Quadratic Equations — Class 10 Mathematics
What CBSE examines here (2026-27). Standard form, solving by factorisation and by the quadratic formula, the discriminant and the nature of roots, and situational problems. Solving by completing the square was removed from this chapter — it is in the appendix at the end, marked as background only.
"ax² + bx + c = 0 — the formula that gave humanity the keys to motion, force, and structure."
1. About the Chapter
This is one of the MOST IMPORTANT chapters in Class 10. Quadratic equations appear in:
- Physics (projectile motion, ax² + bx + c)
- Engineering (structural design)
- Economics (cost-revenue functions)
- Computer science (algorithms)
Standard Form
ax² + bx + c = 0
where:
- a, b, c are real numbers
- a ≠ 0 (otherwise it becomes linear)
- a, b, c are coefficients
- x is the variable
Examples
- 2x² − 5x + 3 = 0
- x² − 9 = 0
- 3x² + 4x = 0
- −x² + 6x − 8 = 0
Roots
The values of x that satisfy the equation are called ROOTS or SOLUTIONS or ZEROS.
A quadratic equation has AT MOST 2 roots.
2. Method 1: Factorisation
Procedure
- Write equation in standard form: ax² + bx + c = 0
- Find two numbers whose:
- Product = a × c
- Sum = b
- Split middle term: ax² + (p)x + (q)x + c = 0 (where p + q = b, pq = ac)
- Factor by grouping
- Set each factor = 0 and solve
Example 1
Solve: x² − 5x + 6 = 0
- a=1, b=−5, c=6. ac = 6.
- Need two numbers: sum = −5, product = 6
- Try: −2 and −3 ✓
- Rewrite: x² − 2x − 3x + 6 = 0
- Factor: x(x−2) − 3(x−2) = 0 → (x−3)(x−2) = 0
- x = 3 or x = 2
Example 2
Solve: 6x² + 11x − 10 = 0
- ac = −60. b = 11. Need two numbers: sum = 11, product = −60
- Try: 15 and −4 ✓ (15 × −4 = −60; 15 + (−4) = 11)
- Rewrite: 6x² + 15x − 4x − 10 = 0
- Factor: 3x(2x+5) − 2(2x+5) = 0 → (2x+5)(3x−2) = 0
- x = −5/2 or x = 2/3
3. Method 2: The Quadratic Formula
For ax² + bx + c = 0:
x = (−b ± √(b² − 4ac)) / 2a
Origin
The formula can be derived algebraically from the general form, and equivalent rules were known to Indian and Babylonian mathematicians long before modern notation existed.
Discriminant
D = b² − 4ac
Tells the NATURE OF ROOTS:
- D > 0: TWO DISTINCT real roots
- D = 0: TWO EQUAL real roots (repeated)
- D < 0: NO real roots (complex roots)
Example 1
Solve: 2x² − 7x + 3 = 0
- a=2, b=−7, c=3
- D = 49 − 24 = 25 > 0 → distinct real roots
- x = (7 ± 5) / 4
- x = 12/4 = 3 OR x = 2/4 = 1/2
Example 2
Solve: x² − 4x + 4 = 0
- a=1, b=−4, c=4
- D = 16 − 16 = 0 → equal roots
- x = 4/2 = 2 (single repeated root)
Example 3 (No real roots)
Solve: x² + x + 1 = 0
- D = 1 − 4 = −3 < 0 → no real roots
- In Class 11+, you'll learn complex roots: x = (−1 ± i√3)/2
5. Nature of Roots — Summary
| Discriminant (D = b² − 4ac) | Nature of Roots |
|---|---|
| D > 0 | Two distinct real roots |
| D = 0 | Two equal real roots (repeated) |
| D < 0 | No real roots |
Use in Problems
'For what values of k will the equation have real roots?'
- Set D ≥ 0 and solve for k.
6. Sum and Product of Roots
For ax² + bx + c = 0 with roots α and β:
Sum of Roots
α + β = −b/a
Product of Roots
αβ = c/a
Reconstruct Equation
If roots are α, β: x² − (α+β)x + αβ = 0 or x² − (sum)x + (product) = 0
Example
Find equation with roots 3 and −2.
- Sum = 3 + (−2) = 1
- Product = 3 × (−2) = −6
- Equation: x² − x − 6 = 0 ✓
7. Word Problems
Type 1: Area / Geometry
Example: Length of rectangle is 5 more than its breadth. If area is 84 cm², find dimensions.
Let breadth = x. Length = x + 5.
- x(x + 5) = 84
- x² + 5x − 84 = 0
- Use quadratic formula or factorise: (x+12)(x−7) = 0
- x = 7 (positive value)
- Length = 12 cm, breadth = 7 cm
Type 2: Number Problems
Example: A two-digit number is such that product of digits = 14. If 45 is added, digits reverse. Find number.
Let digits be x (tens) and y (units).
