b=−5, c=−3; Δ = 25 + 24 = 49 (positive, perfect square!) x = (5 ± 7)/4 → x₁ = 3, x₂ = −0.5
Method 3: Completing the Square
Example: x² + 6x + 5 = 0 x² + 6x = −5; Add (6/2)² = 9: x² + 6x + 9 = 4 (x+3)² = 4 → x+3 = ±2 → x = −1 or x = −5
Simultaneous Linear Equations
Two equations in two unknowns (x and y).
Method 1: Substitution
Solve one equation for one variable, substitute into the other.
Example: x + y = 7 …(1) 2x − y = 5 …(2) From (1): y = 7 − x; Substitute into (2): 2x − (7 − x) = 5 → 3x = 12 → x = 4; y = 3
Method 2: Elimination
Add or subtract equations to eliminate one variable. From (1)+(2): 3x = 12 → x = 4; Substituting: y = 3
Word Problems Leading to Equations
Example: The difference between two numbers is 8, and one number is 3 times the other. Let smaller = x; Larger = 3x; Difference: 3x − x = 8 → 2x = 8 → x = 4; Larger = 12
Example: A rectangle’s perimeter is 60 cm. Length is 5 cm more than width. Let width = w; Length = w+5; Perimeter: 2(2w+5) = 60 → 4w+10 = 60 → w = 12.5 cm; Length = 17.5 cm; Area = 218.75 cm²
Key Post-UTME Exam Facts
- Linear equations: Isolate variable; clear fractions by multiplying by LCD
- Quadratic formula: x = (−b ± √(b²−4ac))/2a — works for ALL quadratics
- Discriminant: Δ = b²−4ac → positive = 2 roots, zero = 1 root, negative = no real roots
- Simultaneous: Substitution or elimination — choose whichever is faster
- ⚡ Exam tip: When a quadratic doesn’t factorize, go straight to the quadratic formula
🟢 Lite — Quick Review (1h–1d)
Rapid summary for last-minute revision before your exam.
- Linear equation: Isolate variable; 3x+7=22 → x=5
- Quadratic: Factor or use x = (−b ± √(b²−4ac))/2a
- Discriminant: Δ = b²−4ac → positive=2 roots, zero=1, negative=none
- Simultaneous: Substitution or elimination
- Equations with fractions: Multiply by LCD to clear denominators
- ⚡ Exam tip: Always verify by substituting your answer back into the original equation
🟡 Standard — Regular Study (2d–2mo)
Standard content for students with a few days to months.
Harder Quadratic Word Problems
Example: Area of rectangle is 60 cm². Length is 4 cm more than width. w(w+4) = 60 → w²+4w−60 = 0 → (w+10)(w−6) = 0 → w = 6 cm, Length = 10 cm
Quadratic Inequalities
Example: Solve x² − x − 12 > 0 Factor: (x−4)(x+3) > 0 The quadratic opens upward. So > 0 when x < −3 OR x > 4. Solution: x ∈ (−∞, −3) ∪ (4, ∞)
🔴 Extended — Deep Study (3mo+)
Comprehensive coverage for students on a longer study timeline.
Quadratic Functions and Graphs
The graph of y = ax² + bx + c is a parabola. Vertex (turning point): x-coordinate = −b/(2a) Axis of symmetry: x = −b/(2a)
Example: y = x² − 6x + 5 Vertex x = −(−6)/(2×1) = 3; y = 9 − 18 + 5 = −4 → Vertex = (3, −4) Opens upward (a=1>0) → minimum point
Sum and Product of Roots
For ax² + bx + c = 0 with roots α and β:
- Sum of roots (α + β) = −b/a
- Product of roots (αβ) = c/a
Example: If roots of 2x² + kx − 6 = 0 are α and β, and α + 2β = 3, find k. α + β = −k/2; αβ = −3 From α + 2β = 3 and α + β = −k/2: subtract → β = 3 + k/2 α = −3 − k (from α = −k/2 − β = −3 − k) Product: (−3−k)(3 + k/2) = −3 → k² + 9k + 12 = 0 → (k+3)(k+4) = 0 → k = −3 or k = −4
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