1. Standard Form
ax² + bx + c = 0
- a, b, c are real numbers
- a ? 0 (if a = 0, it becomes linear)
- Highest power of x is 2
2. Methods to Solve Quadratic Equations
A. Factorisation Method
- Write equation in standard form
- Factorize left side
- Set each factor = 0
- Solve for x
Example: x² - 5x + 6 = 0
? (x-2)(x-3) = 0
? x = 2 or x = 3
B. Completing the Square
- Write as: ax² + bx = -c
- Divide by a: x² + (b/a)x = -c/a
- Add (b/2a)² to both sides
- Write as perfect square
- Take square root
- Solve for x
C. Quadratic Formula
x = [-b ± v(b² - 4ac)] / 2a
D = b² - 4ac is called the discriminant
3. Nature of Roots
D > 0
Two distinct real roots
Roots are:
x = [-b + vD]/2a
x = [-b - vD]/2a
D = 0
Two equal real roots
x = -b/2a
D < 0
No real roots
Imaginary/complex roots
4. Relationship between Roots and Coefficients
If a and ß are roots of ax² + bx + c = 0:
Sum of roots: a + ß = -b/a
Product of roots: aß = c/a
Quadratic with given roots a and ß:
x² - (a+ß)x + aß = 0
5. Discriminant (D) Applications
- D = perfect square: Roots are rational
- D > 0 but not perfect square: Roots are irrational
- D = 0: Roots are real and equal
- D < 0: No real roots (roots are complex)
6. Important Formulas
• a² + ß² = (a+ß)² - 2aß
• a² - ß² = (a+ß)(a-ß)
• a³ + ß³ = (a+ß)³ - 3aß(a+ß)
• a³ - ß³ = (a-ß)³ + 3aß(a-ß)
• (a-ß) = v[(a+ß)² - 4aß]
7. Word Problem Types (NCERT)
- Area problems: Rectangle, triangle areas
- Number problems: Product, sum conditions
- Age problems: Present/past ages
- Speed problems: Time, distance, speed
- Geometry problems: Right triangles, Pythagoras
8. Special Cases
Pure Quadratic Equations
ax² + c = 0
Solution: x = ±v(-c/a)
Quadratic in Other Forms
• ax4 + bx² + c = 0 (let y = x²)
• a(x+p)² + b(x+p) + c = 0 (let y = x+p)
• Reciprocal equations
9. Quick Reference
Standard Form
ax² + bx + c = 0
Quadratic Formula
x = [-b ± vD]/2a
Discriminant
D = b² - 4ac
10. Example (NCERT Style)
Problem: Find roots of 2x² - 7x + 3 = 0
Solution:
Using quadratic formula:
a = 2, b = -7, c = 3
D = (-7)² - 4×2×3 = 49 - 24 = 25
vD = 5
x = [7 ± 5] / 4
x1 = (7+5)/4 = 3, x2 = (7-5)/4 = 1/2