NCERT Solutions Class 11 Mathematics Complex Numbers Exercise 5.3

Question 5:
Solve the equation x² + 3x + 5 = 0

Answer:
The given quadratic equation is x² + 3x + 5 = 0
On comparing the given equation with ax2 + bx + c = 0, we get
a = 1, b = 3, and c = 5
Therefore, the discriminant of the given equation is
D = b² – 4ac = 3² – 4 * 1 * 5 =9 – 20 = –11
Therefore, the required solutions = (-b ± √D)/2a
                                                             = {-3 ± √(-11)}/(2 * 1)
                                                             = {-3 ±√(-1) * √11}/2
                                                             = (-3 ± i√11)/2                       [Since i = √(-1)]

Question 6:
Solve the equation x² – x + 2 = 0

Answer:
The given quadratic equation is x² – x + 2 = 0
On comparing the given equation with ax² + bx + c = 0, we get
a = 1, b = –1, and c = 2
Therefore, the discriminant of the given equation is
D = b² – 4ac = (–1)² – 4 * 1 * 2 = 1 – 8 = –7
Therefore, the required solutions = (-b ± √D)/2a
                                                             = {1 ± √(-7)}/(2 * 1)
                                                             = {1 ±√(-1) * √7}/2
                                                             = (1 ± i√7)/2                       [Since i = √(-1)]

Question 7:
Solve the equation √2x² + x + √2 = 0

Answer:
The given quadratic equation is √2x² + x + √2 = 0
On comparing the given equation with ax² + bx + c = 0, we get
a = √2, b = 1, and c = √2
Therefore, the discriminant of the given equation is
D = b² – 4ac = 1² – 4 * √2 * √2 = 1 – 8 = –7
Therefore, the required solutions = (-b ± √D)/2a
                                                             = {-1 ± √(-7)}/(2 * √2)
                                                             = {-1 ±√(-1) * √7}/2√2
                                                             = (-1 ± i√7)/2√2                       [Since i = √(-1)]

Question 8:
Solve the equation √3x² – √2x + 3√3 = 0

Answer:
The given quadratic equation is √3x² – √2x + 3√3 = 0
On comparing the given equation with ax² + bx + c = 0, we get
a = √3, b = -√2, and c = 3√3
Therefore, the discriminant of the given equation is
D = b² – 4ac = (-√2)² – 4 * √3 * 3√3 = 2 – 36 = –34
Therefore, the required solutions = (-b ± √D)/2a
                                                             = {√2 ± √(-34)}/(2 * √3)
                                                             = {√2 ±√(-1) * √34}/2√3
                                                             = (√2 ± i√34)/2√3                       [Since i = √(-1)]

Question 9:
Solve the equation: x² + x + 1/√2 = 0

Answer:
The given quadratic equation is x² + x + 1/√2 = 0
⇒ √2x² + √2x + 1 = 0
On comparing the given equation with ax² + bx + c = 0, we get
a = √2, b = √2, and c = 1
Therefore, the discriminant of the given equation is
D = b² – 4ac = (√2)² – 4 * √2 * 1 = 2 – 4√2
Therefore, the required solutions = (-b ± √D)/2a
                                                             = [-√2 ± √(2 – 4√2)/(2 * √2)
                                                             = [-√2 ± √{2(1 – 2√2)}/2√2
                                                             = [-√2 ± √2√{2√2 – 1)} * √(-1)]/2√2
                                                             = [-1 ± √ (2√2 – 1)i]/2                             [Since i = √(-1)]
 
 
 

Question 10:
Solve the equation x² + x/√2 + 1 = 0

Answer:
The given quadratic equation is x² + x/√2 + 1 = 0
⇒ √2x² + x + √2 = 0
On comparing the given equation with ax² + bx + c = 0, we get
a = √2, b = 1, and c = √2
Therefore, the discriminant of the given equation is
D = b² – 4ac = 1² – 4 * √2 * √2 = 1 – 8 = –7
Therefore, the required solutions = (-b ± √D)/2a
                                                             = {-1 ± √(-7)}/(2 * √2)
                                                             = {-1 ±√(-1) * √7}/2√2
                                                             = (-1 ± i√7)/2√2                       [Since i = √(-1)]