Algebra

Algebra in RRB and SSC exams is rarely about heavy computation — it is about recognising the right pattern and applying a one-line identity before reaching for the calculator. Master a handful of symmetric identities and the quadratic root relationships, and most algebra questions collapse to single-step answers.

Definition: An algebraic identity is an equation that holds true for all values of its variables (e.g., (a + b)² = a² + 2ab + b²). Unlike an equation, it has no solution to find — it is always true.

Definition: The discriminant of a quadratic ax² + bx + c is D = b² − 4ac. It determines the nature of the roots without solving the equation.

Solving simultaneous linear equations

Two equations in two unknowns can be solved by substitution (express one variable from one equation, plug into the other) or elimination (multiply equations to match a coefficient, then add or subtract).

Question: Solve 2x + 3y = 13 and 4x − y = 5.
Solution:
Step 1: From the second equation, y = 4x − 5 (coefficient of y is −1, so substitution is clean).
Step 2: Substitute into the first: 2x + 3(4x − 5) = 13 → 2x + 12x − 15 = 13 → 14x = 28.
Step 3: x = 2 → y = 4(2) − 5 = 3.
Conclusion: x = 2, y = 3.

When to use which: substitution works best when one variable has coefficient ±1; otherwise elimination is faster.

Sum and product of roots — Vieta's formulas

For any quadratic ax² + bx + c = 0, the roots α and β satisfy:

• α + β = −b/a (sum of roots)
• αβ = c/a (product of roots)

These let you reconstruct a quadratic from its roots: x² − (sum)x + (product) = 0.

Question: The sum of roots is 7 and the product is 12. Find the quadratic and its roots.
Solution:
Step 1: Equation: x² − 7x + 12 = 0.
Step 2: Factorise: (x − 3)(x − 4) = 0.
Conclusion: Roots are 3 and 4.

Question: Find k so that 2x² + kx + 8 = 0 has equal roots.
Solution:
Step 1: Equal roots ⟺ D = 0 → k² − 4(2)(8) = 0 → k² = 64.
Conclusion: k = ±8.

Nature of roots summary:

• D > 0: two distinct real roots.
• D = 0: two equal (repeated) real roots.
• D < 0: two complex conjugate roots (no real solution).

The cubic identity: a³ + b³

The most frequently tested cubic identity is:

a³ + b³ = (a + b)³ − 3ab(a + b)

This lets you find a³ + b³ from just the symmetric quantities (sum and product) without ever finding a and b individually.

Question: If a + b = 5 and ab = 6, find a³ + b³.
Solution:
Step 1: a³ + b³ = (5)³ − 3(6)(5) = 125 − 90 = 35.

Derived identities from the same root:

• a³ − b³ = (a − b)³ + 3ab(a − b).
• a² + b² = (a + b)² − 2ab = 25 − 12 = 13.
• (a − b)² = (a + b)² − 4ab = 25 − 24 = 1, so a − b = 1.

The x + 1/x chain

These appear so often in RRB papers that recognising them instantly is worth real exam marks.

Starting identity: (x + 1/x)² = x² + 2 + 1/x², so x² + 1/x² = (x + 1/x)² − 2.

Chain extension:

• x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x)

Question: If x + 1/x = 4, find x² + 1/x² and x³ + 1/x³.
Solution:
Step 1: x² + 1/x² = 4² − 2 = 14.
Step 2: x³ + 1/x³ = 4³ − 3 × 4 = 64 − 12 = 52.

Similarly, for x − 1/x = k:

• x² + 1/x² = k² + 2.
• x³ − 1/x³ = k³ + 3k.

The a + b + c = 0 shortcut

The complete identity is: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

When a + b + c = 0, the right side vanishes, giving the beautiful result:

a³ + b³ + c³ = 3abc (only valid when a + b + c = 0).

Question: If a + b + c = 0, find (a³ + b³ + c³) / (abc).
Solution: By the identity, a³ + b³ + c³ = 3abc, so the ratio = 3.

Why it matters: These identity-based questions are designed to be solved in under 30 seconds. A candidate who reaches for brute-force algebra wastes 2–3 minutes and risks sign errors. Recognising the pattern converts a seemingly hard question into a one-line answer.

Real-world example: Imagine two coaching batches where fee multipliers satisfy a + b = 5 and ab = 6 (perhaps ₹2 and ₹3 per unit, satisfying 2+3=5 and 2×3=6). A "cubed-impact" metric a³ + b³ = 35 can be computed instantly from the identity, without solving for the individual unknowns. That is exactly the mindset examiners reward — working only with symmetric quantities rather than the individual values.

