Algebra — SSC CGL Study Notes

Overview

Algebra is a recurring and highly scoring section in SSC CGL Tier 1, typically contributing 3–5 questions per paper. This topic tests your ability to manipulate symbols, form and solve equations, and apply polynomial identities quickly. Mastery of algebra not only boosts your score directly but also strengthens your foundation for topics like time-and-work equations, age problems, and certain data interpretation questions.

The SSC CGL syllabus emphasizes three core areas: **linear equations** (single and simultaneous), **quadratic equations** (factorization, formula, and nature of roots), and **polynomial identities** (algebraic expansions and simplifications). Questions are often disguised as word problems or mixed with number system concepts. Speed and algebraic manipulation skills are critical—most algebra questions in SSC CGL must be solved within 60–90 seconds.

Focus on recognizing patterns, applying identities without derivation, and eliminating answer choices strategically. Calculator use is prohibited, so mental math and short-cut techniques are essential.

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Key Concepts

• **Linear equation in one variable**: An equation of the form ax + b = 0 where a ≠ 0. Solution: x = -b/a. Always has exactly one solution.

• **Simultaneous linear equations**: Two equations in two variables (e.g., ax + by = c and dx + ey = f). Solve by substitution, elimination, or cross-multiplication. Unique solution exists when lines intersect (coefficients not proportional).

• **Quadratic equation**: Standard form ax² + bx + c = 0 where a ≠ 0. Maximum power of variable is 2. Can have 0, 1, or 2 real solutions.

• **Discriminant (D)**: D = b² - 4ac determines nature of roots. D > 0: two distinct real roots; D = 0: one repeated real root; D < 0: no real roots (complex roots).

• **Polynomial identities**: Pre-derived expansion formulas that simplify expressions quickly. Memorize the standard identities—deriving them during the exam wastes time.

• **Factorization**: Breaking a quadratic expression into two binomials. Example: x² + 5x + 6 = (x + 2)(x + 3). Used when middle-term splitting is quick.

• **Vieta's formulas**: For ax² + bx + c = 0, sum of roots = -b/a and product of roots = c/a. Useful when questions ask about roots without solving the equation explicitly.

• **Polynomial degree**: Highest power of the variable. Linear = degree 1, quadratic = degree 2, cubic = degree 3. Number of terms doesn't define degree.

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Formulas / Key Facts

**Linear Equations:**

• Solution to ax + b = 0 is x = -b/a
• For simultaneous equations, if a₁/a₂ = b₁/b₂ ≠ c₁/c₂, no solution (parallel lines)

**Quadratic Formula:**

• Roots of ax² + bx + c = 0 are x = [-b ± √(b² - 4ac)] / 2a

**Discriminant:**

• D = b² - 4ac; nature of roots depends on sign of D

**Sum and Product of Roots:**

• Sum α + β = -b/a
• Product αβ = c/a

**Polynomial Identities (must memorize):**

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• a² - b² = (a + b)(a - b)
• (a + b)³ = a³ + 3a²b + 3ab² + b³
• (a - b)³ = a³ - 3a²b + 3ab² - b³
• a³ + b³ = (a + b)(a² - ab + b²)
• a³ - b³ = (a - b)(a² + ab + b²)
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

**Special Result:**

• If a + b + c = 0, then a³ + b³ + c³ = 3abc

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Worked Examples

**Example 1: Linear Equation** Solve: 3x - 7 = 2x + 5

*Solution:* Step 1: Bring variable terms to one side: 3x - 2x = 5 + 7 Step 2: Simplify: x = 12 **Answer: x = 12**

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**Example 2: Simultaneous Equations (Elimination Method)** Solve: 2x + 3y = 13 and 3x - y = 3

*Solution:* Step 1: Multiply second equation by 3: 9x - 3y = 9 Step 2: Add to first equation: (2x + 3y) + (9x - 3y) = 13 + 9 → 11x = 22 → x = 2 Step 3: Substitute x = 2 in second equation: 3(2) - y = 3 → 6 - y = 3 → y = 3 **Answer: x = 2, y = 3**

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**Example 3: Quadratic Equation by Factorization** Solve: x² - 7x + 12 = 0

*Solution:* Step 1: Find two numbers whose sum = -7 and product = 12: -3 and -4 Step 2: Factorize: (x - 3)(x - 4) = 0 Step 3: Set each factor to zero: x = 3 or x = 4 **Answer: x = 3, 4**

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**Example 4: Using Polynomial Identity** If x + 1/x = 5, find x² + 1/x²

*Solution:* Step 1: Square both sides of x + 1/x = 5: (x + 1/x)² = 25 Step 2: Expand using (a + b)² = a² + 2ab + b²: x² + 2(x)(1/x) + 1/x² = 25 x² + 2 + 1/x² = 25 Step 3: Simplify: x² + 1/x² = 25 - 2 = 23 **Answer: 23**

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**Example 5: Nature of Roots** Determine nature of roots of 2x² + 3x + 5 = 0

*Solution:* Step 1: Calculate discriminant: D = b² - 4ac = 3² - 4(2)(5) = 9 - 40 = -31 Step 2: Since D < 0, no real roots exist **Answer: No real roots**

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Common Mistakes

**Mistake 1: Sign errors when transposing** Wrong thinking: In 3x - 7 = 5, moving -7 gives 3x = 5 - 7 Correct fix: When moving -7 to the right side, sign becomes +7: 3x = 5 + 7 = 12

**Mistake 2: Forgetting ± in quadratic formula** Wrong thinking: Using only the + part of ±√D and missing one root Correct fix: Always compute both [-b + √D]/2a and [-b - √D]/2a to get both roots

**Mistake 3: Incorrectly expanding (a - b)²** Wrong thinking: (a - b)² = a² - b² Correct fix: (a - b)² = a² - 2ab + b²; the middle term -2ab is crucial

**Mistake 4: Confusing sum/product of roots formulas** Wrong thinking: For ax² + bx + c = 0, sum of roots = b/a Correct fix: Sum of roots = -b/a (note the negative sign); product = c/a

**Mistake 5: Applying identities without checking conditions** Wrong thinking: Using a³ + b³ + c³ = 3abc for all values Correct fix: a³ + b³ + c³ = 3abc only when a + b + c = 0; otherwise use full identity

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Quick Reference

• Linear equation ax + b = 0 → x = -b/a • Discriminant D = b² - 4ac tells nature of roots (positive/zero/negative) • Sum of roots = -b/a; Product of roots = c/a • (a + b)² = a² + 2ab + b²; (a - b)² = a² - 2ab + b² • a² - b² = (a + b)(a - b) • If a + b + c = 0, then a³ + b³ + c³ = 3abc