Algebra — SSC CHSL Study Notes

Overview

Algebra is a core scoring area in SSC CHSL Tier 1, typically contributing 3–5 direct questions. The exam focuses on three pillars: solving linear and quadratic equations, applying polynomial identities, and factorising expressions. Mastery here is non-negotiable because algebra underpins several other topics like time-speed-distance word problems and profit-loss equations.

SSC CHSL algebra questions are application-based rather than theoretical. You won't be asked to "prove" an identity, but you must recognise which identity applies and use it instantly. Speed matters — most algebra questions should take 45–60 seconds. The syllabus emphasises practical problem-solving: given an equation or expression, transform it using identities or factorisation, then compute the answer. Topics like linear equations in one or two variables, quadratic formula, discriminant-based questions, and standard polynomial identities (a² – b², a³ + b³) appear regularly.

Students often lose marks by misapplying identities or making sign errors during factorisation. The goal of these notes is to give you a mental toolkit — formulas you can recall instantly, the logic behind factorisation, and the common question types you'll face.

Key Concepts

**Linear equation in one variable**: An equation of the form ax + b = 0 where a ≠ 0. Solution is x = –b/a. Always isolate the variable on one side by performing inverse operations.

**Linear equation in two variables**: Form ax + by = c. SSC CHSL typically gives simultaneous equations (a pair) which you solve by substitution, elimination, or cross-multiplication to find unique values of x and y.

**Quadratic equation**: Standard form ax² + bx + c = 0 where a ≠ 0. Solutions come from factorisation, completing the square, or the quadratic formula x = (–b ± √(b² – 4ac)) / 2a.

**Discriminant (D)**: For ax² + bx + c = 0, D = b² – 4ac. If D > 0, two distinct real roots; D = 0, two equal real roots; D < 0, no real roots (complex roots, rarely tested in CHSL).

**Polynomial identities**: Pre-memorised algebraic formulas that let you expand or factorise expressions in one step. SSC CHSL heavily tests the ability to recognise which identity fits a given expression.

**Factorisation**: Expressing a polynomial as a product of simpler polynomials. Methods include taking out common factors, grouping, using identities, and splitting the middle term for quadratics.

**Roots and coefficients**: For ax² + bx + c = 0 with roots α and β, sum of roots α + β = –b/a and product αβ = c/a. Use these for questions asking about root properties without solving the equation.

**Degree of a polynomial**: The highest power of the variable. Linear (degree 1), quadratic (degree 2), cubic (degree 3). CHSL focuses on degrees 1 and 2, occasional degree 3 identities.

Formulas / Key Facts

### Polynomial Identities (Must Memorise)

1. **(a + b)² = a² + 2ab + b²** 2. **(a – b)² = a² – 2ab + b²** 3. **a² – b² = (a + b)(a – b)** 4. **(a + b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a + b)** 5. **(a – b)³ = a³ – 3a²b + 3ab² – b³ = a³ – b³ – 3ab(a – b)** 6. **a³ + b³ = (a + b)(a² – ab + b²)** 7. **a³ – b³ = (a – b)(a² + ab + b²)** 8. **(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca** 9. **a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)** 10. **If a + b + c = 0, then a³ + b³ + c³ = 3abc**

### Quadratic Formula and Properties

**Quadratic formula**: x = (–b ± √(b² – 4ac)) / 2a
**Sum of roots**: α + β = –b/a
**Product of roots**: αβ = c/a
**Discriminant**: D = b² – 4ac

### Useful Derived Relations

**a² + b² = (a + b)² – 2ab = (a – b)² + 2ab**
**a⁴ + a²b² + b⁴ = (a² + ab + b²)(a² – ab + b²)**
**(a + b)² – (a – b)² = 4ab**
**(a + b)² + (a – b)² = 2(a² + b²)**

Worked Examples

**Example 1 — Solve linear equation** Solve: 3x – 7 = 2x + 5

*Solution*: 3x – 2x = 5 + 7 x = 12

**Example 2 — Solve simultaneous equations** 2x + y = 7 x – y = 2

*Solution (elimination)*: Add both equations: (2x + y) + (x – y) = 7 + 2 → 3x = 9 → x = 3 Substitute x = 3 in second equation: 3 – y = 2 → y = 1 Answer: x = 3, y = 1

**Example 3 — Factorise using identity** Factorise: x² – 9

*Solution*: Recognise a² – b² = (a + b)(a – b) where a = x, b = 3 x² – 9 = (x + 3)(x – 3)

**Example 4 — Solve quadratic by factorisation** Solve: x² – 5x + 6 = 0

*Solution*: Find two numbers that multiply to 6 and add to –5: –2 and –3 x² – 5x + 6 = (x – 2)(x – 3) = 0 x – 2 = 0 or x – 3 = 0 x = 2 or x = 3

**Example 5 — Apply a³ + b³ identity** If x + (1/x) = 5, find x³ + (1/x³)

*Solution*: Use identity: a³ + b³ = (a + b)³ – 3ab(a + b) Here a = x, b = 1/x, a + b = 5, ab = x · (1/x) = 1 x³ + (1/x³) = 5³ – 3 · 1 · 5 = 125 – 15 = 110

Common Mistakes

**Mistake**: Confusing (a – b)² with a² – b² *Fix*: (a – b)² = a² – 2ab + b², not a² – b². The middle term –2ab is crucial. a² – b² is the difference of squares identity = (a + b)(a – b).

**Mistake**: Sign errors when expanding (a – b)³ *Fix*: (a – b)³ has alternating signs: a³ – 3a²b + 3ab² – b³. Write it down systematically or use (a – b)³ = a³ – b³ – 3ab(a – b) to avoid errors.

**Mistake**: Incorrect splitting of middle term in quadratics *Fix*: For ax² + bx + c, find two numbers that multiply to a · c and add to b. In x² – 5x + 6, a · c = 6, b = –5, so split –5x = –2x – 3x, then group.

**Mistake**: Applying quadratic formula with wrong signs *Fix*: The formula is x = (–b ± √D) / 2a. The –b has a minus, so if b = –5, then –b = +5. Double-check the sign of b before substituting.

**Mistake**: Forgetting to check D before assuming real roots *Fix*: Always compute D = b² – 4ac. If D < 0, the equation has no real solutions — don't waste time solving further.

Quick Reference

Linear one variable: ax + b = 0 → x = –b/a
Quadratic formula: x = (–b ± √(b² – 4ac)) / 2a
(a + b)² = a² + 2ab + b²; (a – b)² = a² – 2ab + b²
a² – b² = (a + b)(a – b)
a³ + b³ = (a + b)(a² – ab + b²); a³ – b³ = (a – b)(a² + ab + b²)
Sum of roots = –b/a; Product of roots = c/a
If a + b + c = 0, then a³ + b³ + c³ = 3abc

Algebra