# 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. 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 â‰  0.

Solving the problems of this exercise helps students in understanding how a problem on Quadratic Equation should be answered. The exercise contains 10 questions, which the students can solve and practice. They can refer toÂ NCERT Solutions for class 11 Maths for a better understanding of the concepts.

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### Access Other Exercise Solutions of 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 for Class 11 Maths Chapter 5.3 Exercise

Solve each of the following equations:

1. x2 + 3 = 0

Solution:

x2Â + 3 = 0

On comparing it withÂ ax2Â +Â bxÂ +Â cÂ = 0, we have

aÂ = 1,Â bÂ = 0, andÂ cÂ = 3

So, the discriminant of the given equation will be

D =Â b2Â â€“ 4acÂ = 02Â â€“ 4 Ã— 1 Ã— 3 = â€“12

Hence, the required solutions are:

2. 2x2 + x + 1 = 0

Solution:

2x2Â +Â xÂ + 1 = 0

On comparing it withÂ ax2Â +Â bxÂ +Â cÂ = 0, we have

aÂ = 2,Â bÂ = 1, andÂ cÂ = 1

So, the discriminant of the given equation will be

D =Â b2Â â€“ 4acÂ = 12Â â€“ 4 Ã— 2 Ã— 1 = 1 â€“ 8 = â€“7

Hence, the required solutions are:

3. x2 + 3x + 9 = 0

Solution:

x2Â + 3xÂ + 9 = 0

On comparing it withÂ ax2Â +Â bxÂ +Â cÂ = 0, we have

aÂ = 1,Â bÂ = 3, andÂ cÂ = 9

So, the discriminant of the given equation will be

D =Â b2Â â€“ 4acÂ = 32Â â€“ 4 Ã— 1 Ã— 9 = 9 â€“ 36 = â€“27

Hence, the required solutions are:

4. â€“x2Â +Â xÂ â€“ 2 = 0

Solution:

â€“x2Â +Â xÂ â€“ 2 = 0

On comparing it withÂ ax2Â +Â bxÂ +Â cÂ = 0, we have

aÂ = â€“1,Â bÂ = 1, andÂ cÂ = â€“2

So, the discriminant of the given equation will be

D =Â b2Â â€“ 4acÂ = 12Â â€“ 4 Ã— (â€“1) Ã— (â€“2) = 1 â€“ 8 = â€“7

Hence, the required solutions are:

5. x2Â + 3xÂ + 5 = 0

Solution:

x2Â + 3xÂ + 5 = 0

On comparing it withÂ ax2Â +Â bxÂ +Â cÂ = 0, we have

aÂ = 1,Â bÂ = 3, andÂ cÂ = 5

So, the discriminant of the given equation will be

D =Â b2Â â€“ 4acÂ = 32Â â€“ 4 Ã— 1 Ã— 5 =9 â€“ 20 = â€“11

Hence, the required solutions are:

6. x2Â â€“Â xÂ + 2 = 0

Solution:

x2Â â€“Â xÂ + 2 = 0

On comparing it withÂ ax2Â +Â bxÂ +Â cÂ = 0, we have

aÂ = 1,Â bÂ = â€“1, andÂ cÂ = 2

So, the discriminant of the given equation is

D =Â b2Â â€“ 4acÂ = (â€“1)2Â â€“ 4 Ã— 1 Ã— 2 = 1 â€“ 8 = â€“7

Hence, the required solutions are

7. âˆš2x2Â + xÂ + âˆš2 = 0

Solution:

âˆš2x2Â + xÂ + âˆš2 = 0

On comparing it withÂ ax2Â +Â bxÂ +Â cÂ = 0, we have

aÂ = âˆš2,Â bÂ = 1, andÂ cÂ = âˆš2

So, the discriminant of the given equation is

D =Â b2Â â€“ 4acÂ = (1)2Â â€“ 4 Ã— âˆš2 Ã— âˆš2 = 1 â€“ 8 = â€“7

Hence, the required solutions are:

8. âˆš3x2Â â€“ âˆš2xÂ + 3âˆš3 = 0

Solution:

âˆš3x2Â â€“ âˆš2xÂ + 3âˆš3 = 0

On comparing it withÂ ax2Â +Â bxÂ +Â cÂ = 0, we have

aÂ = âˆš3,Â bÂ = -âˆš2, andÂ cÂ = 3âˆš3

So, the discriminant of the given equation is

D =Â b2Â â€“ 4acÂ = (-âˆš2)2Â â€“ 4 Ã— âˆš3 Ã— 3âˆš3 = 2 â€“ 36 = â€“34

Hence, the required solutions are:

9. x2Â + xÂ + 1/âˆš2 = 0

Solution:

x2Â + xÂ + 1/âˆš2 = 0

It can be rewritten as,

âˆš2x2Â + âˆš2xÂ + 1 = 0

On comparing it withÂ ax2Â +Â bxÂ +Â cÂ = 0, we have

aÂ = âˆš2,Â bÂ = âˆš2, andÂ cÂ = 1

So, the discriminant of the given equation is

D =Â b2Â â€“ 4acÂ = (âˆš2)2Â â€“ 4 Ã— âˆš2 Ã— 1 = 2 â€“ 4âˆš2 = 2(1 â€“ 2âˆš2)

Hence, the required solutions are:

10. x2Â + x/âˆš2Â + 1 = 0

Solution:

x2Â + x/âˆš2Â + 1 = 0

It can be rewritten as,

âˆš2x2Â + xÂ + âˆš2 = 0

On comparing it withÂ ax2Â +Â bxÂ +Â cÂ = 0, we have

aÂ = âˆš2,Â bÂ = 1, andÂ cÂ = âˆš2

So, the discriminant of the given equation is

D =Â b2Â â€“ 4acÂ = (1)2Â â€“ 4 Ã— âˆš2 Ã— âˆš2 = 1 â€“ 8 = -7

Hence, the required solutions are: