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Algebra — Concepts, Shortcuts & Practice Bank · Chapter 1

Algebra — Complete Guide

Table of Contents

  1. Introduction to the Topic
  2. Basic Algebraic Identities
  3. Linear Equations (One and Two Variables)
  4. Quadratic Equations
  5. Polynomials and Factorization
  6. Surds and Indices
  7. Sequences and Series (AP, GP)
  8. Special Algebraic Manipulations (Value-Finding Tricks)
  9. Word Problems Converted to Algebra
  10. Common Shortcuts and Tricks
  11. Common Mistakes to Avoid
  12. Solved Examples (25 fully worked problems)
  13. Practice Question Bank (50 questions)
  14. Answer Key with Quick Solutions

1. Introduction to the Topic

Algebra in competitive exams is less about solving complex equations and more about fast identity-based manipulation. Most questions give an expression like “if x + 1/x = 5, find x³ + 1/x³” and expect the answer using standard identities rather than actually solving for x. This chapter also covers linear and quadratic equations, factorization, surds/indices, and arithmetic/geometric progressions — all frequently tested in SSC, RRB, and banking exams (typically 3–6 questions per paper).

The exam-winning approach is: memorize the identity toolbox cold, and practice recognizing which identity an expression is hinting at within seconds.

2. Basic Algebraic Identities

Figure: Visual proof of (a+b)² = a² + 2ab + b².

Squares 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. a² + b² = (a+b)² − 2ab = (a−b)² + 2ab

Cubes 7. (a+b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a+b) 8. (a−b)³ = a³ − 3a²b + 3ab² − b³ = a³ − b³ − 3ab(a−b) 9. a³ + b³ = (a+b)(a² − ab + b²) = (a+b)³ − 3ab(a+b) 10. a³ − b³ = (a−b)(a² + ab + b²) = (a−b)³ + 3ab(a−b)

Three-Variable Identities 11. (a+b+c)² = a² + b² + c² + 2(ab+bc+ca) 12. a² + b² + c² − ab − bc − ca = ½[(a−b)² + (b−c)² + (c−a)²] 13. a³ + b³ + c³ − 3abc = (a+b+c)(a² + b² + c² − ab − bc − ca) 14. Special case: If a+b+c = 0, then a³ + b³ + c³ = 3abc

Reciprocal-type Identities (very frequently tested) 15. If x + 1/x = k, then x² + 1/x² = k² − 2 16. If x + 1/x = k, then x³ + 1/x³ = k³ − 3k 17. If x − 1/x = k, then x² + 1/x² = k² + 2 18. If x − 1/x = k, then x³ − 1/x³ = k³ + 3k 19. If x + 1/x = k, then x⁴ + 1/x⁴ = (k²−2)² − 2 20. If x² + 1/x² = k, then x + 1/x = √(k+2) and x − 1/x = √(k−2)

3. Linear Equations (One and Two Variables)

One variable: Standard form ax + b = 0 ⇒ x = −b/a

Two variables (simultaneous equations): For a₁x + b₁y = c₁ and a₂x + b₂y = c₂, use: - Elimination method: Multiply equations to match coefficients, then add/subtract. - Substitution method: Express one variable in terms of the other and substitute. - Cross-multiplication formula: x / (b₁c₂ − b₂c₁) = y / (c₁a₂ − c₂a₁) = 1 / (a₁b₂ − a₂b₁)

Nature of solutions for two linear equations: - Unique solution: a₁/a₂ ≠ b₁/b₂ - No solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂ - Infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂

4. Quadratic Equations

Figure: Roots of a quadratic placed on the number line.
Figure: Graph of a quadratic equation — roots are where the curve crosses the x-axis.

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

Quadratic formula: x = [−b ± √(b² − 4ac)] / 2a

Discriminant (D) = b² − 4ac: - D > 0: two distinct real roots - D = 0: two equal real roots - D < 0: no real roots (complex/imaginary roots) - D is a perfect square (and D>0): roots are rational

Sum and Product of Roots: For ax²+bx+c=0 with roots α, β: - Sum of roots: α + β = −b/a - Product of roots: αβ = c/a

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

Factorization method (splitting the middle term): For x² + bx + c = 0, find two numbers whose sum is b and product is c.

Common derived expressions using roots: - α² + β² = (α+β)² − 2αβ - (α−β)² = (α+β)² − 4αβ - α³ + β³ = (α+β)³ − 3αβ(α+β) - 1/α + 1/β = (α+β)/αβ

5. Polynomials and Factorization

Degree: The highest power of the variable in a polynomial.

Remainder Theorem: If a polynomial p(x) is divided by (x − a), the remainder is p(a).

Factor Theorem: (x − a) is a factor of p(x) if and only if p(a) = 0.

Common factorization techniques: - Taking out common factors - Grouping terms - Using identities (difference of squares, sum/difference of cubes) - Splitting the middle term for quadratics - Trial roots (±1, ±2, factors of the constant term) for higher-degree polynomials

6. Surds and Indices

Laws of Indices: - aᵐ × aⁿ = aᵐ⁺ⁿ - aᵐ / aⁿ = aᵐ⁻ⁿ - (aᵐ)ⁿ = aᵐⁿ - a⁰ = 1 (a≠0) - a⁻ⁿ = 1/aⁿ - a^(1/n) = ⁿ√a - (ab)ⁿ = aⁿbⁿ - (a/b)ⁿ = aⁿ/bⁿ

Rationalization of Surds: To rationalize 1/(√a + √b), multiply numerator and denominator by (√a − √b): 1/(√a+√b) = (√a−√b)/(a−b)

Common surd simplification trick: If x = √(a + √(a + √(a + …))) to infinity, then x² = a + x ⇒ x² − x − a = 0 ⇒ x = [1 + √(1+4a)]/2

Similarly, if x = √(a × √(a × √(a × …))) to infinity, then x² = ax ⇒ x = a (for a>0, non-trivial case handled via x = a^(2/(2-1))… typically such problems use x = a directly when infinite product nesting of same form, verified via x=√(ax) ⇒x²=ax⇒x=a)

7. Sequences and Series (AP, GP)

Arithmetic Progression (AP): a, a+d, a+2d, a+3d, … - nth term: Tₙ = a + (n−1)d - Sum of n terms: Sₙ = n/2 [2a + (n−1)d] = n/2 [first term + last term]

Geometric Progression (GP): a, ar, ar², ar³, … - nth term: Tₙ = a·rⁿ⁻¹ - Sum of n terms (r≠1): Sₙ = a(rⁿ−1)/(r−1) [if r>1] or a(1−rⁿ)/(1−r) [if r<1] - Sum to infinity (|r|<1): S∞ = a/(1−r)

Useful summation formulas: - Sum of first n natural numbers: n(n+1)/2 - Sum of squares of first n natural numbers: n(n+1)(2n+1)/6 - Sum of cubes of first n natural numbers: [n(n+1)/2]²

8. Special Algebraic Manipulations (Value-Finding Tricks)

These are the “plug and use identity” questions that dominate exam papers.

Pattern: “If x + 1/x = 4, find x² + 1/x².” - Use identity 15: x² + 1/x² = 4² − 2 = 14

Pattern: “If x − 1/x = 3, find x³ − 1/x³.” - Use identity 18: x³ − 1/x³ = 3³ + 3×3 = 27+9 = 36

Pattern: “If a+b+c=0, find a³+b³+c³ in terms of abc.” - Directly: a³+b³+c³ = 3abc (identity 14)

Pattern: “If a² + b² + c² = 50 and a+b+c = 12, find ab+bc+ca.” - (a+b+c)² = a²+b²+c²+2(ab+bc+ca) - 144 = 50 + 2(ab+bc+ca) ⇒ ab+bc+ca = 47

9. Word Problems Converted to Algebra

Word problems on ages, numbers, fractions, and mixtures are very common. The technique is always: assign variables, translate sentences into equations, and solve.

Pattern (ages): “5 years ago, a father was 7 times as old as his son. 5 years hence, the father will be 3 times as old as his son. Find their present ages.” - Let son’s present age = x, father’s = y. - (y−5) = 7(x−5) ⇒ y = 7x−30 - (y+5) = 3(x+5) ⇒ y = 3x+10 - 7x−30 = 3x+10 ⇒ 4x=40 ⇒ x=10, y=40 - Son = 10 years, Father = 40 years

Pattern (two-digit numbers): “A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits get reversed. Find the number.” - Let number = 10a+b. 10a+b = 4(a+b) ⇒ 6a=3b ⇒ b=2a - 10a+b+18 = 10b+a ⇒ 9a+18=9b ⇒ a+2=b (consistent with b=2a only if a=2,b=4) - Number = 24

10. Common Shortcuts and Tricks

  1. Memorize all reciprocal identities (x+1/x type) — these appear in nearly every “find the value” algebra question.
  2. When a question gives x + 1/x, always compute x² + 1/x² first (subtract 2), then x³+1/x³ (using both), building up systematically.
  3. For quadratic roots, don’t always solve fully — sum/product of roots (−b/a, c/a) often answers the question directly without finding α, β individually.
  4. a+b+c=0 special case is a favorite trap/shortcut — instantly gives a³+b³+c³=3abc.
  5. For surd equations with infinite nesting, set up x² = a + x (for nested addition) or recognize x=a (for nested multiplication of same radicand), rather than attempting to expand terms.
  6. In “find unknown constant so that equation has equal roots” problems, just set discriminant = 0 and solve for the constant.
  7. For AP/GP word problems, identify a (first term) and d or r (common difference/ratio) immediately from the given terms before applying any formula.
  8. When solving simultaneous equations under time pressure, prefer elimination when coefficients are simple multiples of each other; use cross-multiplication when there’s no shortcut.
  9. Check answer options by substitution whenever solving takes too long algebraically — plugging in MCQ options is often faster than deriving.
  10. Rationalizing denominators is often just a stepping stone; check if the question wants the simplified surd value or requires you to also add it to another term.

11. Common Mistakes to Avoid

  • Sign errors when squaring (a−b)² — remember the middle term is −2ab, not +2ab.
  • Forgetting the “±” when taking square roots while solving quadratics via extraction.
  • Misapplying the a+b+c=0 identity when the condition doesn’t actually hold.
  • Confusing x+1/x and x−1/x identities (the sign of the “+2” or “+3k” term flips between them — memorize both separately, don’t derive one from memory of the other under pressure).
  • In word problems, defining variables ambiguously (e.g., not being clear whether x is “5 years ago” or “present age”) — always define present age as the variable and shift by addition/subtraction as needed.
  • Errors in expanding (a+b+c)² — forgetting the cross terms 2(ab+bc+ca).
  • Not checking whether both roots of a quadratic are valid in context (e.g., a “speed” or “age” cannot be negative).

12. Solved Examples (25 Fully Worked Problems)

Example 1. If x + 1/x = 5, find x² + 1/x². Solution: x²+1/x² = 5²−2 = 23.

Example 2. If x − 1/x = 4, find x² + 1/x². Solution: x²+1/x² = 4²+2 = 18.

Example 3. If x + 1/x = 3, find x³ + 1/x³. Solution: x³+1/x³ = 3³ − 3(3) = 27−9 = 18.

Example 4. If x − 1/x = 5, find x³ − 1/x³. Solution: x³−1/x³ = 5³ + 3(5) = 125+15 = 140.

Example 5. If a+b = 10 and ab = 21, find a²+b². Solution: a²+b² = (a+b)²−2ab = 100−42 = 58.

Example 6. If a−b = 4 and ab = 12, find a²+b² and (a+b)². Solution: a²+b²=(a−b)²+2ab=16+24=40. (a+b)²=(a−b)²+4ab=16+48=64.

Example 7. Solve for x: 2x² − 7x + 3 = 0. Solution: Splitting middle term: 2x²−6x−x+3=0 ⇒ 2x(x−3)−1(x−3)=0 ⇒ (2x−1)(x−3)=0 ⇒ x=1/2 or x=3.

Example 8. Find the sum and product of the roots of 3x² − 5x + 2 = 0. Solution: Sum = −(−5)/3 = 5/3. Product = 2/3 = 2/3.

Example 9. If one root of x² − 7x + k = 0 is 4, find k and the other root. Solution: Sum of roots=7 ⇒ other root=3. Product=k=4×3=12.

Example 10. For what value of k does x² + kx + 9 = 0 have equal roots? Solution: D=0 ⇒ k²−36=0 ⇒ k=±6.

Example 11. If a+b+c=0, and a=3, b=−5, find a³+b³+c³ in terms of abc, and verify numerically. Solution: c=−(a+b)=2. a³+b³+c³ should equal 3abc. LHS: 27+(−125)+8=−90. RHS: 3×3×(−5)×2=−90. Verified, both = −90.

Example 12. If x=2+√3, find x+1/x and x²+1/x². Solution: 1/x = 1/(2+√3) = (2−√3)/((2+√3)(2−√3)) = (2−√3)/(4−3) = 2−√3. x+1/x = (2+√3)+(2−√3) = 4. x²+1/x² = 4²−2 = 14.

Example 13. Simplify: (a² − b²)/(a − b) + (a² − c²)/(a-c)…

Let’s use a standard cleaner one: Simplify (x³−8)/(x−2). Solution: x³−8 = x³−2³ = (x−2)(x²+2x+4). So (x³−8)/(x−2) = x²+2x+4.

Example 14. If a² + b² + c² = 83 and ab+bc+ca = 71, find (a+b+c)². Solution: (a+b+c)² = a²+b²+c²+2(ab+bc+ca) = 83+142 = 225, so a+b+c = 15.

Example 15. Solve the simultaneous equations: 2x+3y=13, 3x−2y=0. Solution: From second: y=1.5x. Substitute: 2x+3(1.5x)=13 ⇒ 2x+4.5x=13 ⇒6.5x=13⇒x=2. y=1.5×2=3. So x=2, y=3.

Example 16. Find the value of k for which the system 2x+3y=7, kx+9y=21 has infinite solutions. Solution: a₁/a₂=b₁/b₂=c₁/c₂ ⇒ 2/k=3/9=7/21=1/3 ⇒ 2/k=1/3 ⇒ k=6.

Example 17. If x = √5+2, find the value of x − 1/x. Solution: 1/x = 1/(√5+2) = (√5−2)/((√5)²−2²) = (√5−2)/1 = √5−2. x−1/x = (√5+2)−(√5−2) = 4.

Example 18. A number consists of two digits whose sum is 9. If 27 is subtracted from the number, the digits interchange their places. Find the number. Solution: Let digits be a,b with a+b=9, number=10a+b. 10a+b−27=10b+a ⇒9a−9b=27⇒a−b=3. a+b=9, a−b=3 ⇒ a=6,b=3. Number=63.

Example 19. The sum of the first 20 terms of an AP is 610, and the first term is 4. Find the common difference. Solution: Sₙ=n/2[2a+(n−1)d] ⇒ 610=10[8+19d] ⇒ 61=8+19d ⇒19d=53⇒d=53/19≈2.79.

(Let’s pick cleaner numbers for clarity, but keeping as instructive that non-integer answers do occur.)

Example 20. Find the sum of the first 15 terms of the GP: 3, 6, 12, 24, … Solution: a=3, r=2. Sₙ=a(rⁿ−1)/(r−1)=3(2¹⁵−1)/1=3(32768−1)=3×32767=98301.

Example 21. If x²+y²=29 and xy=10, find (x+y)² and (x−y)². Solution: (x+y)²=29+20=49 ⇒x+y=7. (x−y)²=29−20=9⇒x−y=3.

Example 22. Factorize: x²−5x−14. Solution: Need two numbers with sum−5, product−14: −7 and 2. x²−7x+2x−14=x(x−7)+2(x−7)=(x−7)(x+2).

Example 23. If (x−3) is a factor of x³−3x²+kx−12, find k. Solution: By factor theorem, p(3)=0: 27−27+3k−12=0 ⇒3k=12⇒k=4.

Example 24. A father’s age is 3 times his son’s age. 12 years later, the father’s age will be twice the son’s age. Find their present ages. Solution: Father=3x, son=x. (3x+12)=2(x+12)⇒3x+12=2x+24⇒x=12. Son=12 years, Father=36 years.

Example 25. Simplify: √(7+4√3). Solution: 7+4√3 = 4+3+2×2×√3 = 2²+(√3)²+2×2×√3 = (2+√3)². So √(7+4√3) = 2+√3.

13. Practice Question Bank (50 Questions)

Set A — Identities (Q1–Q12) 1. If x+1/x=6, find x²+1/x². 2. If x−1/x=7, find x²+1/x². 3. If x+1/x=4, find x³+1/x³. 4. If x−1/x=6, find x³−1/x³. 5. If a+b=12, ab=27, find a²+b². 6. If a−b=6, a²+b²=68, find ab. 7. If x²+1/x²=27, find x−1/x (positive value). 8. If x²+1/x²=18, find x+1/x (positive value). 9. If a+b+c=15, a²+b²+c²=83, find ab+bc+ca. 10. If a+b+c=0, a²+b²+c²=54, find ab+bc+ca. 11. If a=5, b=−3, c=−2, verify a³+b³+c³=3abc using the a+b+c=0 property. 12. If x+y=8 and x²+y²=34, find xy.

Set B — Quadratic Equations (Q13–Q24) 13. Solve: x²−9x+20=0. 14. Solve: x²+2x−15=0. 15. Find the sum and product of roots of 5x²+3x−2=0. 16. If one root of x²−8x+k=0 is 3, find k and the other root. 17. For what value of m does x²+mx+16=0 have equal roots? 18. Solve: 6x²−x−2=0. 19. If α,β are roots of x²−5x+6=0, find α²+β². 20. If α,β are roots of x²−3x−4=0, find 1/α+1/β. 21. Find the quadratic equation whose roots are 4 and −3. 22. If the roots of x²−px+q=0 are in the ratio 2:3, find the relation between p and q. 23. Solve: x²−(√3+1)x+√3=0. 24. Find the value of k so that the equation kx²+4x+1=0 has real and equal roots.

Set C — Polynomials & Factorization (Q25–Q32) 25. Factorize: x²−11x+30. 26. Factorize: x²+x−12. 27. Factorize: 4x²−9y². 28. Factorize: x³−27. 29. Factorize: 8x³+1. 30. If (x−2) is a factor of x³−2x²+ax−4, find a. 31. Find the remainder when x³−4x²+5x−2 is divided by (x−1). 32. Factorize: x³−6x²+11x−6.

Set D — Surds and Indices (Q33–Q38) 33. Simplify: (2^5 × 23)/24. 34. Simplify: (27)^(2/3). 35. Rationalize: 1/(√7−√5). 36. If x=3+2√2, find x+1/x. 37. Simplify: √(11+6√2). 38. Solve for x: 3^(2x−1) = 27.

Set E — Sequences and Series (Q39–Q44) 39. Find the 15th term of the AP: 3, 7, 11, 15, … 40. Find the sum of the first 25 terms of the AP: 2, 5, 8, 11, … 41. The 4th term of a GP is 24 and the 7th term is 192. Find the first term and common ratio. 42. Find the sum of the first 10 terms of the GP: 5, 10, 20, 40, … 43. Find the sum to infinity of the GP: 8, 4, 2, 1, … 44. Find the sum of the first 30 natural numbers.

Set F — Word Problems (Q45–Q50) 45. The sum of a two-digit number and the number obtained by reversing its digits is 121. If the digits differ by 3, find the number. 46. A father is 4 times as old as his son. After 20 years, he will be twice as old. Find their present ages. 47. Five years ago, a man was 3 times as old as his daughter. Ten years hence, he will be twice as old as her. Find their present ages. 48. The denominator of a fraction is 4 more than the numerator. If 3 is added to both, the fraction becomes 3/4. Find the original fraction. 49. Two numbers are in the ratio 3:5. If 9 is subtracted from each, the ratio becomes 12:23. Find the numbers. 50. The sum of two numbers is 25 and their product is 154. Find the numbers.

14. Answer Key with Quick Solutions

Set A 1. 6²−2=34 2. 7²+2=51 3. 4³−12=64−12=52 4. 6³+18=216+18=234 5. 144−54=90 6. a²+b²=(a−b)²+2ab⇒68=36+2ab⇒ab=16 7. x−1/x=√(27−2)=√25=5 8. x+1/x=√(18+2)=√20=2√5≈4.47 9. 225=83+2s⇒2s=142⇒s=71 10. 0=54+2s⇒s=−27 11. a+b+c=0;3abc=3(5)(−3)(−2)=90;a³+b³+c³=125−27−8=90 ✓ Verified 12. (x+y)²=64=34+2xy⇒2xy=30⇒xy=15

Set B 13. (x−5)(x−4)=0⇒x=5,4 14. (x+5)(x−3)=0⇒x=−5,3 15. sum=−3/5,product=−2/5 16. sum=8⇒other root=5;k=product=3×5=15 17. D=0⇒m²=64⇒m=±8 18. (3x−2)(2x+1)=0⇒x=2/3,−1/2 19. α+β=5,αβ=6;α²+β²=25−12=13 20. (α+β)/αβ=−3/−4… wait αβ=−4(from x²−3x−4,c/a=−4);α+β=3. 1/α+1/β=3/(−4)=−3/4 21. x²−(sum)x+product=x²−(1)x+(−12)=x²−x−12=0 22. roots 2k,3k;sum=5k=p;product=6k²=q⇒k=p/5;q=6p²/25⇒25q=6p² 23. (x−√3)(x−1)=0⇒x=√3,1 24. D=0⇒16−4k=0⇒k=4

Set C 25. (x−5)(x−6) 26. (x+4)(x−3) 27. (2x−3y)(2x+3y) 28. (x−3)(x²+3x+9) 29. (2x+1)(4x²−2x+1) 30. p(2)=0:8−8+2a−4=0⇒2a=4⇒a=2 31. p(1)=1−4+5−2=0 32. (x−1)(x−2)(x−3)

Set D 33. 2(5+3−4)=24=16 34. (27^(1/3))²=3²=9 35. multiply by(√7+√5)/(√7+√5):(√7+√5)/(7−5)=(√7+√5)/2 36. 1/x=1/(3+2√2)=(3−2√2)/(9−8)=3−2√2;x+1/x=(3+2√2)+(3−2√2)=6 37. 11+6√2=9+2+2×3×√2=(3+√2)²⇒3+√2 38. 2x−1=3⇒2x=4⇒x=2

Set E 39. a=3,d=4;T15=3+14×4=3+56=59 40. a=2,d=3;S25=25/2[4+24×3]=25/2×76=950 41. ar³=24,ar⁶=192⇒r³=8⇒r=2;a×8=24⇒a=3,r=2 42. a=5,r=2;S10=5(2^10−1)/1=5×1023=5115 43. a=8,r=1/2;S∞=8/(1−0.5)=16 44. 30×31/2=465

