Algebra (Identities, Linear/Quadratic Equations)

Algebra (Identities, Linear/Quadratic Equations)

🟢 Lite

Key Rule / Formula

Quadratic formula: x = (−b ± √(b²−4ac)) / 2a. For equation ax² + bx + c = 0, sum of roots = −b/a, product = c/a. Key identity: (a+b)² = a² + 2ab + b².

Memory Trick

“Sum = −b/a, Product = c/a” — for ax² + bx + c = 0. If b² − 4ac < 0, roots are imaginary (no real solutions).

1-Sentence Summary

SSC tests algebraic identities for quick simplification and quadratic equations for finding roots, with the discriminant b²−4ac determining whether roots are real, equal, or unequal.

Quick Example

Q: If α and β are roots of x² − 7x + 12 = 0, find α + β and αβ. A: Sum = 7, Product = 12 — Using sum = −(−7)/1 = 7, product = 12/1 = 12.

🟡 Standard

Concept

Algebra in SSC CGL Tier 2 focuses on three areas: algebraic identities (for quick simplification), linear equations (solved by substitution or elimination), and quadratic equations (solved by factorisation or quadratic formula). The identities are the real speed booster — questions that look complex often collapse using the right identity.

Key Identities to Master: (a+b)², (a−b)², a²−b², (a+b+c)² = a²+b²+c²+2(ab+bc+ca), (a+b)³ = a³+3a²b+3ab²+b³, a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca).

Linear Equations in Two Variables: ax + by = c and dx + ey = f. Solve by elimination or substitution. Consistency condition: a/e ≠ b/d for unique solution; a/e = b/d = c/f for infinitely many solutions; a/e = b/d ≠ c/f for no solution.

Quadratic Equations: Convert to standard form ax² + bx + c = 0. Solve by: (1) factorisation if roots are integers/rationals, (2) quadratic formula if not. The discriminant D = b² − 4ac determines nature: D > 0 → two distinct real roots; D = 0 → equal roots; D < 0 → no real roots.

Key Points

• If roots are α, β: α+β = −b/a, αβ = c/a. This is Vieta’s formula.
• If one root is 2+√3, the other is 2−√3 (conjugate pair for rational coefficients).
• For linear equation substitution: if x/a + y/b = 1 and x/c + y/d = 1, solve by cross-multiplication.
• Always rearrange to standard form before identifying a, b, c for quadratic formula.
• (x + 1/x)² = x² + 2 + 1/x² — useful for problems giving x + 1/x and asking for x² + 1/x².

Worked Example

Q: If x + 1/x = 3, find x² + 1/x². Approach: (x + 1/x)² = x² + 2 + 1/x² = 3² = 9. So x² + 1/x² = 9 − 2 = 7. Answer: 7

SSC Pattern / Tips

• Quadratic equations in SSC often have integer or simple fractional roots — try factorisation first.
• For identity-based simplification, always look for patterns: (x+y)² − (x−y)² = 4xy.
• When setting up linear equations from word problems, identify two conditions to form two equations.
• If quadratic has complex roots, they come in conjugate pairs — this doesn’t affect sum/product calculations.

🔴 Extended

Full Concept

Advanced Identities: Beyond basic identities, Tier 2 tests: a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca). When a+b+c = 0, this simplifies to a³+b³+c³ = 3abc — this special case is frequently tested. Another important one: (x−a)(x−b)(x−c)… expansion gives symmetric sums of roots — this is connected to Vieta’s formulas for higher degree polynomials.

Cubic Equations: If α, β, γ are roots of ax³ + bx² + cx + d = 0, then: α+β+γ = −b/a, αβ+βγ+γα = c/a, αβγ = −d/a. For equations that factor into (x−a)(x−b)(x−c) = 0, the roots are simply a, b, c.

Quadratic in Quadratic: Some questions give expressions like x⁴ + 1/x⁴ and ask to find x⁴ + 1/x⁴ given x + 1/x. The chain: x + 1/x → square → x² + 1/x² → square → x⁴ + 1/x⁴. Each step adds 2. So from x + 1/x = n: x² + 1/x² = n² − 2; x⁴ + 1/x⁴ = (n² − 2)² − 2.

Linear Equations with Three Variables: These are solved using substitution or elimination. The condition for consistency (solution exists) for a₁x + b₁y + c₁z = d₁ and similar pairs: the ratios a₁/a₂ = b₁/b₂ = c₁/c₂ must hold for infinite solutions, or two equations must be consistent with the third.

Maximum/Minimum of Quadratic: For ax²+bx+c, vertex is at x = −b/(2a). Maximum value if a < 0, minimum if a > 0. This is useful for word problems about area or cost optimisation.

Condition for Common Roots: If two quadratics share a root, subtract one from the other to eliminate x² term, then solve for x. If they have both roots common, the coefficients are proportional.

SSC CGL Deep Analysis

• Frequency: 1–2 questions per paper. Quadratic equations with Vieta’s formulas appear almost every year. Algebraic identities are embedded in other topics.
• Difficulty: Medium. The conceptually challenging parts are cubic equations with three unknowns and common root conditions.
• Recent trend: Questions combining algebra with geometry (e.g., find the roots of a quadratic representing sides of a triangle satisfying a condition).
• Newer patterns: Problems where the quadratic equation is hidden inside a word problem — “a number exceeds its square root by…” leading to x = √x + something.
• Total weight in Tier 2: Roughly 3–4% of the quant paper.

High-Scoring Strategy

1. For any expression with x + 1/x, square once to get x² + 1/x², square again for x⁴ + 1/x⁴.
2. For cubic roots with a+b+c = 0: a³+b³+c³ = 3abc — use this when a+b+c is given as zero.
3. When two quadratics share a root: If ax²+bx+c = 0 and dx²+ex+f = 0 share root r, then ar²+br+c = 0 and dr²+er+f = 0. Eliminate r² to find r, then find other roots.
4. For maximum/minimum, use −b/(2a) for the vertex x-value, then substitute to find extreme value.
5. In linear three-variable systems, use elimination: eliminate same variable from two pairs, solve resulting two equations.

SSC-Level Practice

Q1: If x² + 1/x² = 14, find x + 1/x (x > 0). Answer: 4 — Working: (x + 1/x)² = x² + 2 + 1/x² = 14 + 2 = 16. So x + 1/x = √16 = 4 (positive since x > 0).

Q2: Find the condition that one root of ax²+bx+c = 0 is double the other. Answer: 2b² = 9ac — Working: Let roots be α, 2α. Sum = 3α = −b/a → α = −b/(3a). Product = 2α² = c/a → 2(b²/9a²) = c/a → 2b²/9a² = c/a → 2b² = 9ac.

Common Traps

• Trap 1: Not rearranging the quadratic to standard form ax²+bx+c = 0 before identifying a, b, c. Using wrong coefficients gives wrong roots.
• Trap 2: Forgetting the sign in Vieta: sum of roots = −b/a (negative!), not b/a. Product = c/a (positive if c and a have same sign).
• Trap 3: Assuming x + 1/x = n always gives x = n/2 ± √(n²−4)/2. This is correct, but only valid when n² ≥ 4 (real x exists). If n² < 4, x is complex.

Content adapted based on your selected roadmap duration.