Quadratic Equations FOR Placement

Meta Description: Master quadratic equations for placement exams 2026. Learn formulas, solving tricks, nature of roots, and practice 23+ solved problems for banking & IT tests.

Introduction

Quadratic Equations form the backbone of the Quantitative Aptitude section in almost every major placement exam in 2026. In banking exams like IBPS PO, SBI Clerk, and RBI Grade B, candidates typically encounter 4–5 direct questions in prelims, often focusing on comparing roots of two equations. In IT corporate placements (TCS NQT, Infosys, Wipro, Accenture), 1–2 conceptual questions test your speed and accuracy with discriminants or Vieta’s relationships. Mastering this topic can significantly boost your sectional score, as questions are highly predictable and formula-driven. This guide covers every essential concept, shortcut, and exam-level problem to ensure you solve quadratic equations confidently and quickly.

Key Formulas & Concepts

Standard Form:
ax² + bx + c = 0 where a ≠ 0

1. Quadratic Formula:
x = [-b ± √(b² - 4ac)] / 2a
Gives exact roots for any quadratic equation.

2. Discriminant (D):
D = b² - 4ac

• D > 0 → Two distinct real roots
• D = 0 → Two equal real roots
• D < 0 → Complex/imaginary roots (no real solution)

3. Vieta’s Formulas (Root Relationships):
If α and β are roots:

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

4. Factorization Method:
Split the middle term bx into two terms whose product equals a × c and sum equals b.

5. Root Comparison Rule:
Solve both equations, list roots in ascending order, then compare every possible pair. If all combinations yield the same inequality, establish a relation. Otherwise, relation cannot be established.

Solved Examples (Basic Level)

1. Solve: 2x² - 7x + 3 = 0
Product = a×c = 6, Sum = -7. Split: -6x - x
2x² - 6x - x + 3 = 0 → 2x(x - 3) - 1(x - 3) = 0
(2x - 1)(x - 3) = 0 → x = 1/2, 3

2. Solve: x² - 5x + 6 = 0
Split: -3x - 2x → x(x - 3) - 2(x - 3) = 0
(x - 3)(x - 2) = 0 → x = 3, 2

3. Use quadratic formula: 3x² + 2x - 1 = 0
a=3, b=2, c=-1, D = 4 + 12 = 16
x = [-2 ± √16]/6 → x = (-2+4)/6 = 1/3, x = (-2-4)/6 = -1

4. Find nature of roots: 4x² + 4x + 1 = 0
D = 16 - 16 = 0 → Roots are real and equal.

5. Find equation if sum of roots = -3, product = -10
Standard form: x² - (sum)x + product = 0
x² + 3x - 10 = 0

Practice Questions (Medium Level)

6. Compare: I. 2x² - 9x + 10 = 0 | II. 4y² - 13y + 10 = 0
I: (2x - 5)(x - 2) = 0 → x = 5/2, 2
II: (4y - 5)(y - 2) = 0 → y = 5/4, 2
Comparing: 2.5 ≥ 1.25, 2.5 ≥ 2, 2 ≥ 1.25, 2 = 2 → x ≥ y

7. If α, β are roots of x² - 8x + 15 = 0, find 1/α + 1/β
α + β = 8, αβ = 15
1/α + 1/β = (α + β)/αβ = 8/15

8. Solve: √3x² + 10x + 7√3 = 0
a=√3, b=10, c=7√3, D = 100 - 84 = 16
x = [-10 ± 4]/(2√3) → x = -6/(2√3) = -√3, x = -14/(2√3) = -7/√3

9. Find k if roots of 2x² + kx + 18 = 0 are equal
Equal roots → D = 0 → k² - 4(2)(18) = 0 → k² = 144 → k = ±12

10. Compare: I. 6x² + 7x + 2 = 0 | II. 15y² - 38y - 40 = 0
I: (3x + 2)(2x + 1) = 0 → x = -2/3, -1/2
II: (5y - 20)(3y + 2) → (5y + 4)(3y - 10) = 0 → y = -4/5, 10/3
-0.66 > -0.8, -0.5 > -0.8, but -0.66 < 3.33 → Mixed → Relation cannot be established

11. One root of 3x² - 5x + p = 0 is 2. Find p & other root.
Substitute x=2: 12 - 10 + p = 0 → p = -2
Equation: 3x² - 5x - 2 = 0 → (3x + 1)(x - 2) = 0 → Other root = -1/3

12. Form quadratic with roots (2 + √5) and (2 - √5)
Sum = 4, Product = 4 - 5 = -1 → x² - 4x - 1 = 0

13. Solve: x² - (√2 + 1)x + √2 = 0
Split: -√2x - x → x(x - √2) - 1(x - √2) = 0 → x = 1, √2

Advanced Questions

14. If α, β are roots of x² - 2x + 3 = 0, find α² + β²
α + β = 2, αβ = 3
α² + β² = (α + β)² - 2αβ = 4 - 6 = -2

15. Compare: I. x² - 7x + 12 = 0 | II. y² - 5y + 6 = 0
I: x = 3, 4 | II: y = 2, 3
3 ≥ 2, 4 > 2, 3 = 3, 4 > 3 → All satisfy x ≥ y → x ≥ y

16. Find range of m for which x² + (m+1)x + 1 = 0 has real roots
Real roots → D ≥ 0 → (m+1)² - 4 ≥ 0 → (m+1)² ≥ 4
m+1 ≥ 2 or m+1 ≤ -2 → m ≥ 1 or m ≤ -3

17. If roots are in ratio 2:3, prove 6b² = 25ac
Let roots be 2k, 3k. Sum = 5k = -b/a → k = -b/5a
Product = 6k² = c/a → Substitute k: 6(b²/25a²) = c/a
Multiply by 25a²: 6b² = 25ac → Proved

18. Solve: (x + 2)/(x - 2) + (x - 2)/(x + 2) = 10/3
Let u = (x+2)/(x-2). Then u + 1/u = 10/3
3u² - 10u + 3 = 0 → (3u - 1)(u - 3) = 0 → u = 1/3, 3
Case 1: (x+2)/(x-2) = 3 → x = 4
Case 2: (x+2)/(x-2) = 1/3 → x = -4 → x = ±4

Common Mistakes to Avoid

• Ignoring the negative sign in sum formula: α + β = -b/a (not +b/a)
• Declaring D < 0 as "no solution" instead of "no real roots/imaginary roots"
• Forgetting to check both + and - when using the quadratic formula
• Incorrectly establishing x > y when roots overlap (e.g., x = {3,5}, y = {2,4} → relation undefined)
• Assuming a = 1 always, leading to wrong factorization splits
• Missing the condition a ≠ 0 in parameter-based questions

Shortcut Tricks

1. Sum-Product Quick Factorization:
Instead of trial & error, find two numbers whose product is a×c and sum is b. Divide by a if needed. Example: 6x² + 7x + 2 = 0 → 12 & 7 → split directly.

2. Discriminant Sign Check (Mental):
For ax² + bx + c, if a & c have same sign, b² must exceed 4ac for real roots. If opposite signs, D is always positive → always real roots.

3. Reciprocal Roots Shortcut:
If roots are reciprocals, c/a = 1 → c = a. Saves time in MCQs.

4. Comparison Table Method:
Arrange roots on a number line: y1 y2 x1 x2 → instantly see overlaps. If entire set of x is right of y, x > y.

5. Equal Roots Parameter Trick:
When asked "roots are equal", directly apply b² = 4ac. Skip full equation solving.

Previous Year Questions

19. (IBPS PO 2023) Compare: I. 3x² - 13x + 12 = 0 | II. 2y² - 15y + 25 = 0
I: (3x - 4)(x - 3) = 0 → x = 4/3, 3
II: (2y - 5)(y - 5) = 0 → y = 2.5, 5
1.33 < 2.5, 3 < 5, but 3 > 2.5 → Mixed → Relation cannot be established

20. (SBI Clerk 2022) If α + β = 10, αβ = 21, find α³ + β³
Formula: α³ + β³ = (α + β)³ - 3αβ(α + β)
= 1000 - 3(21)(10) = 1000 - 630 = 370

21. (TCS NQT 2023) Find roots of 2x² - √5x - 3 = 0
D = 5 + 24 = 29
x = [√5 ± √29]/4 → x = (√5 + √29)/4, (√5 - √29)/4

22. (Infosys 2022) For what k are roots of x² - 4x + k = 0 real and distinct?
Distinct real → D > 0 → 16 - 4k > 0 → 4k < 16 → k < 4

23. (RBI Grade B 2021) Compare: I. x² + 8x + 15 = 0 | II. 2y² + 11y + 14 = 0
I: x = -3, -5
II: y = -2, -3.5
-3 < -2, -3 > -3.5, -5 < -3.5, -5 < -2 → Overlap at different positions → Relation cannot be established

Quick Revision

• Standard form: ax² + bx + c = 0, a ≠ 0
• Roots: x = [-b ± √(b²-4ac)]/2a
• Discriminant D = b² - 4ac dictates nature (>, =, < 0)
• Sum of roots = -b/a, Product = c/a
• For comparison: solve both, list roots, test all 4 pairs
• Equal roots → D = 0; Real roots → D ≥ 0
• Always verify sign of b in -b/a
• Banking exams prefer comparison questions; IT tests focus on discriminant & parameters
• Practice 10–15 comparison sets daily for speed under 90 seconds per set