Common misconception: Many students believe a³ + b³ + c³ = 3abc always. It does not — the shortcut works only when a + b + c = 0. For general a, b, c, the full identity must be used. Applying the shortcut when the sum is nonzero is one of the most common algebraic errors in RRB papers.

:::compare Substitution vs Elimination

Situation	Preferred method	Example
One variable has coefficient ±1	Substitution	4x − y = 5 → y = 4x−5
Both variables have large coefficients	Elimination	3x+4y=10, 5x+2y=12
Three-variable systems	Elimination (two rounds)	Standard
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:::keypoints Key points

• Vieta's formulas: sum of roots = −b/a, product = c/a; build a quadratic as x² − (sum)x + product = 0.
• Equal roots ⟺ discriminant D = b² − 4ac = 0.
• a³ + b³ = (a + b)³ − 3ab(a + b); needs only the sum and product.
• x² + 1/x² = (x + 1/x)² − 2; x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x).
• a³ + b³ + c³ = 3abc only when a + b + c = 0.
• a² + b² = (a + b)² − 2ab; (a − b)² = (a + b)² − 4ab.
• Spot the pattern first — RRB algebra rewards recognition, not computation.
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  :::memory
  "Sum-Product-Cube" — the three-step shortcut: know the Sum, know the Product, compute the Cube using one identity.
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  :::recap
• Simultaneous equations: substitution when coefficient is ±1, elimination otherwise.
• Vieta's formulas let you build or analyse a quadratic without solving it.
• The x + 1/x chain extends to any power using a recursive identity.
• The 3abc shortcut is conditional on a + b + c = 0 — never apply it blindly.
• Pattern recognition, not heavy algebra, is the RRB examiner's design intent.
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Algebra is the language in which most of mathematics is written — and for competitive exams, fluency with linear equations, quadratic equations and algebraic identities is the difference between spending two minutes on a problem and spending twenty seconds.

Definition: Linear equation — an equation in which the variable appears to the first power only; its graph is a straight line; general form: ax + b = 0 (one variable) or ax + by + c = 0 (two variables).

Definition: Quadratic equation — a polynomial equation of degree 2; general form: ax² + bx + c = 0, where a ≠ 0.

Definition: Discriminant — the expression D = b² − 4ac inside the quadratic formula; its sign determines the nature of the roots without actually solving the equation.

Linear Equations in One Variable

The simplest case: ax + b = 0 → x = −b/a

Example: 3x − 12 = 0 → x = 12/3 = 4.

Word problems almost always reduce to this. The skill is translation: "A number increased by 7 equals three times the number" becomes x + 7 = 3x → x = 3.5.

Fractional and decimal equations: Clear fractions by multiplying every term by the LCM of denominators before solving.

Example: x/3 + x/4 = 7 → multiply by 12 → 4x + 3x = 84 → 7x = 84 → x = 12.

Two Variables — Simultaneous Linear Equations

Two equations are needed to solve for two unknowns. Three methods:

Method 1: Substitution

Express one variable in terms of the other from one equation, substitute into the second.
Example: 2x + y = 8 and x − y = 1.
From second equation: x = y + 1. Substitute: 2(y+1) + y = 8 → 3y = 6 → y = 2, x = 3.

Method 2: Elimination

Multiply equations to make coefficients of one variable equal, then add or subtract.
Example: 3x + 2y = 12 and 5x + 3y = 19.
Multiply first by 3: 9x + 6y = 36. Multiply second by 2: 10x + 6y = 38. Subtract: x = 2. Back-substitute: y = 3.

Method 3: Cross-Multiplication (fastest for objective exams)

For equations in standard form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:

x / (b₁c₂ − b₂c₁) = y / (c₁a₂ − c₂a₁) = 1 / (a₁b₂ − a₂b₁)

Learn the pattern once; it saves writing many lines of algebra in speed tests.

Consistency check (important for exam traps):

• Unique solution: a₁/a₂ ≠ b₁/b₂ (lines intersect — consistent and independent).
• Infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂ (lines coincide — consistent and dependent).
• No solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (lines are parallel — inconsistent).

Quadratic Equations — Full Coverage

Standard form: ax² + bx + c = 0 (a ≠ 0)

The Quadratic Formula

x = [−b ± √(b² − 4ac)] / 2a

This always works. The discriminant D = b² − 4ac tells you what to expect before you calculate:

D value	Nature of roots
D > 0	Two distinct real roots
D = 0	One repeated real root: x = −b/2a
D < 0	No real roots (complex roots — not relevant for Class 10 / SSC level)

Factorisation Method (faster when it works)

Split the middle term: find two numbers p and q such that p + q = b and p × q = ac. Then factorise.

