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Algebra Notes for Competitive Exams

Komal Gupta

UPSC Exam Expert & Content Editor

Updated 17 Oct 2026

PDF Notes: Algebra Notes for Competitive Exams

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Algebra is one of the highest-weightage chapters in SSC CGL and CHSL and an important part of CDS, CAPF and bank exams. In UPSC CSAT, algebra appears mostly through linear equations hidden inside word problems on ages, money, numbers and work. Many students fear algebra because of symbols, but competitive exam algebra is limited to a fixed set of identities, simple equations and a few smart tricks. Once you learn them, algebra questions become quicker than arithmetic questions, because many can be solved by simply putting values.

Quick Facts: Algebra at a Glance

  • Core tools: About 15 standard identities cover most exam questions.
  • Linear equations: Used in word problems on numbers, ages, coins and fractions.
  • Quadratic equation: ax² + bx + c = 0; roots = [−b ± √(b² − 4ac)] ÷ 2a.
  • Special family: x + 1/x and x − 1/x questions are very common in SSC.
  • Biggest shortcut: Value-putting (substituting simple values) in symmetric expressions.

1. Important Algebraic Identities

Write these on a sheet and revise them daily until they are automatic.

No. Identity
1 (a + b)² = a² + 2ab + b²
2 (a − b)² = a² − 2ab + b²
3 a² − b² = (a + b)(a − b)
4 (a + b)² + (a − b)² = 2(a² + b²)
5 (a + b)² − (a − b)² = 4ab
6 (a + b)³ = a³ + b³ + 3ab(a + b)
7 (a − b)³ = a³ − b³ − 3ab(a − b)
8 a³ + b³ = (a + b)(a² − ab + b²)
9 a³ − b³ = (a − b)(a² + ab + b²)
10 (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
11 a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
12 If a + b + c = 0, then a³ + b³ + c³ = 3abc

Example: If a + b = 7 and ab = 12, find a² + b². Using identity 1: a² + b² = (a + b)² − 2ab = 49 − 24 = 25.

Example: Find 27³ + (−20)³ + (−7)³. Since 27 − 20 − 7 = 0, the value = 3 × 27 × (−20) × (−7) = 11340.

2. The x + 1/x Family

These questions appear almost every year in SSC exams. Learn the chain:

  • If x + 1/x = k, then x² + 1/x² = k² − 2.
  • x³ + 1/x³ = k³ − 3k.
  • x⁴ + 1/x⁴ = (k² − 2)² − 2.
  • If x − 1/x = m, then x² + 1/x² = m² + 2 and x³ − 1/x³ = m³ + 3m.
  • If x + 1/x = 2, then x = 1, so any power expression like x¹⁰⁰ + 1/x¹⁰⁰ = 2.
  • If x + 1/x = −2, then x = −1.
  • If x + 1/x = 1, then x³ = −1, so x³ + 1/x³ = −2.

Example: If x + 1/x = 4, find x³ + 1/x³. Answer = 4³ − 3 × 4 = 64 − 12 = 52.

Trick: If x² − kx + 1 = 0, divide by x to get x + 1/x = k. Many questions are disguised this way.

3. Linear Equations

A linear equation has variables with power 1. In one variable it looks like ax + b = 0, and in two variables like ax + by = c.

Solving Two Equations in Two Variables

  • Substitution method: Find one variable from one equation and put it in the other.
  • Elimination method: Multiply equations to make one variable's coefficients equal, then add or subtract.

Example: 2x + 3y = 13 and 3x − y = 3. Multiply the second by 3: 9x − 3y = 9. Add to the first: 11x = 22, so x = 2 and y = 3.

Conditions for Number of Solutions

Condition (a1x + b1y = c1, a2x + b2y = c2) Type of solution Graph
a1/a2 ≠ b1/b2 Unique solution Lines intersect
a1/a2 = b1/b2 = c1/c2 Infinite solutions Lines coincide
a1/a2 = b1/b2 ≠ c1/c2 No solution Lines are parallel

Word Problems Using Linear Equations

Example (CSAT type): A purse has ₹5 and ₹10 coins, 25 coins in all, worth ₹180. Let ₹5 coins = x, ₹10 coins = 25 − x. Then 5x + 10(25 − x) = 180 → 250 − 5x = 180 → x = 14. So there are 14 coins of ₹5 and 11 coins of ₹10.

Example: The sum of two numbers is 50 and their difference is 14. Numbers = (50 + 14)/2 = 32 and (50 − 14)/2 = 18.

4. Quadratic Equations

A quadratic equation has the standard form ax² + bx + c = 0, where a ≠ 0. It has two roots, usually written as α and β.

  • Quadratic formula: x = [−b ± √(b² − 4ac)] ÷ 2a.
  • Sum of roots: α + β = −b/a.
  • Product of roots: αβ = c/a.
  • Equation from roots: x² − (sum)x + (product) = 0.

