Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations – Exercise 5.3

Solve each of the following equations:

Question 1. x² + 3 = 0

Solution:

We have,

x² + 3 = 0 or x² + 0 × x + 3 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = 1 , b = 0 and c = 3

D = (0)² – 4*(1)*(3)

D = -12 

Since , x = 

Therefore , 

x = 

x = 

x = ± √3 i

Hence , the solution of x² + 3 = 0 is ± √3 i

Question 2. 2x² + x + 1 = 0

Solution:

We have,

2x² + x + 1 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = 2 , b = 1 and c = 1

D = (1)² – 4*(2)*(1)

D = -7

Since , x = 

Therefore ,

x = 

x = 

Hence , the solution of  2x² + x + 1 = 0  is  .

Question 3. x² + 3x + 9 = 0

Solution:

We have,

x² + 3x + 9 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = 1 , b = 3 and c = 9

D = (3)² – 4*(1)*(9)

D = -27

Since , x = 

Therefore ,

x = 

x = 

Hence , the solution of  x² + 3x + 1 = 0  is  .

Question 4. -x²+ x – 2 = 0

Solution:

We have,

-x² + x – 2 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = -1 , b = 1 and c = -2

D = (1)² – 4*(-1)*(-2)

D = -7

Since , x = 

Therefore ,

x = 

x = 

Hence , the solution of  -x² + x – 2= 0  is  .

Question 5. x² + 3x + 5 = 0

Solution:

We have,

x² + 3x + 5 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = 1 , b = 3 and c = 5

D = (3)² – 4*(1)*(5)

D = -11

Since , x = 

Therefore ,

x = 

x = 

Hence , the solution of  -x² + x – 2= 0  is  .

Question 6. x² – x + 2 = 0

Solution:

We have,

x² – x + 2 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = 1 , b = -1 and c = 2

D = (-1)² – 4*(1)*(2)

D =  -7

Since , x = 

Therefore ,

x = 

x = 

Hence , the solution of  -x² + x – 2= 0  is  .

Question 7. √2x² + x + √2 = 0

Solution:

We have,

√2x² + x + √2 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = √2 , b = 1 and c = √2

D = (1)² – 4*(√2)*(√2)

D =  -7

Since , x = 

Therefore ,

x = 

Hence , the solution of  -x² + x – 2= 0  is  .

Question 8. √3x²  – √2x + 3√3 = 0

Solution:

We have,

√3x² – √2x + 3√3 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = √3 , b = -√2 and c = 3√3

D = (-√2)² – 4*(√3)*(3√3)

D =  -34

Since , x = 

Therefore ,

x = 

x =  

Hence , the solution of  -x² + x – 2= 0  is .

Question 9. x² + x +   = 0

Solution:

We have,

x² + x +  = 0  or  √2x² + √2x + 1 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = √2 , b = √2 and c = 1

D = (√2)² – 4*(√2)*(1)

D =  2 – 4√2 = 2 ( 1 – 2√2 )

Since , x = 

Therefore ,

x = 

x = 

x = 

Hence , the solution of  -x² + x – 2= 0  is .

Question 10. x² +  + 1 = 0

Solution:

We have,

x² +  + 1 = 0  or √2x² + x + √2 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = √2 , b = 1 and c = √2

D = (1)² – 4*(√2)*(√2)

D =  -7

Since , x = 

Therefore ,

x = 

x = 

Hence , the solution of  x² +  + 1 = 0  is .