Example: 6x² + 11x − 10 = 0.
Need p + q = 11, p × q = 6 × (−10) = −60. Try p = 15, q = −4: 15 + (−4) = 11 ✓, 15 × (−4) = −60 ✓.
Rewrite: 6x² + 15x − 4x − 10 = 0 → 3x(2x + 5) − 2(2x + 5) = 0 → (3x − 2)(2x + 5) = 0.
Roots: x = 2/3 and x = −5/2.

Vieta's Formulas (Sum and Product of Roots)

If α and β are the roots of ax² + bx + c = 0:

Sum of roots: α + β = −b/a
Product of roots: αβ = c/a

These are enormously useful for exam problems that ask for expressions like α² + β², α³ + β³, or 1/α + 1/β without requiring you to find α and β individually.

Example: If α + β = −b/a and αβ = c/a, then α² + β² = (α + β)² − 2αβ = b²/a² − 2c/a.

Forming equations from roots: If you know α and β, the equation is:
x² − (α + β)x + αβ = 0.

Algebraic Identities — The Power List

Identities are equations true for all values of variables; they allow rapid simplification:

Squares and Cubes

Identity	Use
(a + b)² = a² + 2ab + b²	Expanding or recognising perfect squares
(a − b)² = a² − 2ab + b²	Same
a² − b² = (a + b)(a − b)	Difference of squares — factor or evaluate
(a + b)³ = a³ + 3a²b + 3ab² + b³	Cube expansions
(a − b)³ = a³ − 3a²b + 3ab² − b³	Cube expansions
a³ + b³ = (a + b)(a² − ab + b²)	Sum of cubes
a³ − b³ = (a − b)(a² + ab + b²)	Difference of cubes

The Golden Identity for Exams

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

Special case: If a + b + c = 0, then a³ + b³ + c³ = 3abc.

This is one of the most frequently tested identities in SSC, RRB and banking exams. Recognise when the three terms sum to zero.

Derived Results from Square Identities

If (a + b)² = a² + 2ab + b², then:

• a² + b² = (a + b)² − 2ab
• (a − b)² = (a + b)² − 4ab
• 2(a² + b²) = (a + b)² + (a − b)²

Laws of Indices (Exponents)

Law	Expression	Example
Product rule	aᵐ × aⁿ = aᵐ⁺ⁿ	2³ × 2⁴ = 2⁷ = 128
Quotient rule	aᵐ ÷ aⁿ = aᵐ⁻ⁿ	5⁶ ÷ 5² = 5⁴ = 625
Power of a power	(aᵐ)ⁿ = aᵐⁿ	(3²)³ = 3⁶ = 729
Zero exponent	a⁰ = 1 (a ≠ 0)	7⁰ = 1
Negative exponent	a⁻ⁿ = 1/aⁿ	2⁻³ = 1/8
Fractional exponent	a^(m/n) = ⁿ√(aᵐ) = (ⁿ√a)ᵐ	8^(2/3) = (∛8)² = 4
Product rule for bases	aᵐ × bᵐ = (ab)ᵐ	2³ × 3³ = 6³ = 216
Quotient rule for bases	aᵐ / bᵐ = (a/b)ᵐ	6⁴ / 3⁴ = 2⁴ = 16

Trap: aᵐ × bⁿ ≠ (ab)ᵐ⁺ⁿ unless m = n. Different bases require different treatment.

Worked Examples

Question 1: If x + y = 7 and xy = 12, find x² + y², x³ + y³, and x − y.

Solution:
Step 1: x² + y² = (x + y)² − 2xy = 49 − 24 = 25.
Step 2: x − y = √[(x + y)² − 4xy] = √[49 − 48] = √1 = 1 (assuming x > y).
Step 3: x³ + y³ = (x + y)(x² − xy + y²) = 7 × (25 − 12) = 7 × 13 = 91.
Conclusion: x² + y² = 25, x³ + y³ = 91, x − y = 1.

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Question 2: Find the discriminant of 2x² − 4x + 3 = 0 and determine the nature of roots.

Solution:
Step 1: a = 2, b = −4, c = 3.
Step 2: D = b² − 4ac = 16 − 24 = −8.
Conclusion: D < 0, so the equation has no real roots (complex roots).

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Question 3: If a + b + c = 0, find the value of (a³ + b³ + c³) / abc.

Solution:
Step 1: Since a + b + c = 0, apply the golden identity: a³ + b³ + c³ = 3abc.
Step 2: (a³ + b³ + c³) / abc = 3abc/abc = 3.
Conclusion: The value is 3 — regardless of specific values of a, b, c (as long as their sum is zero).

