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 foundational school-level mathematics. This review covers linear rearrangements, simultaneous equations, quadratics, inequality intervals, and word problems involving rates, mixtures, and ages without asserting an official PAF frequency or timing.
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.
Practice Focus
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: