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

• Last Updated : 10 Mar, 2021

### Question 1. x2 + 3 = 0

Solution:

We have,

x2 + 3 = 0 or x2 + 0 Ã— 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 = Â± âˆš3 i

Hence , the solution of x2 + 3 = 0 is Â± âˆš3 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 â€“ 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. âˆš2x2 + x + âˆš2 = 0

Solution:

We have,

âˆš2x2 + x + âˆš2 = 0 —(1)

Discriminant, D = b2 – 4ac

from (1) , a = âˆš2 , b = 1 and c = âˆš2

D = (1)2 – 4*(âˆš2)*(âˆš2)

D =  -7

Since , x =

Therefore ,

x =

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

### Question 8. âˆš3x2  – âˆš2x + 3âˆš3 = 0

Solution:

We have,

âˆš3x2 – âˆš2x + 3âˆš3 = 0 —(1)

Discriminant, D = b2 – 4ac

from (1) , a = âˆš3 , b = -âˆš2 and c = 3âˆš3

D = (-âˆš2)2 – 4*(âˆš3)*(3âˆš3)

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  âˆš2x2 + âˆš2x + 1 = 0 —(1)

Discriminant, D = b2 – 4ac

from (1) , a = âˆš2 , b = âˆš2 and c = 1

D = (âˆš2)2 – 4*(âˆš2)*(1)

D =  2 – 4âˆš2 = 2 ( 1 – 2âˆš2 )

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 âˆš2x2 + x + âˆš2 = 0 —(1)

Discriminant, D = b2 – 4ac

from (1) , a = âˆš2 , b = 1 and c = âˆš2

D = (1)2 – 4*(âˆš2)*(âˆš2)

D =  -7

Since , x =

Therefore ,

x =

x =

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

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