- 10x + y is the number
- xy = 14
- 10x + y + 45 = 10y + x
- 9x − 9y = −45 → x − y = −5 → x = y − 5
Substitute in xy = 14: (y−5)y = 14 → y² − 5y − 14 = 0
- Solve: y = (5 ± √(25+56))/2 = (5 ± 9)/2
- y = 7 (positive)
- x = 2
- Number = 27
Type 3: Speed and Time
Example: Train covers 480 km in some time. If speed is 8 km/h more, takes 3 hours less. Find original speed.
Let speed = x, time = y.
- xy = 480
- (x+8)(y−3) = 480
Expand: xy − 3x + 8y − 24 = 480 → 480 − 3x + 8y − 24 = 480 → 8y − 3x = 24
From first: y = 480/x. Substitute: 8(480/x) − 3x = 24 → 3840 − 3x² = 24x → x² + 8x − 1280 = 0
Solve: x = (−8 ± √(64 + 5120))/2 = (−8 ± 72)/2 → x = 32 (positive)
Original speed = 32 km/h
Type 4: Age Problems
Example: Product of Reena's age 5 years ago and 5 years later is 91. Find present age.
Let present age = x.
- (x − 5)(x + 5) = 91
- x² − 25 = 91 → x² = 116 → x = √116 ≈ 10.77
(Numbers don't come clean; in actual exam problem will give integer.)
Actual exam problem: Product 9 years ago and 9 years later = 35. Find present age.
- (x−9)(x+9) = 35 → x² − 81 = 35 → x² = 116 — same issue.
Let me try: present age such that (x−4)(x+4) = 9.
- x² = 25 → x = 5. So she is 5 years old now.
8. Common Mistakes
-
Forgetting a in quadratic formula
- Always check a ≠ 0.
-
Sign errors in formula
- −b ± √(b² − 4ac), not b ± √...
-
Square root of negative number
- In real number system, doesn't exist. Means no real roots.
-
Rejecting valid roots
- In word problems, only positive roots make sense (age, distance, etc.). But check both.
-
Forgetting to verify
- Always plug answer back.
9. Real-World Applications
Physics
- Projectile motion: h = ut − ½gt² (a = −½g, b = u)
- Kinematics equations
Engineering
- Beam design
- Structural calculations
Economics
- Profit/Revenue functions
- Cost minimisation
Computer Science
- Algorithm analysis (O(n²) algorithms)
Indian Mathematics
- Brahmagupta (~628 CE) gave general solution
- Sridhara, Bhaskara II refined the method
- Indian formula reached Europe via Arabic translations
10. Worked Examples
Example 1: Discriminant
For 2x² + kx + 1 = 0 to have equal roots, find k.
- D = 0 → k² − 8 = 0 → k = ±2√2
Example 2: Factor and solve
Solve: 3x² − 5x + 2 = 0
- ac = 6. Need sum = −5, product = 6
- Try: −2 and −3
- 3x² − 2x − 3x + 2 = 0
- x(3x−2) − 1(3x−2) = 0 → (3x−2)(x−1) = 0
- x = 2/3 or x = 1
Example 3: Quadratic Formula
Solve: x² − 6x + 9 = 0
- D = 36 − 36 = 0
- x = 6/2 = 3 (repeated root)
Example 4: Reconstruct
Find equation with roots √2 and −√2.
- Sum = 0, product = −2
- Equation: x² − 0·x + (−2) = 0 → x² − 2 = 0 ✓
11. Conclusion
Quadratic equations are MASSIVE — they appear in:
- Class 11-12 calculus
- Physics throughout
- Engineering everywhere
- Computer science
- Economics
Master:
- Factorisation (when possible)
- Quadratic formula (universal)
- Discriminant for nature of roots
- Word problems (real applications)
In 2026, students get TWO board attempts — use the first as practice if needed. But MASTER this chapter — it's a high-mark, high-impact topic.
Quadratic equations are the gateway to advanced mathematics. Master them now.
Appendix — beyond the current syllabus
Not examinable in CBSE 2026-27. Solving quadratic equations by completing the square was removed from this chapter during rationalisation. Factorisation and the quadratic formula cover everything you are asked to solve. It is kept here because it explains where the quadratic formula comes from, and because older guidebooks still drill it.
Completing the square
- Write the equation in standard form and divide through by a, giving x² + (b/a)x + c/a = 0.
- Move the constant across: x² + (b/a)x = −c/a.
- Add (b/2a)² to both sides — this is the step that "completes" the square.
- The left side is now (x + b/2a)².
- Take square roots of both sides and solve for x.
Example — solve x² − 6x + 5 = 0.
x² − 6x = −5. Half of 6 is 3, and 3² = 9, so add 9 to both sides:
x² − 6x + 9 = 4, that is (x − 3)² = 4.
Then x − 3 = ±2, giving x = 5 or x = 1 — the same roots factorisation gives.
Run this same process on the general equation ax² + bx + c = 0 and what drops out is exactly the quadratic formula. That is the reason the technique is worth seeing once, even though it is no longer examined.
