Olive Defence
Numerical Ability · AFCAT

NA10 — Algebra

🔢 Numerical Ability – NA10 AFCAT Level ☆ Low Priority
📌 AFCAT Focus 2022–2026: Algebra gives 1–2 questions. Most tested: solving simultaneous equations, using identities (a±b)², a²−b², and quadratic roots (sum/product). The discriminant (b²−4ac) tells the nature of roots — frequently tested in AFCAT.

1. Key Algebraic Identities

Fig. 1.1 — Algebraic Identities: The Must-Memorise Set
ALGEBRAIC IDENTITIES — Memorise All Six (a + b)² = a² + 2ab + b² (a − b)² = a² − 2ab + b² a² − b² = (a+b)(a−b) (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) DERIVED SHORTCUTS (a+b)²−(a−b)² = 4ab (a+b)²+(a−b)² = 2(a²+b²) a²+b² = s²−2p

2. Quadratic Equations

Fig. 2.1 — Quadratic ax² + bx + c = 0: Roots, Discriminant & Nature
x = −b ± √(b²−4ac) 2a Roots α and β — Vieta's Formulas Sum: α + β = −b/a Product: αβ = c/a Form equation from roots: x² − (sum)x + (product) = 0 Roots 2 & 3 → x²−5x+6=0 Roots 3 & 4 → x²−7x+12=0 Discriminant D = b² − 4ac D > 0 → 2 distinct real roots D = 0 → 2 equal real roots D < 0 → No real roots (complex/imaginary roots) AFCAT tests real root cases only Example: x²+x+1=0 → D=−3 → none
POLYNOMIALS & SURDS

3. Polynomials — Remainder & Factor Theorem

🔢 Remainder Theorem

  • If polynomial P(x) is divided by (x − a), the remainder = P(a)
  • Example: P(x) = x³ − 2x + 1 divided by (x − 2):
    Remainder = P(2) = 8 − 4 + 1 = 5
  • If (x + a) is divisor: substitute x = −a
  • No need to do long division — just substitute!

🔢 Factor Theorem

  • (x − a) is a factor of P(x) if and only if P(a) = 0
  • Example: Is (x − 1) a factor of x³ − 6x² + 11x − 6?
    P(1) = 1 − 6 + 11 − 6 = 0 → Yes, it is a factor
  • Converse: if P(a) = 0, then (x − a) is a factor
  • Use to find factors of cubic polynomials in AFCAT
Topic CQuadratic Factorisation MethodAFCAT Direct
Split Middle
For ax²+bx+c=0: find two numbers that multiply to a×c and add to b, then split the middle term.
Example: x²+5x+6=0 → find two numbers: product=6, sum=5 → (2,3) → x²+2x+3x+6 → x(x+2)+3(x+2) → (x+2)(x+3)=0 → x = −2 or −3
Perfect Square
x²+6x+9 = (x+3)² → root x = −3 (repeated). Recognise: middle term = 2×√(first)×√(last).
Difference of Squares
x²−16 = (x+4)(x−4) → roots x = ±4. Whenever c is negative and b=0, use a²−b² = (a+b)(a−b).

4. Surds & Rationalisation

√ Laws of Surds

  • √a × √b = √(ab)
  • √a ÷ √b = √(a/b)
  • (√a)² = a
  • √a + √b ≠ √(a+b) — cannot be added directly
  • Order of surd: ⁿ√a = a^(1/n)
  • Mixed surd: a√b (rational × irrational)

√ Rationalisation

  • Multiply by conjugate to remove surd from denominator
  • 1/(a + √b) = (a − √b) / (a² − b)
  • 1/(√a + √b) = (√a − √b) / (a − b)
  • Example: 1/(2+√3) = (2−√3)/(4−3) = 2 − √3
  • AFCAT usually asks to simplify or rationalise the denominator

📐 Formula Sheet — NA10

Core Identities
(a+b)² = a²+2ab+b²
(a−b)² = a²−2ab+b²
a²−b² = (a+b)(a−b)
(a+b)²−(a−b)² = 4ab
Cubic Identities
(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)
Quadratic Roots
x = [−b ± √(b²−4ac)] / 2a
Sum = −b/a | Product = c/a
D>0: real & distinct
D=0: equal | D<0: no real roots
Simultaneous Equations
Elimination: make coefficients equal
Substitution: express one, sub in other
Cross-multiply: a₁x+b₁y=c₁ and a₂x+b₂y=c₂
x/(b₁c₂−b₂c₁) = y/(c₁a₂−c₂a₁) = 1/(a₁b₂−a₂b₁)
Laws of Exponents
aᵐ×aⁿ = aᵐ⁺ⁿ | aᵐ÷aⁿ = aᵐ⁻ⁿ
(aᵐ)ⁿ = aᵐⁿ | a⁰ = 1
a⁻ⁿ = 1/aⁿ | a^(1/n) = ⁿ√a
(ab)ⁿ = aⁿbⁿ
Surds (Rationalisation)
1/(a+√b) = (a−√b)/(a²−b)
√a × √b = √(ab)
√a / √b = √(a/b)
Rationalise: multiply by conjugate

📝 Topic-Wise PYQs — NA10

Q1. If x + 1/x = 5, find x² + 1/x². AFCAT PYQ
(a) 23(b) 25(c) 27(d) 21
✔ Answer: (a) 23
Square both sides: (x+1/x)² = 25 → x²+2+1/x² = 25 → x²+1/x² = 23. Using (a+b)² = a²+2ab+b².
Q2. Sum of roots of 2x²−5x+3=0 is: AFCAT PYQ
(a) 5/2(b) 3/2(c) −5/2(d) 2/5
✔ Answer: (a) 5/2
Sum = −b/a = −(−5)/2 = 5/2. Product = c/a = 3/2.
Q3. The discriminant of x²−4x+5=0 is: ⚡ Tricky
(a) 4(b) −4(c) 36(d) 0
✔ Answer: (b) −4
D = b²−4ac = (−4)²−4(1)(5) = 16−20 = −4. Since D<0, no real roots.
Q4. Solve: 3x+2y=12 and x−y=1. Find x. AFCAT PYQ
(a) 2(b) 3(c) 4(d) 14/5
✔ Answer: (d) 14/5
From eq2: x = y+1. Sub in eq1: 3(y+1)+2y=12 → 5y+3=12 → y=9/5. x=9/5+1 = 14/5.

🧠 Quick Memory Chart — NA10

🔢 Identities
  • (a+b)²= a²+2ab+b²
  • (a−b)²= a²−2ab+b²
  • a²−b²= (a+b)(a−b)
  • (a+b)²−(a−b)² = 4ab
🔢 Quadratic
  • Sum = −b/a
  • Product = c/a
  • D>0: distinct real
  • D=0: equal | D<0: none
🔢 Exponents
  • aᵐ×aⁿ = aᵐ⁺ⁿ
  • (aᵐ)ⁿ = aᵐⁿ
  • a⁰ = 1
  • a^(1/n) = ⁿ√a
📝 Mock Tests 🎯 Subject Quiz ✈️ Telegram
This material is for personal AFCAT exam preparation only. Unauthorised reproduction or distribution is prohibited.
All rights reserved  ·  ODEA.Classes@gmail.com  ·  OliveDefence.com
Q5. Factorise x² − 7x + 12 and find its roots. AFCAT PYQ
(a) (x−3)(x−4); roots 3,4(b) (x+3)(x+4); roots −3,−4(c) (x−3)(x+4); roots 3,−4(d) (x−6)(x−2); roots 6,2
✔ Answer: (a) (x−3)(x−4); roots 3 and 4
Find two numbers: product = 12, sum = −7 → −3 and −4. So x²−7x+12 = (x−3)(x−4). Roots: x=3, x=4. Check: 3+4=7=−(−7)/1=7 ✓. 3×4=12=12/1=12 ✓.
Q6. Rationalise the denominator: 1/(√5 + √3) AFCAT PYQ
(a) (√5+√3)/2(b) (√5−√3)/2(c) (√5−√3)/8(d) √5−√3
✔ Answer: (b) (√5−√3)/2
Multiply by conjugate (√5−√3)/(√5−√3): Numerator = (√5−√3). Denominator = (√5)²−(√3)² = 5−3 = 2. Result = (√5−√3)/2.