Quick Concept Summary
Most aptitude algebra is not about solving for unknowns — it is about manipulating identities. Questions hand you symmetric information (a + b and ab, or a − b) and ask for a²+b², a³+b³, or 1/a + 1/b. If you know the standard identities, these become one-line substitutions.
Must-Know Identities
| Identity | Form |
|---|---|
| (a + b)² | a² + 2ab + b² |
| (a − b)² | a² − 2ab + b² |
| a² − b² | (a − b)(a + b) |
| a² + b² | (a + b)² − 2ab |
| (a + b)³ | a³ + b³ + 3ab(a + b) |
| a³ + b³ | (a + b)(a² − ab + b²) |
| a³ − b³ | (a − b)(a² + ab + b²) |
Common Derived Results
- a − b = √((a + b)² − 4ab)
- 1/a + 1/b = (a + b) / ab
- a/b + b/a = (a² + b²) / ab
- If x + 1/x = k, then x² + 1/x² = k² − 2 and x³ + 1/x³ = k³ − 3k
Quadratic Equations
For ax² + bx + c = 0:
- Roots: x = (−b ± √(b² − 4ac)) / 2a
- Sum of roots = −b/a, Product of roots = c/a
- Discriminant b² − 4ac: > 0 two real roots, = 0 equal roots, < 0 no real roots
Worked Examples
Example 1: If a + b = 7 and ab = 12, find a² + b². (a+b)² − 2ab = 49 − 24 = 25.
Example 2: If x + 1/x = 4, find x² + 1/x². k² − 2 = 16 − 2 = 14.
Example 3: Factorise and solve x² − 5x + 6 = 0. (x − 2)(x − 3) = 0 → x = 2 or 3.
Example 4: If a − b = 3 and ab = 10, find a + b. a + b = √((a−b)² + 4ab) = √(9 + 40) = √49 = 7.
Common Mistakes
- Forgetting the 2ab term when expanding (a ± b)².
- Mixing up a³+b³ and a³−b³ factor signs (middle term: a²∓ab+b²).
- Using a − b = √((a+b)² − 4ab) but forgetting it can be ±.
Exam Tips
- When given a + b and ab, expect a²+b² or a³+b³ — substitute directly.
- For x + 1/x problems, memorise: square → −2, cube → −3k.
- Check the discriminant to predict root nature before solving fully.