Algebra

Algebra is not just a Class 10 chapter — it is the skeleton of every quantitative-aptitude section on RRB and SSC exams, powering speed, work, mixture, and data-interpretation problems alike. This lesson builds the core machinery: linear equations, quadratic equations, useful identities, and index laws.

Definition: A linear equation contains variables only to the first power (e.g. 3x + 5 = 11). A quadratic equation contains a squared term as the highest degree: ax² + bx + c = 0, where a ≠ 0.

Solving linear equations — one variable

Any linear equation in one variable reduces to ax + b = 0, giving x = −b/a.

Question: Solve 5x − 3 = 2x + 9.
Solution:
Step 1: Bring x terms to the left: 5x − 2x = 9 + 3 → 3x = 12.
Step 2: x = 12/3.
Conclusion: x = 4.

Two equations in two variables

Elimination method: scale one or both equations so that the coefficient of one variable matches, then subtract to cancel it.

Question: Solve the system: 2x + 3y = 13 and 5x − y = 7.
Solution:
Step 1: Multiply the second equation by 3: 15x − 3y = 21.
Step 2: Add to the first: (2x + 15x) + (3y − 3y) = 13 + 21 → 17x = 34 → x = 2.
Step 3: Substitute x = 2 in the first: 4 + 3y = 13 → y = 3.
Conclusion: x = 2, y = 3.

Cross-multiplication shortcut for a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:

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

This is fast once memorised but carry-mistakes are common — use elimination in time-pressure unless you are very comfortable with the formula.

Quadratic equations and the discriminant

The general formula for roots of ax² + bx + c = 0:

x = (−b ± √D) / 2a, where D = b² − 4ac is the discriminant.

The discriminant alone tells you the nature of the roots before you calculate anything:

:::compare Discriminant and nature of roots

Value of D	Nature of roots	Example
D > 0	Two distinct real roots	x² − 5x + 6 = 0 → D = 25 − 24 = 1
D = 0	Two equal (repeated) real roots	x² − 6x + 9 = 0 → D = 36 − 36 = 0
D < 0	No real roots (complex/imaginary)	x² + x + 1 = 0 → D = 1 − 4 = −3
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Sum and product of roots: if α and β are the roots of ax² + bx + c = 0,

• α + β = −b/a
• αβ = c/a

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

Question: One root of 3x² − 10x + k = 0 is 2. Find k and the other root.
Solution:
Step 1: Sum of roots = 10/3. If one root is 2, the other is 10/3 − 2 = 4/3.
Step 2: Product of roots = k/3. Product = 2 × 4/3 = 8/3 → k = 8.
Conclusion: k = 8, other root is 4/3.

Why it matters: The sum/product shortcut bypasses long division or guessing; checking the sign of D answers "how many real solutions?" in seconds without touching a square root — both are huge time-savers in a 60-second exam question.

Algebraic identities to memorise

These transform complicated expressions into factorisable or manageable forms:

• (a + b)² = a² + 2ab + b²
• (a − b)² = a² − 2ab + b²
• a² − b² = (a + b)(a − b) ← most frequently tested
• (a + b)³ = a³ + 3a²b + 3ab² + b³
• a³ + b³ = (a + b)(a² − ab + b²)
• a³ − b³ = (a − b)(a² + ab + b²)
• 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

Question: If x + y = 5 and xy = 6, find x² + y².
Solution:
Step 1: x² + y² = (x + y)² − 2xy.
Step 2: = 25 − 12.
Conclusion: x² + y² = 13.

Question: If a + b + c = 0, find (a³ + b³ + c³) / abc.
Solution:
Step 1: Since a + b + c = 0, the identity gives a³ + b³ + c³ = 3abc.
Step 2: Ratio = 3abc / abc.
Conclusion: 3.

Real-world example: A shopkeeper knows two products' prices add to ₹50 and their product is ₹600. Those are exactly the sum and product of roots of x² − 50x + 600 = 0. Factor: (x − 20)(x − 30) = 0. The prices are ₹20 and ₹30 — algebra recovers both unknowns from two clues, with no trial and error.

Laws of indices (exponents)

Law	Expression
Multiply same base	aᵐ × aⁿ = aᵐ⁺ⁿ
Divide same base	aᵐ ÷ aⁿ = aᵐ⁻ⁿ
Power of a power	(aᵐ)ⁿ = aᵐⁿ
Zero exponent	a⁰ = 1 (a ≠ 0)
Negative exponent	a⁻ⁿ = 1/aⁿ
Fractional exponent	a^(m/n) = ⁿ√(aᵐ)
Product base	(ab)ⁿ = aⁿbⁿ

Question: Simplify: (2³ × 2⁴) ÷ 2⁵.
Solution:
Step 1: Numerator = 2^(3+4) = 2⁷.
Step 2: 2⁷ ÷ 2⁵ = 2^(7−5) = 2².
Conclusion: 4.

Common misconception: Students waste time computing √D when a question only asks about the nature of roots — "how many real solutions does this equation have?" Just check whether b² − 4ac is positive, zero, or negative. No calculation beyond that is needed. Checking the sign is often the entire solution.

