1. Quadratic Equation – Introduction
A quadratic equation is an equation of the form:
ax² + bx + c = 0, where a ≠ 0
x is the unknown, and a, b, c are constants.
Examples:
x² − 5x + 6 = 0
2x² + 3x − 2 = 0
2. Types of Quadratic Equations
Standard Form: ax² + bx + c = 0
Factorable Form: Can be written as (px + q)(rx + s) = 0
Incomplete Quadratic: Missing either bx or c term, e.g.,
x² + 5x = 0
x² − 9 = 0
3. Methods to Solve Quadratic Equations
(a) Factorization Method
Split middle term or factor common terms.
Set each factor = 0 to find x.
Example:
Q: x² − 5x + 6 = 0
Solution:
x² − 5x + 6 = (x − 2)(x − 3) = 0
x − 2 = 0 → x = 2
x − 3 = 0 → x = 3
(b) Using Quadratic Formula
For ax² + bx + c = 0,
x = [−b ± √(b² − 4ac)] / 2a
Discriminant (D): D = b² − 4ac
D > 0 → Two real and distinct roots
D = 0 → Two real and equal roots
D < 0 → No real roots (imaginary roots)
Example:
Q: 2x² − 3x − 2 = 0
Solution:
a = 2, b = −3, c = −2
D = (−3)² − 4 × 2 × (−2) = 9 + 16 = 25
x = [3 ± √25] / 4
x = (3 + 5)/4 = 8/4 = 2
x = (3 − 5)/4 = −2/4 = −0.5
Roots: x = 2, x = −0.5
(c) Completing the Square
Make the equation in the form: (x + p)² = q
Then take square root both sides
Steps:
Divide by a if a ≠ 1
Move c to the other side
Add (b/2)² to both sides
Factor left as square, solve for x
Example:
Q: x² + 6x − 7 = 0
Solution:
x² + 6x = 7
Add (6/2)² = 9 to both sides → x² + 6x + 9 = 16
(x + 3)² = 16
x + 3 = ±4
x = 1 or x = −7
4. Relation Between Roots and Coefficients
If quadratic is ax² + bx + c = 0 and roots are α, β
Sum of roots: α + β = −b / a
Product of roots: α × β = c / a
Example:
Q: 2x² − 5x + 3 = 0
Solution:
Sum = −(−5)/2 = 5/2
Product = 3/2
5. Nature of Roots
Discriminant (D) = b² − 4ac
| D | Nature of Roots |
|---|---|
| D > 0 and perfect square | Two real and rational roots |
| D > 0 and not perfect square | Two real and irrational roots |
| D = 0 | Two real and equal roots |
| D < 0 | Two imaginary roots |
6. Word Problems / Real-Life Applications
Age problem: Use x² equation to find age.
Area problem: Form x² + bx + c = 0 using dimensions.
Profit/Loss problem: Solve quadratic to find cost or price.
Example:
Q: Area of rectangle = 60, length = x + 2, breadth = x − 3. Find x.
Solution:
(x + 2)(x − 3) = 60
x² − x − 6 = 60
x² − x − 66 = 0
Factor: (x − 11)(x + 6) = 0
x = 11 (length positive)
7. Quick Tips / Tricks
Always check for factorable quadratics first.
Use quadratic formula if factorization is hard.
Use discriminant to check nature before solving.
Keep relation of roots in mind for shortcuts.
Important Questions – Quadratic Equations
Q1: Solve x² − 5x + 6 = 0
Solution:
Factor: (x − 2)(x − 3) = 0
x − 2 = 0 → x = 2
x − 3 = 0 → x = 3
Q2: Solve 2x² − 3x − 2 = 0
Solution:
a = 2, b = −3, c = −2
D = b² − 4ac = 9 + 16 = 25
x = [3 ± √25] / 4
x = 2, x = −0.5
Q3: Solve x² + 6x − 7 = 0 by completing square
Solution:
x² + 6x = 7
Add (6/2)² = 9 → x² + 6x + 9 = 16
(x + 3)² = 16
x + 3 = ±4
x = 1, x = −7
Q4: Solve 3x² + 10x − 8 = 0
Solution:
D = 10² − 4 × 3 × (−8) = 100 + 96 = 196
x = [−10 ± √196] / (2 × 3)
x = (−10 + 14)/6 = 4/6 = 2/3
x = (−10 − 14)/6 = −24/6 = −4
Q5: Solve x² − 2x − 15 = 0
Solution:
Factor: (x − 5)(x + 3) = 0
x = 5, x = −3
Q6: Sum and product problem: Sum of roots = 5, product = 6
Solution:
Equation: x² − (sum)x + product = 0
x² − 5x + 6 = 0
Factor: (x − 2)(x − 3) = 0
x = 2, x = 3
Q7: Word problem – Rectangle area 60, sides x + 2 and x − 3
Solution:
(x + 2)(x − 3) = 60
x² − x − 6 = 60
x² − x − 66 = 0
Factor: (x − 11)(x + 6) = 0
x = 11 (positive)
Q8: Solve 4x² + 12x + 9 = 0
Solution:
D = 12² − 4 × 4 × 9 = 144 − 144 = 0
x = −12 / (2 × 4) = −12 / 8 = −3/2
Q9: Solve 2x² − 7x + 3 = 0
Solution:
D = (−7)² − 4 × 2 × 3 = 49 − 24 = 25
x = [7 ± √25] / 4
x = (7 + 5)/4 = 12/4 = 3
x = (7 − 5)/4 = 2/4 = 0.5
Q10: Solve x² − x − 12 = 0
Solution:
Factor: (x − 4)(x + 3) = 0
x = 4, x = −3
Q11: Solve 5x² − 20x + 15 = 0
Solution:
Divide whole equation by 5 → x² − 4x + 3 = 0
Factor: (x − 1)(x − 3) = 0
x = 1, x = 3
Q12: Solve 3x² − 2x − 8 = 0
Solution:
D = (−2)² − 4 × 3 × (−8) = 4 + 96 = 100
x = [2 ± √100] / 6
x = (2 + 10)/6 = 12/6 = 2
x = (2 − 10)/6 = −8/6 = −4/3
Q13: Solve x² + 4x + 3 = 0
Solution:
Factor: (x + 3)(x + 1) = 0
x = −3, x = −1
Q14: Solve 2x² + 5x − 3 = 0
Solution:
D = 25 + 24 = 49
x = [−5 ± 7] / 4
x = (−5 + 7)/4 = 2/4 = 0.5
x = (−5 − 7)/4 = −12/4 = −3
Q15: Solve 6x² − x − 2 = 0
Solution:
D = (−1)² − 4 × 6 × (−2) = 1 + 48 = 49
x = [1 ± 7]/12
x = 8/12 = 2/3
x = −6/12 = −1/2
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