### Solve each of the following equations:

**Question **1. x^{2} + 3 = 0

**Solution:**

We have,

x

^{2}+ 3 = 0 or x^{2}+ 0 × x + 3 = 0 —(1)Discriminant, D = b

^{2}– 4acfrom (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 x

^{2}+ 3 = 0 is ± √3 i

**Question **2. 2x^{2} + x + 1 = 0

**Solution:**

We have,

2x

^{2}+ x + 1 = 0 —(1)Discriminant, D = b

^{2}– 4acfrom (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 2x

^{2}+ x + 1 = 0 is .

**Question **3. x^{2} + 3x + 9 = 0

**Solution:**

We have,

x

^{2}+ 3x + 9 = 0 —(1)Discriminant, D = b

^{2}– 4acfrom (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 x

^{2}+ 3x + 1 = 0 is .

**Question **4. -x^{2}+ x – 2 = 0

**Solution:**

We have,

-x

^{2}+ x – 2 = 0 —(1)Discriminant, D = b

^{2}– 4acfrom (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 -x

^{2}+ x – 2= 0 is .

**Question **5. x^{2} + 3x + 5 = 0

**Solution:**

We have,

x

^{2}+ 3x + 5 = 0 —(1)Discriminant, D = b

^{2}– 4acfrom (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 -x

^{2}+ x – 2= 0 is .

**Question **6. x^{2} – x + 2 = 0

**Solution:**

We have,

x

^{2}– x + 2 = 0 —(1)Discriminant, D = b

^{2}– 4acfrom (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 -x

^{2}+ x – 2= 0 is .

**Question **7. √2x^{2} + x + √2 = 0

**Solution:**

We have,

√2x

^{2}+ x + √2 = 0 —(1)Discriminant, D = b

^{2}– 4acfrom (1) , a = √2 , b = 1 and c = √2

D = (1)

^{2}– 4*(√2)*(√2)D = -7

Since , x =

Therefore ,

x =

Hence , the solution of -x

^{2}+ x – 2= 0 is .

**Question **8. √3x^{2 }– √2x + 3√3 = 0

**Solution:**

We have,

√3x

^{2}– √2x + 3√3 = 0 —(1)Discriminant, D = b

^{2}– 4acfrom (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 -x

^{2}+ x – 2= 0 is .

**Question **9. x^{2 }+ x + = 0

**Solution:**

We have,

x

^{2}+ x + = 0 or √2x^{2 }+ √2x + 1 = 0 —(1)Discriminant, D = b

^{2}– 4acfrom (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 -x

^{2}+ x – 2= 0 is .

**Question **10. x^{2 }+ + 1 = 0

**Solution:**

We have,

x

^{2}+ + 1 = 0 or √2x^{2 }+ x + √2 = 0 —(1)Discriminant, D = b

^{2}– 4acfrom (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 x

^{2}+ + 1 = 0 is .