Algebra — Core

Equations are the language algebra uses to pin down the unknown, and a handful of identities can turn an ugly multi-line expansion into a single-step answer. Mastering the discriminant and the sum/product rules for quadratics means you often don't have to solve for the roots at all.

Definition: A linear equation has variables only to the first power (e.g., ax + b = 0). Its graph is a straight line, and it has exactly one solution (provided a ≠ 0).

Definition: A quadratic equation takes the form ax² + bx + c = 0, where a ≠ 0. Its graph is a parabola and it can have 0, 1 or 2 real roots depending on the discriminant.

Definition: The discriminant D = b² − 4ac is the expression under the square root in the quadratic formula. It determines the nature of the roots without solving the equation.

Linear Equations

Single variable: ax + b = 0 → x = −b/a. Every linear equation in one variable has exactly one solution.

Two variables — two equations needed: Given two linear equations in x and y, solve by:

• Substitution: express one variable from one equation, substitute into the other.
• Elimination: multiply equations by constants to match a coefficient, then subtract.
• Cross-multiplication shortcut for the system ax + by + c = 0 and a'x + b'y + c' = 0:

x / (bc' − b'c) = y / (a'c − ac') = 1 / (ab' − a'b)

This is fast in exams when asked to find x + y or just one variable.

Special cases:

• If the two equations are multiples of each other (same line), there are infinitely many solutions.
• If they are parallel (same slope, different intercept), there is no solution.

Quadratic Equations

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

Quadratic formula: x = (−b ± √D) / 2a, where D = b² − 4ac.

Discriminant guide:

D value	Nature of roots
D > 0	Two distinct real roots
D = 0	One repeated real root (equal roots)
D < 0	No real roots (two complex conjugate roots)

Common misconception: Many students think D = 0 means "no solution". In fact D = 0 gives a repeated real root — a single value that satisfies the equation. It is D < 0 that yields no real solution. This distinction is directly tested in Class 10 board exams and SSC tier-1.

Sum and product of roots (Vieta's formulas): For roots α and β:

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

These are enormously useful: if a question says "the sum of roots is 7 and product is 12, write the equation" → x² − 7x + 12 = 0. Or if "one root is 3, find the other" and you know the product is 12 → other root = 4.

Forming the quadratic from roots:
x² − (sum of roots)x + (product of roots) = 0.

Algebraic Identities — the Complete Set

These must be memorised and instantly recognisable:

Identity	Use case
(a+b)² = a² + 2ab + b²	Expand or factorise a perfect square
(a−b)² = a² − 2ab + b²	Same; remember the middle term is negative
a² − b² = (a+b)(a−b)	Difference of two squares — instant factorisation
(a+b)³ = a³ + 3a²b + 3ab² + b³	Cube of a sum
(a−b)³ = a³ − 3a²b + 3ab² − b³	Cube of a difference
a³ + b³ = (a+b)(a² − ab + b²)	Sum of two cubes
a³ − b³ = (a−b)(a² + ab + b²)	Difference of two cubes
a³+b³+c³ − 3abc = (a+b+c)(a²+b²+c²−ab−bc−ca)	Three-variable factorisation
If a+b+c = 0, then a³+b³+c³ = 3abc	The most frequently tested special case

Real-world example: The discriminant tells a thrown cricket ball whether it will ever reach a given height. Set the height equation equal to that target height → the discriminant says: D > 0 means the ball passes that height twice (going up and coming down), D = 0 means it barely grazes that height at its peak, D < 0 means it never gets there.

Laws of Indices

Law	Expression
Multiplication of same base	aᵐ × aⁿ = aᵐ⁺ⁿ
Division of 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	(ab)ⁿ = aⁿbⁿ
Quotient	(a/b)ⁿ = aⁿ/bⁿ

The fractional-exponent law a^(1/2) = √a and a^(1/3) = ∛a is the bridge between powers and roots, and it unlocks simplification of many index-based expressions.

Worked Examples

Question: If α and β are roots of x² − 5x + 6 = 0, find α² + β².
Solution:
Step 1: α + β = 5 (−b/a = −(−5)/1); αβ = 6 (c/a).
Step 2: α² + β² = (α+β)² − 2αβ = 25 − 12.
Conclusion: 13.

Question: Solve: 2x + 3y = 8 and 3x − y = 5.
Solution:
Step 1: From equation 2: y = 3x − 5. Substitute into equation 1: 2x + 3(3x − 5) = 8 → 2x + 9x − 15 = 8 → 11x = 23.
Step 2: x = 23/11 ≈ 2.09. Then y = 3(23/11) − 5 = (69 − 55)/11 = 14/11 ≈ 1.27.
Conclusion: x = 23/11, y = 14/11.

Question: If a + b + c = 0, find the value of a³ + b³ + c³.
Solution: Directly apply the identity: if a + b + c = 0, then a³ + b³ + c³ = 3abc.

