 # NCERT Solutions for Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations Exercise 5.3

To learn the method of solving problems, at first, students should get an idea of how the problems given in the exercises are solved. To know the solving process of these problems, understanding the concept is a must. Chapter 5 Complex Numbers and Quadratic Equations of Class 11 Maths, is categorised under the CBSE Syllabus for 2022-23. Exercise 5.3 of NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations is based on Quadratic Equations.

In algebra, a quadratic equation is any equation having the form ax2+bx+c=0, where x represents an unknown, and a, b and c represent known numbers, with a ≠ 0.

Solving the problems of this exercise helps students understand how a problem on Quadratic Equation should be answered. The exercise contains 10 questions, which the students can solve and practise. They can refer to NCERT Solutions for Class 11 Maths for a better understanding of the concepts and excel in the board exam.

### Download the PDF of NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations Exercise 5.3    ### Access Other Exercise Solutions to Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations

Exercise 5.1 Solutions 14 Questions

Exercise 5.2 Solutions 8 Questions

Miscellaneous Exercise on Chapter 5 Solutions 20 Questions

#### Access Solutions to Class 11 Maths Chapter 5 Exercise 5.3

Solve each of the following equations:

1. x2 + 3 = 0

Solution:

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:

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:

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:

x2 + – 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:

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:

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:

√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:

√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:

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:

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: 