:::keypoints Key points

• Linear: x = −b/a; for two variables, use elimination or substitution.
• Quadratic roots: x = (−b ± √D) / 2a; the sign of D = b² − 4ac tells nature before you compute roots.
• D > 0 → two real roots; D = 0 → one repeated root; D < 0 → no real roots.
• Sum of roots = −b/a; product of roots = c/a — use these to find unknown coefficients.
• a² − b² = (a+b)(a−b) and the cube identities are tested directly and embedded in harder problems.
• If a + b + c = 0, then a³ + b³ + c³ = 3abc — a frequent one-liner answer.
• Index laws: add exponents when multiplying, subtract when dividing, multiply when raising a power.
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:::memory
"B-squared minus four-AC determines the doors" — D = b² − 4ac; positive opens two doors (distinct roots), zero opens one door (repeated root), negative opens no real door (complex roots).
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:::recap

• One discriminant sign tells you how many real roots exist, with no further calculation.
• Sum/product of roots let you rebuild equations and find missing coefficients.
• Standard identities — especially a² − b² and a³ + b³ + c³ = 3abc — replace long factoring.
• Index laws reduce complicated power expressions to simple arithmetic on the exponents.
• Two-variable linear systems: match one coefficient by scaling, then subtract (elimination).
• The sum/product shortcut: always set up x² − (sum)x + (product) = 0 to write the equation from roots.
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RRB and SSC algebra rewards the student who recognises the pattern in the first five seconds and writes the answer in the next ten — and that speed comes entirely from knowing a handful of identities and worked examples so well that applying them is reflex, not reasoning.

Definition: An algebraic identity is an equation that holds true for all values of its variables — unlike an equation, which is only true for specific values — allowing you to replace slow calculation with a known shortcut.

Definition: The discriminant of a quadratic ax² + bx + c = 0 is D = b² − 4ac. Its sign determines the nature of roots: D > 0 (two distinct real roots), D = 0 (two equal real roots), D < 0 (no real roots).

Essential Identities to Memorise

Before the worked examples, internalise this identity sheet. Every example below is just an application of one of these:

Identity	Formula
Square of sum	(a+b)² = a² + 2ab + b²
Square of difference	(a−b)² = a² − 2ab + b²
Difference of squares	a² − b² = (a+b)(a−b)
Cube of sum	(a+b)³ = a³ + 3a²b + 3ab² + b³
Sum of cubes	a³ + b³ = (a+b)(a² − ab + b²) = (a+b)³ − 3ab(a+b)
Difference of cubes	a³ − b³ = (a−b)(a² + ab + b²)
Three-variable sum	(a+b+c)² = a²+b²+c²+2(ab+bc+ca)
Special: if a+b+c=0	a³+b³+c³ = 3abc
Reciprocal square	(x + 1/x)² = x² + 1/x² + 2
Reciprocal cube	(x + 1/x)³ = x³ + 1/x³ + 3(x + 1/x)

Why it matters: These identities appear in RRB NTPC, SSC CGL, and banking exams year after year. Students who memorise the identity sheet solve these in 15–20 seconds; those who don't take 3–4 minutes of algebra.

Worked Examples

Example 1 — Two-variable simultaneous equations

Question: Solve 2x + 3y = 13 and 4x − y = 5.

Solution:
Step 1: From the second equation, isolate y: y = 4x − 5.
Step 2: Substitute into the first: 2x + 3(4x − 5) = 13 → 2x + 12x − 15 = 13 → 14x = 28 → x = 2.
Step 3: Substitute back: y = 4(2) − 5 = 3.
Conclusion: x = 2, y = 3.

Real-world example: Two fruit vendors. One sells 2 mangoes and 3 oranges for ₹13; another sells 4 mangoes and 1 orange for ₹5. This is exactly a 2-variable system — a real application of simultaneous equations in everyday Indian market pricing.

Example 2 — Quadratic from sum and product of roots

Question: The sum of the roots of a quadratic equation is 7 and the product of the roots is 12. Find the equation and its roots.

Solution:
Step 1: A quadratic with roots α and β is x² − (α+β)x + αβ = 0.
Step 2: Substitute: x² − 7x + 12 = 0.
Step 3: Factorise: (x − 3)(x − 4) = 0.
Conclusion: The quadratic is x² − 7x + 12 = 0; roots are 3 and 4.

Common misconception: Students often write x² + (sum)x + product = 0 (wrong sign). The correct form is x² − (sum)x + product = 0.

Example 3 — Sum of cubes using identity

Question: If a + b = 5 and ab = 6, find a³ + b³.

Solution:
Step 1: Use the identity: a³ + b³ = (a+b)³ − 3ab(a+b).
Step 2: = (5)³ − 3(6)(5) = 125 − 90.
Conclusion: a³ + b³ = 35.