Nature of Roots (Discriminant D = b² − 4ac)

Value of D Nature of roots
D > 0 and a perfect square Real, unequal and rational
D > 0 but not a perfect square Real, unequal and irrational
D = 0 Real and equal
D < 0 No real roots (imaginary)

Example: Solve x² − 7x + 12 = 0. Factor: (x − 3)(x − 4) = 0, so x = 3 or 4. Check: sum = 7 = −b/a and product = 12 = c/a.

Example: For what value of k does x² + kx + 9 = 0 have equal roots? D = k² − 36 = 0, so k = ±6.

Comparing Roots (Bank Exam Style)

Bank exams ask you to solve two quadratics (one in x, one in y) and compare. Find all roots, then check: if every x is greater than every y, then x > y; if some overlap, the relation cannot be established. Using the sign rule saves time: if the equation is x² + bx + c with both b and c positive, both roots are negative.

5. The Value-Putting Trick

When an expression is symmetric and options are numbers, put simple values that satisfy the condition.

Example: If a + b + c = 0, find (a² + b² + c²) ÷ (ab + bc + ca). Put a = 1, b = 1, c = −2. Numerator = 1 + 1 + 4 = 6; denominator = 1 − 2 − 2 = −3. Answer = −2.

Always pick values that are allowed (avoid zero if there is division) and verify with a second set if options are close.

6. Polynomials: Remainder and Factor Theorem

  • Remainder theorem: When p(x) is divided by (x − a), the remainder is p(a).
  • Factor theorem: If p(a) = 0, then (x − a) is a factor of p(x).

Example: Remainder when x³ − 2x² + 4 is divided by (x − 2) = 8 − 8 + 4 = 4.

Common Mistakes to Avoid

  1. Writing (a + b)² = a² + b², forgetting the 2ab term.
  2. Using wrong signs in sum of roots: it is −b/a, not b/a.
  3. Forgetting the condition a ≠ 0 in quadratic questions.
  4. Putting values that make a denominator zero in value-putting.
  5. Not checking whether a word problem needs a positive answer (age or count cannot be negative).

How to Prepare Algebra in 10 Days

  • Days 1–2: Learn identities and solve 40 direct questions.
  • Days 3–4: Master the x + 1/x family with 30 questions.
  • Days 5–6: Practise linear equations and word problems.
  • Days 7–8: Quadratic equations, nature of roots and comparison questions.
  • Days 9–10: Previous year SSC and CDS questions with a timer.

Algebra rewards pattern recognition. After solving enough questions, you will start to see which identity a question is built on within a few seconds of reading it.

Where Algebra Appears in Different Exams

In SSC CGL Tier 1 and Tier 2, algebra usually contributes several direct questions on identities, the x + 1/x family and simple equations. CDS and NDA papers test quadratic equations, roots and polynomials in more depth. Bank prelims focus on comparing the roots of two quadratic equations. UPSC CSAT uses algebra inside word problems, so the skill being tested is forming the right equation from a sentence. Adjust your practice to the exam you are targeting.

Frequently Asked Questions

Which algebraic identities are most important for SSC exams?

(a + b)², (a − b)², a² − b², (a + b)³, a³ + b³, a³ − b³ and a³ + b³ + c³ − 3abc are the most frequently used identities.

What is the formula for x² + 1/x² if x + 1/x is given?

If x + 1/x = k, then x² + 1/x² = k² − 2.

What is the formula for x³ + 1/x³?

If x + 1/x = k, then x³ + 1/x³ = k³ − 3k.

What is the quadratic formula?

For ax² + bx + c = 0, x = [−b ± √(b² − 4ac)] ÷ 2a.

What is the discriminant of a quadratic equation?

The discriminant is D = b² − 4ac. It tells whether the roots are real and equal, real and unequal, or not real.

What is the sum and product of roots?

Sum of roots = −b/a and product of roots = c/a for ax² + bx + c = 0.

If a + b + c = 0, what is a³ + b³ + c³?

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

When do two linear equations have no solution?

When a1/a2 = b1/b2 ≠ c1/c2. The lines are parallel and never meet.

What is the value-putting method in algebra?

It means putting simple numbers that satisfy the given condition into the expression to find the answer quickly, especially in symmetric expressions.

Is algebra asked in UPSC CSAT?

CSAT rarely asks pure identities, but linear equations are used in many word problems on numbers, ages, money and work.

What is the remainder theorem?

When a polynomial p(x) is divided by (x − a), the remainder equals p(a).

How can I get faster at algebra?

Memorise identities, practise the x + 1/x family, use value-putting wherever possible and solve previous year questions with a timer.

About the Author

Komal Gupta

UPSC Exam Expert & Content Editor

Komal is a senior UPSC exam strategist with 8+ years of experience helping aspirants crack CSE, CDS and NDA exams. She writes in-depth notification breakdowns, exam pattern guides and preparation strategies.