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Question 4: Solve the system: 3x + 4y = 10 and 4x − 3y = 5.

Solution (Elimination):
Step 1: Multiply first equation by 3 and second by 4: 9x + 12y = 30 and 16x − 12y = 20.
Step 2: Add: 25x = 50 → x = 2.
Step 3: Back-substitute: 3(2) + 4y = 10 → 4y = 4 → y = 1.
Conclusion: x = 2, y = 1.

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Question 5: If one root of x² − 5x + k = 0 is 2, find k and the other root.

Solution:
Step 1: x = 2 is a root, so 4 − 10 + k = 0 → k = 6.
Step 2: Sum of roots = 5 (from −b/a = 5/1). Other root = 5 − 2 = 3.
Conclusion: k = 6, other root = 3.

Common Misconceptions Corrected

Misconception 1: "x² = 9 has only one solution, x = 3."
Correction: x² = 9 → x = ±3. Both +3 and −3 satisfy the equation. Never forget the negative root.

Misconception 2: "(a + b)² = a² + b²."
Correction: The middle term 2ab is missing. This is the most common algebraic error in Class 9–10 and competitive exams.

Misconception 3: "a³ + b³ = (a + b)³."
Correction: (a + b)³ = a³ + 3a²b + 3ab² + b³. The factorisation of a³ + b³ is (a + b)(a² − ab + b²).

Misconception 4: "If D = 0, there are no roots."
Correction: D = 0 means one repeated real root (x = −b/2a), not no roots. D < 0 means no real roots.

Misconception 5: "a⁰ = 0."
Correction: Any non-zero number to the power 0 equals 1. This is a definition that makes the index laws consistent.

Real-World Indian Context

GST and price equations: "A shopkeeper adds 18% GST to his cost price and sells at ₹1,888. Find the cost price." This is a one-variable linear equation: CP × 1.18 = 1888 → CP = 1600. The ability to set up and solve such equations quickly is tested in bank exams (SBI PO, IBPS) as part of simplification sections.

Physics problems in NEET and JEE: Quadratic equations arise naturally in kinematics (s = ut + ½at²), optics (mirror formula) and circuit analysis. The Vieta formulas allow physics teachers to find "the other root" without solving from scratch — a skill rewarded in multiple-choice settings.

:::compare Quadratic: Factorisation vs. Formula vs. Vieta

Method	When to use	Speed	Always works?
Factorisation	When factors are integers and easy to spot	Fastest	No (rational roots only easy to spot)
Quadratic formula	Always, especially when D is not a perfect square	Medium	Yes
Vieta formulas	When asked for sum/product/derived expressions without needing individual roots	Fastest for such questions	Yes (gives relations, not individual roots)
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:::keypoints Key points

• Linear equation: ax + b = 0 → x = −b/a; two variables require two equations (substitution, elimination, or cross-multiplication).
• Quadratic formula: x = [−b ± √D] / 2a where D = b² − 4ac.
• Discriminant D: >0 two distinct real roots; =0 one repeated root; <0 no real roots.
• Vieta's formulas: sum of roots = −b/a; product of roots = c/a.
• Golden identity: if a + b + c = 0, then a³ + b³ + c³ = 3abc.
• (a + b)² = a² + 2ab + b² — the 2ab term is the most commonly forgotten piece.
• a³ + b³ = (a+b)(a²−ab+b²); a³ − b³ = (a−b)(a²+ab+b²).
• Index laws: aᵐ × aⁿ = aᵐ⁺ⁿ; (aᵐ)ⁿ = aᵐⁿ; a⁰ = 1; a⁻ⁿ = 1/aⁿ; a^(m/n) = ⁿ√(aᵐ).
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:::memory
For quadratic roots mnemonic — "Sum is Negative B on A, Product is C on A":
α + β = −b/a, αβ = c/a.
For the golden identity, remember "Sum Zero → Cubes Three": if sum = 0, then a³+b³+c³ = 3abc.
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:::recap

• Linear equations in one variable solve directly; two-variable systems need two equations solved by substitution, elimination or cross-multiplication.
• The quadratic formula always works; the discriminant predicts the nature of roots before calculation.
• Vieta's formulas provide sum and product of roots without solving — powerful for "find a²+b²" type questions.
• The identity a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca), with its special case for a+b+c = 0, is a high-frequency exam shortcut.
• Index laws must be applied correctly; the most common errors involve zero exponents, negative exponents and fractional exponents.
• Avoid the three classic blunders: (a+b)² ≠ a²+b²; x²=9 has two roots ±3; D=0 means one repeated root (not zero roots).
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