Quadratic Equations
Board: GSEB | Class: Std 10
Comprehensive study notes for Quadratic Equations by Ajay Yadav (Math King of Katargam). Master every concept with clear explanations, solved examples, and practice problems.
Key Concepts
Standard Form
A quadratic equation in x is of the form ax² + bx + c = 0 where a ≠ 0. The highest power of the variable is 2.
Solving by Factorization
Write ax² + bx + c as (px + q)(rx + s) = 0. Then set each factor to zero: px + q = 0 or rx + s = 0. This gives the roots.
Solving by Completing the Square
Rewrite ax² + bx + c = 0 as (x + b/2a)² = (b² – 4ac)/4a². Take square root on both sides. This method derives the quadratic formula.
Quadratic Formula
x = (-b ± √b² – 4ac)/2a. The expression D = b² – 4ac is called the discriminant.
Nature of Roots
D > 0: Two distinct real roots. D = 0: One real root (repeated). D < 0: No real roots (complex conjugates).
Word Problems
Quadratic equations model many real-world situations: area problems, number problems, time-speed-distance, age problems, and profit-loss scenarios.
Important Formulas
| Standard Form | ax² + bx + c = 0, a ≠ 0 |
| Quadratic Formula | x = (-b ± √D)/2a where D = b² - 4ac |
| Discriminant | D = b² - 4ac |
| D > 0 | Two distinct real roots |
| D = 0 | One real root (repeated) |
| D < 0 | No real roots |
Solved Examples
Example 1: Solve x² – 5x + 6 = 0 by factorization.
Solution: x² – 5x + 6 = (x-2)(x-3) = 0. x = 2 or x = 3.
Example 2: Solve 2x² – 5x + 3 = 0 using the formula.
Solution: a=2, b=-5, c=3. D = 25 – 24 = 1. x = (5 ± 1)/4. x = 3/2 or x = 1.
Example 3: Find the discriminant of x² + 4x + 5 = 0.
Solution: D = 16 – 20 = -4. Since D < 0, no real roots exist.
Practice Questions
- Solve: 2x² – 7x + 3 = 0.
- Find k such that x² + kx + 4 = 0 has equal roots.
- The area of a rectangle is 240 cm². Its length is 8 cm more than width. Find dimensions.
- Solve: 9x² + 7x – 2 = 0 using the quadratic formula.
- If the sum of two numbers is 15 and sum of squares is 113, find the numbers.
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