﻿ Class 11 NCERT Math Solution
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TOPICS
Exercise - 5.3

Question-1 :- Solve x2 + 3 = 0.

Solution :-
```  We have, x2 + 3 = 0
x2 = -3
x = ±√-3 = ±√3i
```

Question-2 :-  Solve 2x2 + x + 1 = 0.

Solution :-
```  We have, 2x2 + x + 1 = 0
Now, a = 2; b = 1; c = 1
b2 - 4ac = 12 - 4 x 2 x 1 = 1 - 8 = -7
Therefore, the solutions are given by x = (-1 ± √-7)/4 = (-1 ± √7i)/4
```

Question-3 :- Solve x2 + 3x + 9 = 0.

Solution :-
```  We have, x2 + 3x + 9 = 0
Now, a = 1; b = 3; c = 9
b2 - 4ac = 32 - 4 x 1 x 9 = 9 - 36 = -27
Therefore, the solutions are given by x = (-3 ± √-27)/2 = (-3 ± 3√3i)/2
```

Question-4 :-  Solve -x2 + x - 2 = 0.

Solution :-
```  We have, -x2 + x - 2 = 0
Now, a = -1; b = 1; c = -2
b2 - 4ac = 12 - 4 x (-1) x (-2) = 1 - 8 = -7
Therefore, the solutions are given by x = (-1 ± √-7)/(-2) = (-1 ± √7i)/(-2)
```

Question-5 :-  Solve x2 + 3x + 5 = 0.

Solution :-
```  We have, x2 + 3x + 5 = 0
Now, a = 1; b = 3; c = 5
b2 - 4ac = 32 - 4 x 1 x 5 = 9 - 20 = -11
Therefore, the solutions are given by x = (-3 ± √-11)/2 = (-3 ± √11i)/2
```

Question-6 :-  Solve x2 - x + 2 = 0.

Solution :-
```  We have, x2 - x + 2 = 0
Now, a = 1; b = -1; c = 2
b2 - 4ac = (-1)2 - 4 x 1 x 2 = 1 - 8 = -7
Therefore, the solutions are given by x = (1 ± √-7)/2 = (1 ± √7i)/2
```

Question-7 :-  Solve √2x2 + x + √2 = 0.

Solution :-
```  We have, √2x2 + x + √2 = 0
Now, a = √2; b = 1; c = √2
b2 - 4ac = 12 - 4 x √2 x √2 = 1 - 8 = -7
Therefore, the solutions are given by x = (-1 ± √-7)/2√2 = (-1 ± √7i)/2√2
```

Question-8 :-  Solve √3x2 - √2x + 3√3 = 0.

Solution :-
```  We have, √3x2 - √2x + 3√3 = 0
Now, a = √3; b = -√2; c = 3√3
b2 - 4ac = (-√2)2 - 4 x √3 x 3√3 = 2 - 36 = -34
Therefore, the solutions are given by x = (√2 ± √-34)/2√3 = (√2 ± √34i)/2√3
```

Question-9 :-  Solve x2 + x + 1/√2 = 0.

Solution :-
```  We have, x2 + x + 1/√2 = 0
√2x2 + √2x + 1 = 0
Now, a = √2; b = √2; c = 1
b2 - 4ac = (√2)2 - 4 x √2 x 1 = 2 - 4√2 = -2(2√2 - 1)
Therefore, the solutions are given by x = [-√2 ± √-2(2√2 - 1)]/2√2
= [-√2 ± √2√(2√2 - 1)i]/2√2
= [-1 ± √(2√2 - 1)i]/2
```

Question-10 :-  Solve x2 + x/√2 + 1 = 0.

Solution :-
```  We have, x2 + x/√2 + 1 = 0
√2x2 + x + √2 = 0
Now, a = √2; b = 1; c = √2
b2 - 4ac = 12 - 4 x √2 x √2 = 1 - 8 = -7
Therefore, the solutions are given by x = (-1 ± √-7)/2√2 = (-1 ± √7i)/2√2
```
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