Verification: Note that ab = 6 and a+b = 5 means a and b are roots of x² − 5x + 6 = 0, i.e., a = 2, b = 3. Check: 2³ + 3³ = 8 + 27 = 35. ✓

Example 4 — The x + 1/x reciprocal pattern

This pattern is an RRB staple. Once you recognise it, it takes under 10 seconds.

Question: If x + 1/x = 4, find (a) x² + 1/x² and (b) x³ + 1/x³.

Solution:
Step 1 (for x² + 1/x²): Square both sides: (x + 1/x)² = x² + 2·x·(1/x) + 1/x² = x² + 2 + 1/x².
So 16 = x² + 1/x² + 2 → x² + 1/x² = 14.

Step 2 (for x³ + 1/x³): Use the cube identity: (x + 1/x)³ = x³ + 3·x·(1/x)·(x + 1/x) + 1/x³ = x³ + 1/x³ + 3(x + 1/x).
So 4³ = x³ + 1/x³ + 3(4) → 64 = x³ + 1/x³ + 12 → x³ + 1/x³ = 52.
Conclusion: x² + 1/x² = 14; x³ + 1/x³ = 52.

Common misconception: When squaring (x + 1/x), the middle term is +2 (because 2·x·(1/x) = 2), not 0. Forgetting this is the single most common error in this class of problem.

Example 5 — Discriminant and equal roots

Question: Find the value of k for which 2x² + kx + 8 = 0 has two equal roots.

Solution:
Step 1: Equal roots require discriminant D = 0.
Step 2: D = k² − 4(2)(8) = k² − 64.
Step 3: Set D = 0: k² = 64 → k = ±8.
Conclusion: k = +8 or −8.

Why the ± matters: Both values of k give valid quadratics. For k = 8: 2x² + 8x + 8 = 0 → x² + 4x + 4 = 0 → (x+2)² = 0 → x = −2 (double root). For k = −8: similarly x = +2.

Example 6 — Three-variable identity when a + b + c = 0

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

Solution:
Step 1: Recall the identity: a³ + b³ + c³ − 3abc = (a+b+c)(a² + b² + c² − ab − bc − ca).
Step 2: Since a + b + c = 0, the right-hand side = 0 × (anything) = 0.
Step 3: Therefore a³ + b³ + c³ = 3abc.
Conclusion: (a³ + b³ + c³) / (abc) = 3.

Real-world verification: Let a = 1, b = 2, c = −3 (so a+b+c = 0). Check: 1³ + 2³ + (−3)³ = 1 + 8 − 27 = −18. And 3abc = 3(1)(2)(−3) = −18. ✓

Speed Pattern Recognition — Exam Triage

When you see an algebra question in an exam, spend 5 seconds on this triage:

1. Two-variable linear? → Substitution or elimination; 4 steps.
2. Sum and product given? → Build quadratic with x² − (S)x + P = 0.
3. a + b and ab given, asking for cubes? → Use a³ + b³ = (a+b)³ − 3ab(a+b).
4. x + 1/x given? → Square it (add 2 for x² + 1/x²); cube it (add 3×(x+1/x) for x³ + 1/x³).
5. Equal roots? → D = 0 → b² − 4ac = 0.
6. a + b + c = 0? → a³ + b³ + c³ = 3abc instantly.

:::compare Sum vs Difference — Cube Identity

Identity	Formula
a³ + b³	(a+b)(a² − ab + b²) OR (a+b)³ − 3ab(a+b)
a³ − b³	(a−b)(a² + ab + b²)
a³ + b³ + c³ (when a+b+c=0)	3abc
a³ + b³ + c³ (general)	(a+b+c)(a²+b²+c²−ab−bc−ca) + 3abc
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:::keypoints Key points

• Build a quadratic from roots as x² − (sum)x + (product) = 0 — note the minus before sum.
• a³ + b³ = (a+b)³ − 3ab(a+b) — the two-variable cube identity.
• (x + 1/x)² = x² + 1/x² + 2 — never forget the +2 middle term.
• (x + 1/x)³ = x³ + 1/x³ + 3(x + 1/x) — the cube reciprocal identity.
• Equal roots require discriminant D = b² − 4ac = 0.
• When a + b + c = 0, a³ + b³ + c³ = 3abc — no calculation needed.
• Recognise the pattern before doing algebra — saves 2–3 minutes per question.
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:::memory
"Sum-Product Quadratic, Cube-Triple, Reciprocal-Plus-Two, Zero-Sum-Triple" — the four identity families: (1) x²−Sx+P=0, (2) a³+b³=(a+b)³−3ab(a+b), (3) (x+1/x)²=x²+1/x²+2, (4) a+b+c=0→a³+b³+c³=3abc.
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:::recap

• RRB/SSC algebra is pattern-matching, not heavy computation — the moment you identify the type, the answer is seconds away.
• Sum-and-product instantly gives you the quadratic and its roots.
• The cube and reciprocal identities are the highest-frequency "surprise" questions — master both.
• The discriminant condition D = 0 is the only concept needed for "equal roots" questions.
• Practise triage: read the problem, identify which of the six pattern types it is, then execute the identity.
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