5.2 Equations, Inequalities & Quadratics
Key Takeaways
- Linear ax + b = 0 has unique root x = −b/a (a ≠ 0); systems are solved by substitution, elimination, or matrices later.
- Quadratic ax² + bx + c = 0: discriminant D = b² − 4ac decides two real, one real, or complex conjugate roots.
- Sum of roots = −b/a and product = c/a (Vieta)—faster than solving when MCQs ask symmetric data.
- Inequality direction flips when multiplying/dividing by a negative; critical points split the number line for sign charts.
- Complete the square or formula under time pressure; factoring wins only when integers are obvious.
Why Equations Dominate the Maths Bank
Algebraic equations and inequalities are among the highest-frequency FSc topics in engineering entry maths. On a commonly reported PAF CAE-style paper (~50 maths MCQs / ~25 minutes—hedge: verify your induction notice), expect linear rearrangements, simultaneous equations, quadratics, and inequality intervals mixed with word problems (rates, mixtures, ages).
Linear Equations and Systems
One variable: ax + b = 0 ⇒ x = −b/a if a ≠ 0.
Worked. 5(2x − 3) = 3x + 4 → 10x − 15 = 3x + 4 → 7x = 19 → x = 19/7.
Two equations (elimination). Solve:
2x + 3y = 13
3x − y = 3
Multiply the second by 3: 9x − 3y = 9. Add to the first: 11x = 22 ⇒ x = 2. Then 3(2) − y = 3 ⇒ 6 − y = 3 ⇒ y = 3. Check: 2(2) + 3(3) = 4 + 9 = 13.
Substitution tip. If one equation is already y = …, substitute immediately—often faster than elimination under MCQ timing.
Absolute Value Equations
|x − a| = b (b ≥ 0) ⇒ x − a = b or x − a = −b. If b < 0, no solution.
Worked. |2x − 5| = 7 ⇒ 2x − 5 = 7 or 2x − 5 = −7 ⇒ x = 6 or x = −1.
Quadratic Equations
Standard form: ax² + bx + c = 0, a ≠ 0.
| Method | When to use | Notes |
|---|---|---|
| Factoring | Integer roots obvious | (px + q)(rx + s) = 0 |
| Quadratic formula | Always works | x = (−b ± √D)/(2a) |
| Completing square | Vertex / perfect-square prompts | x² + bx = (x + b/2)² − (b/2)² |
| Vieta only | Sum/product asked | α + β = −b/a, αβ = c/a |
Discriminant D = b² − 4ac:
| D | Roots |
|---|---|
| D > 0 | Two distinct real roots |
| D = 0 | One real repeated root |
| D < 0 | Complex conjugate roots (no distinct real roots) |
Worked formula. Solve 2x² − 5x − 3 = 0.
D = 25 − 4(2)(−3) = 25 + 24 = 49.
x = [5 ± 7]/4.
x = 12/4 = 3 or x = −2/4 = −1/2.
Factor check: (2x + 1)(x − 3) = 2x² − 6x + x − 3 = 2x² − 5x − 3.
Worked Vieta (no solving). For x² − 7x + 10 = 0, sum of roots = 7, product = 10. Roots are 2 and 5. If an MCQ asks “sum of squares of roots,” use α² + β² = (α + β)² − 2αβ = 49 − 20 = 29 without finding each root if you already trust the sum/product.
Completing the square (vertex form). x² + 6x + 5 = 0 → (x + 3)² − 9 + 5 = 0 → (x + 3)² = 4 → x + 3 = ±2 → x = −1 or x = −5. Vertex form y = a(x − h)² + k is useful when the stem asks minimum/maximum of a quadratic function.
Nature of Roots Without Full Solving
For ax² + bx + c with a > 0:
- D > 0 and perfect square → distinct rational roots (if a, b, c rational).
- D > 0 not square → distinct irrational roots.
- Always state assumptions: coefficients rational as in FSc stems.
Worked nature. 3x² − 2x + 5 = 0: D = 4 − 60 = −56 < 0 → complex roots. No real x satisfies it.
Linear Inequalities
Solve like equations, but reverse the inequality when multiplying or dividing by a negative number.
Worked. −3x + 6 > 12 → −3x > 6 → divide by −3 and flip: x < −2.
Interval notation: (−∞, −2).
Quadratic Inequalities (Sign Chart)
Solve ax² + bx + c ≥ 0 (or >, <, ≤) by:
- Find roots (critical points).
- Plot on a number line.
- Test a point in each interval (or use parabola opening direction).
Worked. Solve x² − 5x + 6 ≤ 0.
Factor: (x − 2)(x − 3) ≤ 0. Roots 2 and 3. The product is ≤ 0 between the roots (including endpoints for ≤). Solution: [2, 3].
Worked strict. x² − 5x + 6 > 0 ⇒ (−∞, 2) ∪ (3, ∞).
Rational Inequalities (Critical Extra)
For (x − 1)/(x + 2) ≥ 0, critical points are x = 1 (numerator zero) and x = −2 (undefined—exclude). Sign chart on (−∞, −2), (−2, 1), (1, ∞). Solution typically (−∞, −2) ∪ [1, ∞)—always exclude where denominator is zero.
Word-Problem Pattern (Engineering Flavour)
A jet covers 600 km with the wind in 2 hours and returns against the wind in 3 hours. Find wind speed and still-air speed.
Let u = still-air speed, v = wind. Then (u + v)·2 = 600 ⇒ u + v = 300. And (u − v)·3 = 600 ⇒ u − v = 200. Add: 2u = 500 ⇒ u = 250 km/h, v = 50 km/h.
On the Exam
Prioritize: discriminant sign, Vieta sum/product, inequality flip errors, and interval endpoints (open vs closed). If factoring stalls after 15 seconds, jump to the formula—speed beats elegance on e-testing.
For 2x² − 5x − 3 = 0, what are the roots?
If the roots of x² − 7x + 10 = 0 are α and β, what is α² + β²?
The solution set of x² − 5x + 6 ≤ 0 is:
Solving −3x + 6 > 12 yields: