Question 1.
Find the square root of the following complex numbers:
(i) -8 – 6i
Solution:
(ii) 7 + 24i
Solution:
![Maharashtra Board Class 11 Maths Part 2 Chapter 1 Complex Numbers Ex 1.2 Solution 4 word image 21963 4](https://mhboardsolutions.xyz/wp-content/uploads/2022/03/word-image-21963-4.png)
![Maharashtra Board Class 11 Maths Part 2 Chapter 1 Complex Numbers Ex 1.2 Solution 5 word image 21963 5](https://mhboardsolutions.xyz/wp-content/uploads/2022/03/word-image-21963-5.jpeg)
(iii) 1 + 4√3 i
Solution:
(iv) 3 + 2√10 i
Solution:
(v) 2(1 – √3 i)
Solution:
Question 2.
Solve the following quadratic equations:
(i) 8 + 2x + 1 = 0
Solution:
Given equation is 8 + 2x + 1 = 0
Comparing with a + bx + c = 0, we get
a = 8, b = 2, c = 1
Discriminant = – 4ac
= (2 – 4 × 8 × 1
= 4 – 32
= -28 < 0
So, the given equation has complex roots.
These roots are given by
(ii) 2 – √3 x + 1 = 0
Solution:
Given equation is 2 – √3 x + 1 = 0
Comparing with a + bx + c = 0, we get
a = 2, b = -√3, c = 1
Discriminant = – 4ac
= (-√3 – 4 × 2 × 1
= 3 – 8
= -5 < 0
So, the given equation has complex roots.
These roots are given by
(iii) 3 – 7x + 5 = 0
Solution:
Given equation is 3 – 7x + 5 = 0
Comparing with a + bx + c = 0, we get
a = 3, b = -7, c = 5
Discriminant = – 4ac
= (-7 – 4 × 3 × 5
= 49 – 60
= -11 < 0
So, the given equation has complex roots.
These roots are given by
(iv) – 4x + 13 = 0
Solution:
Given equation is – 4x + 13 = 0
Comparing with a + bx + c = 0, we get
a = 1, b = -4, c = 13
Discriminant = – 4ac
= (-4 – 4 × 1 × 13
= 16 – 52
= -36 < 0
So, the given equation has complex roots.
These roots are given by
∴ The roots of the given equation are 2 + 3i and 2 – 3i.
Question 3.
Solve the following quadratic equations:
(i) + 3ix + 10 = 0
Solution:
Given equation is + 3ix + 10 = 0
Comparing with a + bx + c = 0, we get
a = 1, b = 3i, c = 10
Discriminant = – 4ac
= (3i – 4 × 1 × 10
= 9 – 40
= -9 – 40 ……[∵ = -1]
= -49 < 0
So, the given equation has complex roots.
These roots are given by
∴ x = 2i or x = -5i
∴ The roots of the given equation are 2i and -5i.
(ii) 2 + 3ix + 2 = 0
Solution:
Given equation is 2 + 3ix + 2 = 0
Comparing with ax + bx + c = 0, we get
a = 2, b = 3i, c = 2
Discriminant = – 4ac
= (3i – 4 × 2 × 2
= 9 – 16
= -9 – 16 …..[∵ = -1]
= -25 < 0
So, the given equation has complex roots.
These roots are given by
∴ The roots of the given equation are i and -2i.
(iii) + 4ix – 4 = 0
Solution:
Given equation is + 4ix – 4 = 0
Comparing with a + bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = – 4ac
= (4i – 4 × 1 × (-4)
= 16 + 16
= -16 + 16 …..[∵ = -1]
= 0
So, the given equation has equal roots.
These roots are given by
∴ x = -2i
∴ The root of the given equation is -2i.
(iv) i – 4x – 4i = 0
Solution:
i – 4x – 4i = 0
Multiplying throughout by i, we get
i2 – 4ix – 4 = 0
– – 4ix + 4 = 0 …[∵ = -1]
+ 4ix – 4 = 0
Comparing with a + bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = – 4ac
= (4i – 4 × 1 × (-4)
= 16 + 16
= -16 + 16
= 0
So, the given equation has equal roots.
These roots are given by
∴ The root of the given equation is -2i
Question 4.
Solve the following quadratic equations:
(i) – (2 + i) x – (1 – 7i) = 0
Solution:
Given equation is – (2 + i)x – (1 – 7i) = 0
Comparing with a + bx + c = 0, we get
a = 1, b = -(2 + i), c = -(1 – 7i)
Discriminant = – 4ac
= [-(2 + i) – 4 × 1 × -(1 – 7i)
= 4 + 4i + + 4 – 28i
= 4 + 4i – 1 + 4 – 28i …..[∵ = – 1]
= 7 – 24i
So, the given equation has complex roots.
These roots are given by
So, the given equation has complex roots.
These roots are given by
So, the given equation has complex roots.
These roots are given by
So, the given equation has complex roots.
These roots are given by
![Maharashtra Board Class 11 Maths Part 2 Chapter 1 Complex Numbers Ex 1.2 Solution 35 word image 21963 35](https://mhboardsolutions.xyz/wp-content/uploads/2022/03/word-image-21963-35.png)
![Maharashtra Board Class 11 Maths Part 2 Chapter 1 Complex Numbers Ex 1.2 Solution 36 word image 21963 36](https://mhboardsolutions.xyz/wp-content/uploads/2022/03/word-image-21963-36.jpeg)
![Maharashtra Board Class 11 Maths Part 2 Chapter 1 Complex Numbers Ex 1.2 Solution 37 word image 21963 37](https://mhboardsolutions.xyz/wp-content/uploads/2022/03/word-image-21963-37.jpeg)
Question 5.
Solution:
Solution:
![Maharashtra Board Class 11 Maths Part 2 Chapter 1 Complex Numbers Ex 1.2 Solution 48 word image 21963 50](https://mhboardsolutions.xyz/wp-content/uploads/2022/03/word-image-21963-50.png)
![Maharashtra Board Class 11 Maths Part 2 Chapter 1 Complex Numbers Ex 1.2 Solution 49 word image 21963 51](https://mhboardsolutions.xyz/wp-content/uploads/2022/03/word-image-21963-51.jpeg)
![Maharashtra Board Class 11 Maths Part 2 Chapter 1 Complex Numbers Ex 1.2 Solution 50 word image 21963 52](https://mhboardsolutions.xyz/wp-content/uploads/2022/03/word-image-21963-52.png)
(v) 2 + 5 + 7 – x + 41, if x = -2 – √3i
Solution: