Quadratic Equation (द्विघात समीकरण) – Comparison Rules, Exam Hacks & Speed Tricks | Part 2

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Quadratic Equation (द्विघात समीकरण)

Part 2 — Exam Methods, Comparison Rules & Speed Hacks | परीक्षा पद्धतियाँ एवं तुलना नियम

💡 Mathnija Mantra: रटना नहीं, समझना है!

15 Standard Exam Options | 5 मानक विकल्प

In banking exams, two equations (in terms of x and y) are given. You need to establish the relationship between their roots from these 5 options:

बैंकिंग परीक्षाओं में दो समीकरण (x और y के पदों में) दिए जाते हैं। आपको उनके मूलों की तुलना करके इन 5 विकल्पों में से सही संबंध चुनना होता है:

The five standard root-comparison options used in quadratic equation exam questions

Option	Mathematical Relation	Condition (तुलना की स्थिति)
(A)	x > y	Every root of x is strictly greater than every root of y
(B)	x < y	Every root of x is strictly smaller than every root of y
(C)	x ≥ y	Roots of x are greater than or equal to roots of y
(D)	x ≤ y	Roots of x are smaller than or equal to roots of y
(E)	x = y OR CND	Roots are equal OR relationship cannot be determined (contradiction)

2The 4-Step Comparison Matrix | 4-चरणीय तुलना विधि

Always compare every value of x with every value of y (Total 4 comparisons):

समीकरण के दोनों मूल प्राप्त होने के बाद x के प्रत्येक मान की तुलना y के प्रत्येक मान से करें (कुल 4 तुलनाएँ):

Let Roots be: x = (x₁, x₂) and y = (y₁, y₂)

• Compare x₁ with y₁  ➜  Result 1
• Compare x₁ with y₂  ➜  Result 2
• Compare x₂ with y₁  ➜  Result 3
• Compare x₂ with y₂  ➜  Result 4

⚠️ CND Warning Rule / विरोधाभास नियम: If at any step you get both (>) and (<) together, the answer is directly CND (Relationship Cannot Be Established).

32-Second Elimination Speed Hacks | 2-सेकंड शॉर्टकट ट्रिक्स

Hack 1: Both Constant Terms Negative (c₁ < 0, c₂ < 0)

If the constant term in both equations is negative, their root signs will be (+, −) and (+, −). They always contradict each other.
Direct Result: x = y OR CND (No pen needed!)

Hack 2: Sign Matrix Dominance

If Equation I is (−, +) ➜ Roots: (+, +)
If Equation II is (+, +) ➜ Roots: (−, −)
Positive is always greater than negative.
Direct Result: x > y

4Avoid Decimals: Cross-Multiplication Trick | दशमलव से बचने की विधि

When coefficients of x² and y² (a₁ and a₂) are greater than 1, avoid dividing by coefficients. Instead, cross-multiply the factors:

जब x² और y² के गुणांक (a₁ और a₂) 1 से बड़े हों, तो भाग देकर दशमलव में बदलने के बजाय क्रॉस-गुणा विधि अपनाएँ:

Example: 2x² − 7x + 6 = 0  and  3y² − 11y + 10 = 0

• Equation I (a₁ = 2): Factors = +4, +3
• Equation II (a₂ = 3): Factors = +6, +5
• Cross Multiply: Multiply x factors by a₂ (3)  ➜  x' = 12, 9
• Cross Multiply: Multiply y factors by a₁ (2)  ➜  y' = 12, 10
• Comparison: 12 = 12, 12 > 10, 9 < 12, 9 < 10  ➜  CND (x = y or relation cannot be established)

5Square Root (√k) Equations Hack | करणी वाले समीकरण

For equations containing terms like √k (e.g., x² − 7√3x + 36 = 0):

यदि मध्य पद में करणी (√k) दी गई हो:

3-Step Resolution:

• Step 1: Divide the constant term (c) by the number inside root (k).  [ 36 ÷ 3 = 12 ]
• Step 2: Find factors of this quotient for middle term number.  [ Factors of 12 for 7 ➜ 4, 3 ]
• Step 3: Attach the root (√k) back to factors with sign table.  ➜  x = +4√3, +3√3

🎯 15 Exam-Level Solved Questions | परीक्षा स्तरीय प्रश्न

Q1. (Easy - Sign Hack)

I. x² − 13x + 40 = 0
II. y² + 15y + 56 = 0

Q2. (Easy - Both Constant Negative)

I. x² + 2x − 35 = 0
II. y² − 5y − 24 = 0

Q3. (Standard Factorisation)

I. x² − 17x + 72 = 0
II. y² − 19y + 90 = 0

Q4. (Standard Factorisation)

I. x² + 11x + 30 = 0
II. y² + 7y + 12 = 0

Q5. (Coefficients a > 1)

I. 2x² − 9x + 10 = 0
II. 2y² − 13y + 21 = 0

Q6. (Coefficients a > 1)

I. 3x² + 14x + 15 = 0
II. 2y² + 11y + 14 = 0

Q7. (Power 2 vs Root Trap)

I. x² = 144
II. y = √144

Q8. (Square Root Form)

I. x² − 7√3x + 36 = 0
II. y² − 9√3y + 60 = 0

Q9. (Square Root Form)

I. x² − 5√2x + 12 = 0
II. y² − 3√2y + 4 = 0

Q10. (Moderate Coefficients)

I. 6x² + 7x + 2 = 0
II. 4y² + 12y + 9 = 0

Q11. (Large Numbers)

I. x² − 28x + 195 = 0
II. y² − 30y + 221 = 0

Q12. (Fractional / Linear Base)

I. x² − 25 = 0
II. y² − 10y + 25 = 0

Q13. (Mixed Signs)

I. 2x² + 11x + 14 = 0
II. 4y² + 16y + 15 = 0

Q14. (Moderate Level)

I. 5x² − 18x + 9 = 0
II. 3y² + 5y − 2 = 0

Q15. (High Level / Mains Prep)

I. x² − 8√3x + 45 = 0
II. y² − √2y − 24 = 0

🔑 View Answer Key & Step-by-Step Breakdown (उत्तर व व्याख्या)

Q1: x = +8, +5  |  y = −8, −7  ➜  x > y
Q2: x = +5, −7  |  y = +8, −3  ➜  CND (x = y or relation cannot be established)
Q3: x = +9, +8  |  y = +10, +9  ➜  x ≤ y
Q4: x = −6, −5  |  y = −4, −3  ➜  x < y
Q5: x = +2.5, +2  |  y = +3.5, +3  ➜  x < y
Q6: x = −1.67, −3  |  y = −2, −3.5  ➜  CND (Contradiction)
Q7: x = +12, −12  |  y = +12  ➜  x ≤ y
Q8: x = +4√3, +3√3  |  y = +5√3, +4√3  ➜  x ≤ y
Q9: x = +3√2, +2√2  |  y = +2√2, +√2  ➜  x ≥ y
Q10: x = −0.5, −0.67  |  y = −1.5, −1.5  ➜  x > y
Q11: x = +15, +13  |  y = +17, +13  ➜  CND (15 < 17 but 15 > 13 — contradiction)
Q12: x = +5, −5  |  y = +5, +5  ➜  x ≤ y
Q13: x = −2, −3.5  |  y = −1.5, −2.5  ➜  CND
Q14: x = +3, +0.6  |  y = +0.33, −2  ➜  x > y
Q15: x = +5√3, +3√3  |  y = +4√2, −3√2  ➜  CND (Values overlap: 5√3 ≈ 8.66, 3√3 ≈ 5.19 vs 4√2 ≈ 5.65)

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