How to Solve Quadratic Equations (Factorisation & Formula) – Complete Guide

How to Solve Quadratic Equations (Factorisation & Formula) – Complete Guide

For SSC / Bank / Railway Exams

Introduction

"Master quadratic equations, and you unlock 5–10 marks in every SSC, Bank, or Railway exam—questions that test factorisation and the quadratic formula appear in every paper, often as easy scoring opportunities. Miss them, and you’re leaving marks on the table."

What You Need To Know First

1. What is a quadratic equation? An equation of the form ax² + bx + c = 0, where a, b, and c are numbers, and a ≠ 0.
2. How to expand brackets? Example: (x + 2)(x + 3) = x² + 5x + 6.
3. Basic algebra rules: Adding, subtracting, and multiplying terms.

Key Vocabulary

Formulas To Know

1. Standard Form of a Quadratic Equation ax² + bx + c = 0
2. a = coefficient of x² (must not be zero)
3. b = coefficient of x

4. c = constant term

5. Factorisation Method (When Possible) (px + q)(rx + s) = 0

6. Find two numbers that multiply to a × c and add to b.

7. MEMORISE THIS: If a = 1, look for two numbers that multiply to c and add to b.

8. Quadratic Formula (Always Works) x = [–b ± √(b² – 4ac)] / (2a)

9. ± means "plus or minus" (two solutions).
10. b² – 4ac is the discriminant (tells us the nature of roots).

11. MEMORISE THIS: The formula is given on most exam sheets, but you must know how to plug in values.

12. Discriminant Rules

13. If b² – 4ac > 0 → Two distinct real roots.
14. If b² – 4ac = 0 → One real root (repeated).
15. If b² – 4ac < 0 → No real roots (imaginary).

Step-by-Step Method

Method 1: Factorisation (When Possible)

Use this when a = 1 or the equation can be easily split into two brackets.

1. Write the equation in standard form: ax² + bx + c = 0.
2. Check if a = 1. If yes, skip to Step 3. If no, try to factor out a first.
3. Find two numbers that:
4. Multiply to c (if a = 1).
5. Add to b.
6. Write the equation as: (x + first number)(x + second number) = 0.
7. Set each bracket to zero and solve for x.
8. Write both solutions.

Example (Using Steps): Solve x² – 5x + 6 = 0.

1. Already in standard form: x² – 5x + 6 = 0.
2. a = 1, so proceed.
3. Find two numbers that multiply to 6 and add to –5 → –2 and –3.
4. Write: (x – 2)(x – 3) = 0.
5. Set each bracket to zero:
6. x – 2 = 0 → x = 2
7. x – 3 = 0 → x = 3
8. Solutions: x = 2 and x = 3.

Method 2: Quadratic Formula (Always Works)

Use this when factorisation is difficult or impossible.

1. Write the equation in standard form: ax² + bx + c = 0.
2. Identify a, b, and c.
3. Calculate the discriminant: D = b² – 4ac.
4. Check the discriminant:
5. If D < 0, no real solutions (stop here).
6. If D ≥ 0, proceed.
7. Plug into the quadratic formula: x = [–b ± √D] / (2a)
8. Simplify to find two solutions (if D > 0) or one solution (if D = 0).

Example (Using Steps): Solve 2x² + 4x – 6 = 0.

1. Standard form: 2x² + 4x – 6 = 0.
2. a = 2, b = 4, c = –6.
3. Discriminant: D = (4)² – 4(2)(–6) = 16 + 48 = 64.
4. D = 64 > 0, so two real roots.
5. Plug into formula: x = [–4 ± √64] / (2 × 2) = [–4 ± 8] / 4
6. Two solutions:
7. x = (–4 + 8)/4 = 4/4 = 1
8. x = (–4 – 8)/4 = –12/4 = –3
9. Solutions: x = 1 and x = –3.

Worked Examples

Example 1 – Basic (Factorisation)

Solve: x² + 7x + 12 = 0

1. Standard form: x² + 7x + 12 = 0 (a = 1).
2. Find two numbers that multiply to 12 and add to 7 → 3 and 4.
3. Write: (x + 3)(x + 4) = 0.
4. Set each bracket to zero:
5. x + 3 = 0 → x = –3
6. x + 4 = 0 → x = –4
7. Solutions: x = –3 and x = –4.

What we did and why: We used factorisation because a = 1, making it easy to split into two brackets. Always check for simple factorisation first—it’s faster than the formula.

Example 2 – Medium (Quadratic Formula)

Solve: 3x² – 5x – 2 = 0

1. Standard form: 3x² – 5x – 2 = 0.
2. a = 3, b = –5, c = –2.
3. Discriminant: D = (–5)² – 4(3)(–2) = 25 + 24 = 49.
4. D = 49 > 0, so two real roots.
5. Plug into formula: x = [5 ± √49] / 6 = [5 ± 7] / 6
6. Two solutions:
7. x = (5 + 7)/6 = 12/6 = 2
8. x = (5 – 7)/6 = –2/6 = –1/3
9. Solutions: x = 2 and x = –1/3.

What we did and why: Factorisation was tricky here (a ≠ 1), so we used the quadratic formula. Always calculate the discriminant first—it tells you if solutions exist.

Example 3 – Exam-Style (Disguised Quadratic)

Solve: x(2x – 3) = 5

1. Expand first: 2x² – 3x = 5.
2. Bring to standard form: 2x² – 3x – 5 = 0.
3. a = 2, b = –3, c = –5.
4. Discriminant: D = (–3)² – 4(2)(–5) = 9 + 40 = 49.
5. D = 49 > 0, so two real roots.
6. Plug into formula: x = [3 ± √49] / 4 = [3 ± 7] / 4
7. Two solutions:
8. x = (3 + 7)/4 = 10/4 = 5/2
9. x = (3 – 7)/4 = –4/4 = –1
10. Solutions: x = 5/2 and x = –1.

What we did and why: The equation wasn’t in standard form initially. Always expand and rearrange to ax² + bx + c = 0 before solving. Examiners love disguising quadratics—watch for brackets!

Common Mistakes

Exam Traps

1-Minute Recap (Night Before Exam)

"Listen up—quadratic equations are easy marks if you follow these steps:
1. Always write in standard form first: ax² + bx + c = 0.
2. If a = 1, try factorisation first. Find two numbers that multiply to c and add to b.
3. If factorisation is hard, use the quadratic formula: x = [–b ± √(b² – 4ac)] / (2a). Memorise it!
4. Check the discriminant (b² – 4ac): - D > 0 → two real roots. - D = 0 → one real root. - D < 0 → no real roots.
5. Watch for traps: Expand brackets, avoid sign errors, and don’t assume all quadratics factorise.
6. Practice 3–4 problems tonight. You’ve got this!