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

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2

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