Set F 45. digits a,b:10a+b+10b+a=121⇒11(a+b)=121⇒a+b=11;a−b=3(or b−a=3)⇒a=7,b=4 or a=4,b=7. Number=74 or 47 46. 4x=father,x=son.(4x+20)=2(x+20)⇒4x+20=2x+40⇒2x=20⇒x=10.Son=10,Father=40 47. (y−5)=3(x−5);(y+10)=2(x+10).y=3x−10;y=2x+10.3x−10=2x+10⇒x=20,y=50.Daughter=20,Father=50 48. Let numerator=x,denom=x+4.(x+3)/(x+7)=3/4⇒4x+12=3x+21⇒x=9.Fraction=9/13 49. 3x−9)/(5x−9)=12/23⇒23(3x−9)=12(5x−9)⇒69x−207=60x−108⇒9x=99⇒x=11.Numbers=33 and 55 50. x+y=25,xy=154.x,y roots of t²−25t+154=0.Discriminant=625−616=9;t=(25±3)/2=14 or11.Numbers=14 and 11

15. Advanced Level (SSC CGL Tier-2 / RRB JE / Toughest Competitive Level)

15.1 Advanced Concepts

Higher-power reciprocal identities: Beyond x³+1/x³, exams at this level test x⁵+1/x⁵, x⁶+1/x⁶, and similar — built by successive multiplication of lower-power identities.

Multi-variable symmetric expressions: Expressions in a, b, c requiring simultaneous use of (a+b+c)², a³+b³+c³−3abc, and a²+b²+c²−ab−bc−ca together.

Conditional quadratic root problems: Questions requiring you to find a relationship between coefficients given a condition on the roots (e.g., “one root is the square of the other,” “roots differ by a given amount,” “roots are in a given ratio”).

Higher-degree polynomial division and remainder chains: Using the remainder theorem iteratively, or finding unknown coefficients from multiple given factor/remainder conditions simultaneously.

Nested surd equations and rationalization chains: Multi-step rationalization where the denominator itself contains a nested surd expression.

Simultaneous equations with three variables: Systems of 3 linear equations in 3 unknowns, solved via elimination — a step up from the standard 2-variable systems.

15.2 Advanced Solved Examples

Advanced Example 1. If x+1/x=5, find x⁵+1/x⁵. Solution: x²+1/x²=25−2=23. x³+1/x³=125−15=110. x⁵+1/x⁵ = (x²+1/x²)(x³+1/x³) − (x+1/x) = 23×110 − 5 = 2530−5 = 2525.

Advanced Example 2. If x+1/x=3, find x⁶+1/x⁶. Solution: x²+1/x²=9−2=7. x³+1/x³=27−9=18. x⁶+1/x⁶=(x³+1/x³)²−2=18²−2=324−2=322.

Advanced Example 3. If a+b+c=9 and a²+b²+c²=35, and abc=15, find a³+b³+c³. Solution: ab+bc+ca = [(a+b+c)²−(a²+b²+c²)]/2 = (81−35)/2=23. a³+b³+c³−3abc=(a+b+c)(a²+b²+c²−ab−bc−ca)=9×(35−23)=9×12=108. a³+b³+c³=108+3(15)=108+45=153.

Advanced Example 4. If the roots of x²−px+q=0 are α and α² (one root is the square of the other), find the relation between p and q. Solution: Sum: α+α²=p. Product: α×α²=α³=q ⇒ α=q^(1/3). Substituting into sum: q(1/3)+q(2/3)=p. Cubing approach (standard result): p³ = q(1+q^(1/3))³ is messy; the cleaner standard relation derived by eliminating α is: p³ − 3pq = q² + q is NOT standard; the commonly tested clean version instead gives: since α+α²=p and α³=q, we get α²+α=p ⇒ α(α+1)=p, and α³=q. The relation is typically left as: q(1/3)(1+q(1/3)) = p, i.e., p = q^(1/3) + q^(2/3). (This illustrates the substitution method; exam versions usually give specific numbers rather than asking for the general symbolic relation.)

Advanced Example 5. If one root of x²+px+q=0 is double the other, find the relation between p and q. Solution: Let roots be α and 2α. Sum=3α=−p ⇒ α=−p/3. Product=2α²=q ⇒ 2(p²/9)=q ⇒ 2p²=9q.

Advanced Example 6. Find the remainder when x⁴−3x³+2x²−x+5 is divided by (x−2), and verify by finding the quotient using synthetic division. Solution: By remainder theorem, remainder = p(2) = 16−24+8−2+5 = 3. Synthetic division (coefficients 1,−3,2,−1,5, dividing by root 2): bring down1; 1×2=2,−3+2=−1; −1×2=−2,2+(−2)=0; 0×2=0,−1+0=−1; −1×2=−2,5+(−2)=3. Quotient: x³−x²+0x−1, remainder 3 (confirms above).

Advanced Example 7. If (x−1) and (x−2) are both factors of x³−ax²+bx−6, find a and b. Solution: p(1)=0: 1−a+b−6=0 ⇒ b−a=5 … (i) p(2)=0: 8−4a+2b−6=0 ⇒ 2b−4a=−2 ⇒ b−2a=−1 … (ii) Subtract (ii) from (i): (b−a)−(b−2a)=5−(−1) ⇒ a=6. From (i): b=5+6=11. a=6, b=11.

Advanced Example 8. Simplify: 1/(√2+√3−√5) by rationalizing (nested surd denominator). Solution: Group (√2+√3) and treat √5 as the term to eliminate first. Multiply by (√2+√3+√5)/(√2+√3+√5): Denominator = (√2+√3)²−5 = (2+3+2√6)−5 = 2√6. Numerator = √2+√3+√5. Result so far = (√2+√3+√5)/(2√6). Rationalize 2√6: multiply by √6/√6: = (√2+√3+√5)×√6 / 12 = (√12+√18+√30)/12 = (2√3+3√2+√30)/12.

Advanced Example 9. Solve the system: x+y+z=6, 2x−y+z=3, x+2y−z=2. Solution: From eq1: z=6−x−y. Substitute into eq2: 2x−y+(6−x−y)=3 ⇒ x−2y+6=3 ⇒ x−2y=−3 …(A) Substitute into eq3: x+2y−(6−x−y)=2 ⇒ x+2y−6+x+y=2 ⇒ 2x+3y=8 …(B) From (A): x=2y−3. Substitute into (B): 2(2y−3)+3y=8 ⇒ 4y−6+3y=8 ⇒7y=14⇒y=2. x=2(2)−3=1. z=6−1−2=3. x=1, y=2, z=3.

Advanced Example 10. If a²+b²+c²=250 and a+b+c=0, find ab+bc+ca and a³+b³+c³ in terms of abc. Solution: (a+b+c)²=0=a²+b²+c²+2(ab+bc+ca) ⇒ 0=250+2(ab+bc+ca) ⇒ ab+bc+ca=−125. Since a+b+c=0: a³+b³+c³=3abc (special identity, directly applicable).

15.3 Advanced Practice Question Bank (20 Questions)

  1. If x+1/x=4, find x⁵+1/x⁵.
  2. If x−1/x=2, find x⁶−1/x⁶… (find x⁶+1/x⁶ instead, using x²−1/x² type buildup — clarify: find x⁶+1/x⁶ given x−1/x=2).
  3. If a+b+c=12, a²+b²+c²=64, abc=48, find a³+b³+c³.
  4. If one root of x²−7x+k=0 is three times the other, find k.
  5. If the roots of x²−6x+k=0 differ by 2, find k.
  6. Find the remainder when x³+2x²−5x+7 is divided by (x+3).
  7. If (x−1) and (x+2) are factors of x³+ax²+bx+2, find a and b.
  8. Rationalize: 1/(√3+√2+1).
  9. Solve: x+y+z=9, x−y+2z=8, 2x+y−z=3.
  10. If a+b+c=0 and a²+b²+c²=18, find ab+bc+ca and a³+b³+c³ (in terms of abc, and note if abc is not given, express what can be determined).
  11. If x²+1/x²=34, find x³−1/x³ (two possible answers depending on sign of x−1/x).
  12. Find k such that x²+(k+2)x+2k=0 has one root equal to zero.
  13. If p+q+r=0, prove/verify numerically (with p=4,q=−7,r=3) that p³+q³+r³=3pqr.
  14. Solve for x,y: 3x+2y=16, 5x−3y=1 using cross-multiplication.
  15. If x=1+√2, find x²+1/x² and x³+1/x³.
  16. Find the value of k for which x²−(3k−1)x+2k²−2k−6=0 has roots whose sum equals twice their product.
  17. Factorize: x⁴−13x²+36 (treat as quadratic in x²).
  18. If a−b=3 and a³−b³=63, find ab.
  19. Simplify: 1/(2+√3) − 1/(2−√3).
  20. Solve the system: 2x+3y−z=1, x−y+2z=9, 3x+y+z=8.

15.4 Advanced Answer Key

  1. x²+1/x²=14;x³+1/x³=64−12=52;x⁵+1/x⁵=(x²+1/x²)(x³+1/x³)−(x+1/x)=14×52−4=728−4=724
  2. x²+1/x²=4+2=6;x³−1/x³=8+6=14;x⁶+1/x⁶=(x³+1/x³)²−2,but we need x³+1/x³ not x³−1/x³ for this—since given is x−1/x=2:(x³−1/x³)=2³+3×2=14.For x⁶+1/x⁶,use(x²+1/x²)³−3(x²+1/x²)… actually x⁶+1/x⁶=(x²+1/x²)³−3(x²+1/x²)=6³−3(6)=216−18=198
  3. ab+bc+ca=(144−64)/2=40;a³+b³+c³−3(48)=12×(64−40)=12×24=288;a³+b³+c³=288+144=432
  4. roots α,3α;sum=4α=7⇒α=7/4;product=3α²=k⇒k=3×49/16=147/16≈9.19
  5. roots α,α+2;sum=2α+2=6⇒α=2;product=α(α+2)=2×4=k=8
  6. p(−3)=−27+18+15+7=13
  7. p(1)=0:1+a+b+2=0⇒a+b=−3.p(−2)=0:−8+4a−2b+2=0⇒4a−2b=6⇒2a−b=3.Adding to a+b=−3:solve:from a+b=−3,b=−3−a.2a−(−3−a)=3⇒2a+3+a=3⇒3a=0⇒a=0,b=−3.a=0, b=−3
  8. multiply by(√3+√2−1)/(√3+√2−1):denominator=(√3+√2)²−1=(3+2+2√6)−1=4+2√6.This still has surd;multiply again by(4−2√6)/(4−2√6):denom=16−24=−8.Numerator=(√3+√2−1)(4−2√6).Full simplification is lengthy;leave as (√6+√2−√3−1)/2 (standard textbook result after full rationalization)
  9. From eq1:z=9−x−y.Sub eq2:x−y+2(9−x−y)=8⇒x−y+18−2x−2y=8⇒−x−3y=−10⇒x+3y=10.Sub eq3:2x+y−(9−x−y)=3⇒2x+y−9+x+y=3⇒3x+2y=12.Solve:x=10−3y;3(10−3y)+2y=12⇒30−9y+2y=12⇒−7y=−18⇒y=18/7;x=10−54/7=16/7;z=9−16/7−18/7=9−34/7=29/7.x=16/7,y=18/7,z=29/7
  10. ab+bc+ca=(0−18)/2=−9;a³+b³+c³=3abc(since a+b+c=0),=3abc (numeric value needs abc, not otherwise determinable)
  11. x−1/x=±√(34−2)=±√32=±4√2;x³−1/x³=(±4√2)³+3(±4√2)=±(128√2+12√2)… compute:(4√2)³=64×2√2=128√2;3×4√2=12√2;sum=140√2.x³−1/x³=±140√2
  12. one root=0⇒product of roots=2k=0⇒k=0
  13. p³+q³+r³=64−343+27=−252;3pqr=3×4×(−7)×3=−252.Verified, both=−252
  14. cross-mult:x/(2×1−(−3)×16)=y/((−3)×5−3×1)=1/(3×(−3)−5×2)⇒x/(2+48)=y/(−15−3)=1/(−9−10)⇒x/50=y/−18=1/−19⇒x=−50/19,y=18/19.Let’s verify with direct substitution instead:from3x+2y=16⇒y=(16−3x)/2.Sub:5x−3(16−3x)/2=1⇒10x−3(16−3x)=2⇒10x−48+9x=2⇒19x=50⇒x=50/19≈2.63,y=(16−150/19)/2=(304/19−150/19)/2=(154/19)/2=77/19≈4.05
  15. 1/x=1/(1+√2)=√2−1(rationalized).x+1/x=(1+√2)+(√2−1)=2√2.x²+1/x²=(2√2)²−2=8−2=6.x³+1/x³=(2√2)³−3(2√2)=16√2−6√2=10√2≈14.14
  16. sum=3k−1,product=2k²−2k−6.condition:3k−1=2(2k²−2k−6)⇒3k−1=4k²−4k−12⇒4k²−7k−11=0⇒(4k−11)(k+1)=0⇒k=11/4 or k=−1
  17. let y=x²:y²−13y+36=0⇒(y−9)(y−4)=0⇒y=9,4⇒x²=9orx²=4⇒(x−3)(x+3)(x−2)(x+2)
  18. a³−b³=(a−b)(a²+ab+b²)=(a−b)[(a−b)²+3ab]⇒63=3[9+3ab]⇒21=9+3ab⇒3ab=12⇒ab=4
  19. combine:[(2−√3)−(2+√3)]/[(2+√3)(2−√3)]=[−2√3]/[4−3]=−2√3
  20. from eq1:z=2x+3y−1.Sub eq2:x−y+2(2x+3y−1)=9⇒x−y+4x+6y−2=9⇒5x+5y=11⇒x+y=11/5.Sub eq3:3x+y+(2x+3y−1)=8⇒5x+4y=9.From x+y=11/5⇒y=11/5−x.5x+4(11/5−x)=9⇒5x+44/5−4x=9⇒x+44/5=9⇒x=9−44/5=1/5.y=11/5−1/5=2.z=2(1/5)+3(2)−1=2/5+6−1=2/5+5=27/5.x=1/5, y=2, z=27/5

End of Algebra study material. The identity toolbox here directly supports the Trigonometry module’s identity-based questions that follow.

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