Quadratic Equations Class 10 – Solving Methods, Roots & Discriminant Explained
Quadratic Equations are one of the most important chapters in Class 10 Mathematics (CBSE). They carry significant weightage in board examinations and form the foundation for higher algebra.
In this detailed guide, you will learn:
- Standard form of quadratic equations
- Methods to solve quadratic equations
- Roots and their meaning
- Discriminant and nature of roots
- Step-by-step solved problems
1. What is a Quadratic Equation?
A quadratic equation is a polynomial equation of degree 2. The standard form is:
ax² + bx + c = 0
Where:
- a ≠ 0
- a, b, c are real numbers
- x is the variable
The highest power of x is 2, which makes it quadratic.
2. Methods of Solving Quadratic Equations
There are three main methods used in Class 10:
- Factorization Method
- Completing the Square Method
- Quadratic Formula Method
A. Factorization Method
This method is used when the quadratic expression can be factorized easily.
Example 1: Solve x² − 5x + 6 = 0
Step 1: Split middle term
x² − 2x − 3x + 6 = 0
Step 2: Factorize
x(x − 2) − 3(x − 2) = 0
(x − 2)(x − 3) = 0
Roots: x = 2, x = 3
B. Completing the Square Method
Used when factorization is not simple.
Example 2: Solve x² + 6x + 5 = 0
x² + 6x = −5
Add (6/2)² = 9 to both sides:
x² + 6x + 9 = 4
(x + 3)² = 4
x + 3 = ±2
x = −1 or x = −5
C. Quadratic Formula Method
For equation ax² + bx + c = 0:
x = [-b ± √(b² − 4ac)] / 2a
3. Discriminant and Nature of Roots
The expression inside the square root is called the discriminant.
D = b² − 4ac
- D > 0 → Two distinct real roots
- D = 0 → Equal roots
- D < 0 → No real roots
Example 3: Find nature of roots of 2x² − 4x + 2 = 0
D = (-4)² − 4(2)(2)
D = 16 − 16 = 0
Since D = 0, roots are real and equal.
4. Solved Problems (Exam-Oriented)
Problem 1: Solve 3x² − 5x − 2 = 0
a = 3, b = -5, c = -2
D = (-5)² − 4(3)(-2)
D = 25 + 24 = 49
x = [5 ± √49] / 6
x = (5 ± 7) / 6
x = 2 or x = -1/3
Problem 2: Find two consecutive positive integers whose squares sum to 365.
Let first integer = x
Second integer = x + 1
x² + (x + 1)² = 365
x² + x² + 2x + 1 = 365
2x² + 2x − 364 = 0
x² + x − 182 = 0
(x + 14)(x − 13) = 0
x = 13
Numbers: 13 and 14
Problem 3: Solve 4x² + 4x + 1 = 0
D = 4² − 4(4)(1)
D = 16 − 16 = 0
x = [-4] / 8
x = -1/2
Roots are equal.
Problem 4: Find nature of roots of x² + 4x + 8 = 0
D = 4² − 4(1)(8)
D = 16 − 32 = -16
Since D < 0, no real roots.
Problem 5: Find quadratic equation whose roots are 3 and -2
Sum = 3 + (-2) = 1
Product = 3 × (-2) = -6
Equation: x² − (Sum)x + Product = 0
x² − x − 6 = 0
5. Word Problems Based on Quadratic Equations
Word problems test application skills. Follow these steps:
- Assume variable clearly
- Frame equation correctly
- Solve using appropriate method
- Verify answer
6. Important Exam Tips
✔ Use factorization only if splitting is easy
✔ Be careful with negative signs
✔ Check final answers by substitution
Common Mistakes Students Make
- Forgetting ± sign in quadratic formula
- Calculation mistakes in discriminant
- Incorrect splitting of middle term
- Not writing final roots clearly
Final Revision Strategy for Boards
Practice at least 15–20 quadratic equations daily. Focus on:
- Nature of roots questions
- Word problems
- Formation of quadratic equation
Mastering this chapter ensures strong performance in Class 10 board exams.
