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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 × 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 .

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