Question: Factorise x² − 25.
Solution: Recognise as difference of squares: (x+5)(x−5).

Question: Find the discriminant of 3x² − 5x + 2 = 0 and state the nature of roots.
Solution:
Step 1: D = b² − 4ac = (−5)² − 4(3)(2) = 25 − 24 = 1.
Step 2: D > 0 → two distinct real roots.
Conclusion: Two distinct real rational roots (since D is a perfect square).

Why it matters: These identities and discriminant rules appear across Class 10 boards, RRB NTPC, SSC CGL and state PSC exams. The a+b+c = 0 → a³+b³+c³ = 3abc shortcut and the sum/product of roots formulas regularly appear as one-mark questions that take 30 seconds if known, and 3 minutes if not.

In exams like RRB and SSC, algebra rewards pattern recognition far more than brute-force calculation — the candidate who spots "this is the a³ + b³ form" finishes in seconds while others grind. This lesson drills the high-yield algebraic identities and equation tricks through worked examples, building both speed and accuracy.

Definition: An algebraic identity is an equation that holds true for all values of its variables (e.g., (a+b)² = a² + 2ab + b²); it is used to transform or simplify expressions without solving from scratch.
Definition: The discriminant of ax² + bx + c = 0 is D = b² − 4ac; its sign classifies the roots: D > 0 (two distinct real roots), D = 0 (two equal real roots), D < 0 (complex roots).

Foundation identities to memorise

These must be instant recall:

Identity	Expanded form
(a + b)²	a² + 2ab + b²
(a − b)²	a² − 2ab + b²
(a + b)(a − b)	a² − b²
(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)

The shortcut form of a³ + b³ that avoids expanding cubes: a³ + b³ = (a+b)³ − 3ab(a+b).

Solving simultaneous equations by substitution

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

Solution:

Step 1: Make one variable the subject from the simpler equation. From 4x − y = 5: y = 4x − 5.

Step 2: Substitute into the first equation: 2x + 3(4x − 5) = 13 → 2x + 12x − 15 = 13 → 14x = 28.

Step 3: x = 2. Back-substitute: y = 4(2) − 5 = 3.

Conclusion: x = 2, y = 3. Always verify: 2(2) + 3(3) = 4 + 9 = 13 ✓ and 4(2) − 3 = 5 ✓.

Why choose substitution vs elimination? Substitution is fastest when one equation already has a coefficient of 1 (as above). Elimination (adding/subtracting multiples of equations) is faster when coefficients are equal.

Building a quadratic from its roots

Every quadratic x² + bx + c = 0 with roots α and β satisfies:

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

So you can reverse-engineer: x² − (sum)x + (product) = 0.

Question: A quadratic has root-sum 7 and root-product 12. Find it, and identify the roots.

Solution:

Step 1: The quadratic is x² − 7x + 12 = 0.

Step 2: Factor: find two numbers that multiply to 12 and add to 7 → (3, 4). So (x − 3)(x − 4) = 0.

Conclusion: The quadratic is x² − 7x + 12 = 0, with roots 3 and 4.

The identity shortcuts — the real time-savers

Pattern 1 — Sum of cubes with shortcut:

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

Solution:

Step 1: Use a³ + b³ = (a+b)³ − 3ab(a+b).

Step 2: = (5)³ − 3 × 6 × 5 = 125 − 90.

Conclusion: a³ + b³ = 35. (Factoring to find individual values of a and b would take much longer.)

Pattern 2 — The x + 1/x family:

This appears so often in RRB/SSC that it deserves its own section.

Starting from x + 1/x = 4:

• Square it: (x + 1/x)² = x² + 2 + 1/x² → 16 = x² + 2 + 1/x² → x² + 1/x² = 14.
• Cube it: x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x) = 64 − 12 = 52.
• x⁴ + 1/x⁴: square the result for x² + 1/x²: (14)² − 2 = 194.

The chain is: use x + 1/x → get x² + 1/x² (subtract 2 after squaring) → get x³ + 1/x³ (subtract 3 times the original after cubing).

Pattern 3 — Three-variable zero-sum:

If a + b + c = 0, then a³ + b³ + c³ = 3abc. So (a³ + b³ + c³)/abc = 3.

This is a one-second answer once you recognise the sum-zero condition.

Pattern 4 — (a² + b² + c²) from (a + b + c):

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca). So a² + b² + c² = (a+b+c)² − 2(ab+bc+ca).

Equal roots via the discriminant

Question: Find k so that 2x² + kx + 8 = 0 has equal roots.

Solution:

Step 1: Equal roots ⟺ D = 0, where D = b² − 4ac.

Step 2: Here a = 2, b = k, c = 8: k² − 4(2)(8) = 0 → k² = 64.

Conclusion: k = ±8.

You can verify: with k = 8, the equation is 2x² + 8x + 8 = 0, or x² + 4x + 4 = (x+2)² = 0, giving the double root x = −2. ✓

Why it matters: Time is the scarcest resource in competitive exams. Recognising that a question is "really" one of a handful of standard identities lets you skip pages of algebra and avoid sign errors, directly boosting both speed and accuracy on the quantitative section.

Real-world example: Think of these identities like keyboard shortcuts. A new computer user clicks through five menus to copy text; an expert hits Ctrl+C. Similarly, an aspirant who memorises a³ + b³ = (a+b)³ − 3ab(a+b) "copies" the answer in one line, while a peer expands everything the long way and runs out of time.

Common misconception: Many students believe x² + 1/x² = (x + 1/x)². Squaring x + 1/x actually gives x² + 2 + 1/x² (there is a middle term of 2 × x × 1/x = 2). So you must subtract 2 to isolate x² + 1/x². Dropping that "+2" is one of the most frequent and costly slips in these problems. Similarly, (x + 1/x)³ = x³ + 3x + 3/x + 1/x³, so x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x) — always subtract 3 times the original.

In RRB ALP CBT-1 you answer 75 questions in 60 minutes; Mathematics carries 20 questions (with Reasoning 25, General Science 20, General Awareness 10). Algebra typically contributes 2–4 questions built almost entirely on standard identities — so knowing the identities cold turns each into a 20-second recall, not a page of working. CBT-2 Part A raises Maths to 25 questions, so this pays twice.

Must-know identities

• (a+b)² = a² + 2ab + b² and (a−b)² = a² − 2ab + b²
• a² − b² = (a+b)(a−b)
• (a+b)³ = a³ + b³ + 3ab(a+b); (a−b)³ = a³ − b³ − 3ab(a−b)
• 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)
• If a+b+c = 0 ⇒ a³+b³+c³ = 3abc

The x + 1/x family (very high yield)

Given	Result
x + 1/x = k	x² + 1/x² = k² − 2
x − 1/x = k	x² + 1/x² = k² + 2
x + 1/x = k	x³ + 1/x³ = k³ − 3k
x + 1/x = k	x⁴ + 1/x⁴ = (k²−2)² − 2

Exam Tricks & Tips

• 🎯 Spot "a+b+c = 0" — the answer is almost always 3abc; setters love this shortcut.
• 🎯 For x+1/x problems, never solve for x; jump straight to k²−2 or k³−3k.
• 🎯 a²+b²+c² − ab−bc−ca = ½[(a−b)²+(b−c)²+(c−a)²] — always ≥ 0, useful for "find minimum" items.
• 🎯 Factor a²−b² on sight; it collapses ugly numbers (e.g. 51²−49² = 100×2 = 200).
• 🎯 Memorise (a+b)² and (a−b)² so you can mentally square numbers like 102² = 10000+400+4 = 10404.
• ❌ Common mistake: writing (a+b)³ = a³+b³ (forgetting the +3ab(a+b) term) — this loses easy marks.

Expected exam pattern

Direct identity substitution ("If x+1/x = 3, find x³+1/x³"), simplify-and-factor, or "find the value" where a+b+c = 0. Numbers are chosen to be clean once the right identity is applied.

Quick recap

Master the seven core identities and the x+1/x table. Look for a+b+c = 0 and a²−b² patterns first — they convert most Algebra questions into one-line answers.

Cover the answer, recall, then check. 12 cards on the algebra identities RRB ALP loves.

Q1. Expand (a+b)².
A1. a² + 2ab + b².

Q2. Factorise a² − b².
A2. (a+b)(a−b).

Q3. If a+b+c = 0, what is a³+b³+c³?
A3. 3abc.

Q4. If x + 1/x = 4, find x² + 1/x².
A4. k² − 2 = 16 − 2 = 14.

Q5. If x + 1/x = 3, find x³ + 1/x³.
A5. k³ − 3k = 27 − 9 = 18.

Q6. If x − 1/x = 2, find x² + 1/x².
A6. k² + 2 = 4 + 2 = 6.

Q7. Expand (a+b)³.
A7. a³ + b³ + 3ab(a+b) = a³ + 3a²b + 3ab² + b³.

Q8. Factorise a³ + b³.
A8. (a+b)(a² − ab + b²).

Q9. Factorise a³ − b³.
A9. (a−b)(a² + ab + b²).

Q10. Compute 51² − 49² using a factor trick.
A10. (51+49)(51−49) = 100 × 2 = 200.

Q11. Give the identity for a³+b³+c³−3abc.
A11. (a+b+c)(a²+b²+c²−ab−bc−ca).

Q12. Compute 102² mentally using (a+b)².
A12. 100² + 2·100·2 + 2² = 10000 + 400 + 4 = 10404.