11th Commerce Maths 1 Chapter 3 Exercise 3.2 Answers Maharashtra Board

Complex Numbers Class 11 Commerce Maths 1 Chapter 3 Exercise 3.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.2 Questions and Answers.

Std 11 Maths 1 Exercise 3.2 Solutions Commerce Maths

Question 1.
Find the square root of the following complex numbers:
(i) -8 – 6i
Solution:
Let \(\sqrt{-8-6 i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
-8 – 6i = (a + bi) ²
-8 – 6i = a ² + b ² i ² + 2abi
-8 – 6i = (a ² – b ² ) + 2abi …..[∵ i ² = -1]
Equating real and imaginary parts, we get

(ii) 7 + 24i
Solution:
Let \(\sqrt{7+24 i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
7 + 24i = (a + bi) ²
7 + 24i = a ² + b ² i ² + 2abi
7 + 24i = (a ² – b ² ) + 2abi …..[∵ i ² = -1]
Equating real and imaginary parts, we get

(iii) 1 + 4√3i
Solution:
Let \(\sqrt{1+4 \sqrt{3} i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
1 + 4√3i = (a + bi) ²
1 + 4√3i = a ² + b ² i ² + 2abi
1 +4√3i = (a ² – b ² ) + 2abi ……[∵ i ² = -1]
Equating real and imaginary parts, we get

(iv) 3 + 2√10i
Solution:
Let \(\sqrt{3+2 \sqrt{10}} i\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
3 + 2√10i = (a + bi) ²
3 + 2√10i = a ² + b ² i ² + 2abi
3 + 2√10i = (a ² – b ² ) + 2abi …..[∵ i ² = -1]
Equating real and imaginary parts, we get
a ² – b ² = 3 and 2ab = 2√10
a ² – b ² = 3 and b = \(\frac{\sqrt{10}}{a}\)

(v) 2(1 – √3i)
Solution:
Let \(\sqrt{2(1-\sqrt{3} i)}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
2(1 – √3i) = (a + bi) ²
2(1 – √3i) = a ² + b ² i ² + 2abi
2 – 2√3i = (a ² – b ² ) + 2abi …..[∵ i ² = -1]
Equating real and imaginary parts, we get

Question 2.
Solve the following quadratic equations.
(i) 8x ² + 2x + 1 = 0
Solution:
Given equation is 8x ² + 2x + 1 = 0
Comparing with ax ² + bx + c = 0, we get
a = 8, b = 2, c = 1
Discriminant = b ² – 4ac
= (2) ² – 4 × 8 × 1
= 4 – 32
= -28 < 0
So, the given equation has complex roots.
These roots are given by

∴ the roots of the given equation are \(\frac{-1+\sqrt{7} \mathrm{i}}{8}\) and \(\frac{-1-\sqrt{7} \mathrm{i}}{8}\)

(ii) 2x ² – √3x + 1 = 0
Solution:
Given equation is 2x ² – √3x + 1 = 0
Comparing with ax ² + bx + c = 0, we get
a = 2, b = -√3, c = 1
Discriminant = b ² – 4ac
= (-√3) ² – 4 × 2 × 1
= 3 – 8
= -5 < 0
So, the given equation has complex roots.
These roots are given by

∴ the roots of the given equation are \(\frac{\sqrt{3}+\sqrt{5} i}{4}\) and \(\frac{\sqrt{3}-\sqrt{5} i}{4}\)

(iii) 3x ² – 7x + 5 = 0
Solution:
Given equation is 3x ² – 7x + 5 = 0
Comparing with ax ² + bx + c = 0, we get
a = 3, b = -7, c = 5
Discriminant = b ² – 4ac
= (-7) ² – 4 × 3 × 5
= 49 – 60
= -11 < 0
So, the given equation has complex roots.
These roots are given by

∴ the roots of the given equation are \(\frac{7+\sqrt{11} i}{6}\) and \(\frac{7-\sqrt{11} i}{6}\)

(iv) x ² – 4x + 13 = 0
Solution:
Given equation is x ² – 4x + 13 = 0
Comparing with ax ² + bx + c = 0, we get
a = 1, b = -4, c = 13
Discriminant = b ² – 4ac
= (-4) ² – 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) x ² + 3ix + 10 = 0
Solution:
Given equation is x ² + 3ix + 10 = 0
Comparing with ax ² + bx + c = 0, we get
a = 1, b = 3i, c = 10
Discriminant = b ² – 4ac
= (3i) ² – 4 × 1 × 10
= 9i ² – 40
= -9 – 40 …..[∵ i ² = -1]
= -49
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.
Check:
If x = 2i and x = -5i satisfy the given equation, then our answer is correct.
L.H.S. = x ² + 3ix + 10
= (2i) ² + 3i(2i) + 10i
= 4i ² + 6i ² + 10
= 10i ² + 10
= -10 + 10 ……[∵ i ² = -1]
= 0
= R.H.S.
L.H.S. = x ² + 3ix + 10
= (-5i) ² + 3i(-5i) + 10
= 25i ² – 15i ² + 10
= 10i ² + 10
= -10 + 10 …..[∵ i ² = -1]
= 0
= R.H.S.
Thus, our answer is correct.

(ii) 2x ² + 3ix + 2 = 0
Solution:
Given equation is 2x ² + 3ix + 2 = 0
Comparing with ax ² + bx + c = 0, we get
a = 2, b = 3i, c = 2
Discriminant = b ² – 4ac
= (3i) ² – 4 × 2 × 2
= 9i ² – 16
= -9 – 16
= -25 < 0
So, the given equation has complex roots.
These roots are given by

∴ the roots of the given equation are \(\frac{1}{2}\)i and -2i.

(iii) x ² + 4ix – 4 = 0
Solution:
Given equation is x ² + 4ix – 4 = 0
Comparing with ax ² + bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = b ² – 4ac
= (4i) ² – 4 × 1 × -4
= 16i ² + 16
= -16 + 16 …..[∵ i ² = -1]
= 0
So, the given equation has equal roots.
These roots are given by

∴ the roots of the given equation are -2i and -2i.

(iv) ix ² – 4x – 4i = 0
Solution:
ix ² – 4x – 4i = 0
Multiplying throughout by i, we get
i ² x ² – 4ix – 4i ² = 0
∴ -x ² – 4ix + 4 = 0 ……[∵ i ² = -1]
∴ x ² + 4ix – 4 = 0
Comparing with ax ² + bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = b ² – 4ac
= (4i) ² – 4 × 1 × -4
= 16i ² + 16
= -16 + 16 …..[∵ i ² = -1]
= 0
So, the given equation has equal roots.
These roots are given by

∴ the roots of the given equation are -2i and -2i.

Question 4.
Solve the following quadratic equations.
(i) x ² – (2 + i) x – (1 – 7i) = 0
Solution:
Given equation is x ² – (2 + i)x – (1 – 7i) = 0
Comparing with ax ² + bx + c = 0, we get
a = 1, b = -(2 + i), c = -(1 – 7i)
Discriminant = b ² – 4ac
= [-(2 + i)] ² – 4 × 1 × -(1 – 7i)
= 4 + 4i + i ² + 4 – 28i
= 4 + 4i – 1 + 4 – 28i …….[∵ i ² = -1]
= 7 – 24i
So, the given equation has complex roots.
These roots are given by

(ii) x ² – (3√2 + 2i) x + 6√2i = 0
Solution:
Given equation is x ² – (3√2 + 2i) x + 6√2i = 0
Comparing with ax ² + bx + c = 0, we get
a = 1, b = -(3√2 + 2i), c = 6√2i
Discriminant = b ² – 4ac
= [-(3√2 + 2i)] ² – 4 × 1 × 6√2i
= 18 + 12√2i + 4i ² – 24√2i
= 18 – 12√2i – 4 …..[∵ i ² = -1]
= 14 – 12√2i
So, the given equation has complex roots.
These roots are given by

(iii) x ² – (5 – i) x + (18 + i) = 0
Solution:
Given equation is x ² – (5 – i)x + (18 + i) = 0
Comparing with ax ² + bx + c = 0, we get
a = 1, b = -(5 – i), c = 18 + i
Discriminant = b ² – 4ac
= [-(5 – i)] ² – 4 × 1 × (18 + i)
= 25 – 10i + i ² – 72 – 4i
= 25 – 10i – 1 – 72 – 4i …..[∵ i ² = -1]
= -48 – 14i
So, the given equation has complex roots.
These roots are given by

(iv) (2 + i) x ² – (5 – i) x + 2(1 – i) = 0
Solution:
Given equation is
(2 + i) x ² – (5 – i) x + 2(1 – i) = 0
Comparing with ax ² + bx + c = 0, we get
a = 2 + i, b = -(5 – i), c = 2(1 – i)
Discriminant = b ² – 4ac
= [-(5 – i)] ² – 4 × (2 + i) × 2(1 – i)
= 25 – 10i + i ² – 8(2 + i)(1 – i)
= 25 – 10i + i ² – 8(2 – 2i + i – i ² )
= 25 – 10i – 1 – 8(2 – i + 1) …..[∵ i ² = -1]
= 25 – 10i – 1 – 16 + 8i – 8
= -2i
So, the given equation has complex roots.
These roots are given by