NCERT Solutions Class 11th Maths Chapter – 5 Complex Numbers and Quadratic Equations Exercise 5.3
| Textbook | NCERT |
| class | Class – 11th |
| Subject | Mathematics |
| Chapter | Chapter – 5 |
| Chapter Name | Complex Numbers and Quadratic Equations |
| grade | Class 11th Maths solution |
| Medium | English |
| Source | last doubt |
NCERT Solutions Class 11th Maths Chapter – 5 Complex Numbers and Quadratic Equations Exercise 5.3
?Chapter – 5?
✍Complex Numbers and Quadratic Equations✍
?Exercise 5.3?
Solve each of the following equations:
1. x2 + 3 = 0
Solution:
Given quadratic equation,
x2 + 3 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 1, b = 0, and c = 3
So, the discriminant of the given equation will be
D = b2 – 4ac = 02 – 4 × 1 × 3 = –12
Hence, the required solutions are:
2. 2x2 + x + 1 = 0
Solution:
Given quadratic equation,
2x2 + x + 1 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 2, b = 1, and c = 1
So, the discriminant of the given equation will be
D = b2 – 4ac = 12 – 4 × 2 × 1 = 1 – 8 = –7
Hence, the required solutions are:
3. x2 + 3x + 9 = 0
Solution:
Given quadratic equation,
x2 + 3x + 9 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 1, b = 3, and c = 9
So, the discriminant of the given equation will be
D = b2 – 4ac = 32 – 4 × 1 × 9 = 9 – 36 = –27
Hence, the required solutions are:
4. –x2 + x – 2 = 0
Solution:
Given quadratic equation,
–x2 + x – 2 = 0
On comparing it with ax2 + bx + c = 0, we have
a = –1, b = 1, and c = –2
So, the discriminant of the given equation will be
D = b2 – 4ac = 12 – 4 × (–1) × (–2) = 1 – 8 = –7
Hence, the required solutions are:
5. x2 + 3x + 5 = 0
Solution:
Given quadratic equation,
x2 + 3x + 5 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 1, b = 3, and c = 5
So, the discriminant of the given equation will be
D = b2 – 4ac = 32 – 4 × 1 × 5 =9 – 20 = –11
Hence, the required solutions are:
6. x2 – x + 2 = 0
Solution:
Given quadratic equation,
x2 – x + 2 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 1, b = –1, and c = 2
So, the discriminant of the given equation is
D = b2 – 4ac = (–1)2 – 4 × 1 × 2 = 1 – 8 = –7
Hence, the required solutions are
7. √2x2 + x + √2 = 0
Solution:
Given quadratic equation,
√2x2 + x + √2 = 0
On comparing it with ax2 + bx + c = 0, we have
a = √2, b = 1, and c = √2
So, the discriminant of the given equation is
D = b2 – 4ac = (1)2 – 4 × √2 × √2 = 1 – 8 = –7
Hence, the required solutions are:
8. √3x2 – √2x + 3√3 = 0
Solution:
Given quadratic equation,
√3x2 – √2x + 3√3 = 0
On comparing it with ax2 + bx + c = 0, we have
a = √3, b = -√2, and c = 3√3
So, the discriminant of the given equation is
D = b2 – 4ac = (-√2)2 – 4 × √3 × 3√3 = 2 – 36 = –34
Hence, the required solutions are:
9. x2 + x + 1/√2 = 0
Solution:
Given quadratic equation,
x2 + x + 1/√2 = 0
It can be rewritten as,
√2x2 + √2x + 1 = 0
On comparing it with ax2 + bx + c = 0, we have
a = √2, b = √2, and c = 1
So, the discriminant of the given equation is
D = b2 – 4ac = (√2)2 – 4 × √2 × 1 = 2 – 4√2 = 2(1 – 2√2)
Hence, the required solutions are:
10. x2 + x/√2 + 1 = 0
Solution:
Given quadratic equation,
x2 + x/√2 + 1 = 0
It can be rewritten as,
√2x2 + x + √2 = 0
On comparing it with ax2 + bx + c = 0, we have
a = √2, b = 1, and c = √2
So, the discriminant of the given equation is
D = b2 – 4ac = (1)2 – 4 × √2 × √2 = 1 – 8 = -7
Hence, the required solutions are:
