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.
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**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.
*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)