RD Sharma Solutions for Class 11 Chapter 14 - Quadratic Equations Exercise 14.2

Exercise 14.2 of Chapter 14 discusses problems based on quadratic equations with complex coefficients. Our expert faculty team have prepared the solutions to help students in their exam preparation to obtain good marks in Maths. If you wish to secure an excellent score then solving RD Sharma Class 11 Solutions is an utmost necessity. Practising the textbook questions will help you in analyzing your level of preparation and build a strong command over the subject. Students can easily download the pdf from the links given below.

Download the pdf of RD Sharma Solutions for Class 11 Maths Exercise 14.2 Chapter 14 – Quadratic Equations

 

rd sharma class 11 maths ch 14 ex 2 01
rd sharma class 11 maths ch 14 ex 2 02
rd sharma class 11 maths ch 14 ex 2 03
rd sharma class 11 maths ch 14 ex 2 04
rd sharma class 11 maths ch 14 ex 2 05
rd sharma class 11 maths ch 14 ex 2 06
rd sharma class 11 maths ch 14 ex 2 07
rd sharma class 11 maths ch 14 ex 2 08
rd sharma class 11 maths ch 14 ex 2 09
rd sharma class 11 maths ch 14 ex 2 10
rd sharma class 11 maths ch 14 ex 2 11
rd sharma class 11 maths ch 14 ex 2 12
rd sharma class 11 maths ch 14 ex 2 13
rd sharma class 11 maths ch 14 ex 2 14

 

Also, access another exercise of RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations

Exercise 14.1 Solutions

Access answers to RD Sharma Solutions for Class 11 Maths Exercise 14.2 Chapter 14 – Quadratic Equations

1. Solving the following quadratic equations by factorization method:

(i) x2 + 10ix – 21 = 0

(ii) x2 + (1 – 2i)x – 2i = 0

(iii) x2 – (2√3 + 3i) x + 6√3i = 0

(iv) 6x2 – 17ix – 12 = 0

Solution:

(i) x2 + 10ix – 21 = 0

Given: x2 + 10ix – 21 = 0

x2 + 10ix – 21 × 1 = 0

We know, i2 = –1 ⇒ 1 = –i2

By substituting 1 = –i2 in the above equation, we get

x2 + 10ix – 21(–i2) = 0

x2 + 10ix + 21i2 = 0

x2 + 3ix + 7ix + 21i2 = 0

x(x + 3i) + 7i(x + 3i) = 0

(x + 3i) (x + 7i) = 0

x + 3i = 0 or x + 7i = 0

x = –3i or –7i

∴ The roots of the given equation are –3i, –7i

(ii) x2 + (1 – 2i)x – 2i = 0

Given: x2 + (1 – 2i)x – 2i = 0

x2 + x – 2ix – 2i = 0

x(x + 1) – 2i(x + 1) = 0

(x + 1) (x – 2i) = 0

x + 1 = 0 or x – 2i = 0

x = –1 or 2i

∴ The roots of the given equation are –1, 2i

(iii) x2 – (2√3 + 3i) x + 6√3i = 0

Given: x2 – (2√3 + 3i) x + 6√3i = 0

x2 – (2√3x + 3ix) + 6√3i = 0

x2 – 2√3x – 3ix + 6√3i = 0

x(x – 2√3) – 3i(x – 2√3) = 0

(x – 2√3) (x – 3i) = 0

(x – 2√3) = 0 or (x – 3i) = 0

x = 2√3 or x = 3i

∴ The roots of the given equation are 2√3, 3i

(iv) 6x2 – 17ix – 12 = 0

Given: 6x2 – 17ix – 12 = 0

6x2 – 17ix – 12 × 1 = 0

We know, i2 = –1 ⇒ 1 = –i2

By substituting 1 = –i2 in the above equation, we get

6x2 – 17ix – 12(–i2) = 0

6x2 – 17ix + 12i2 = 0

6x2 – 9ix – 8ix + 12i2 = 0

3x(2x – 3i) – 4i(2x – 3i) = 0

(2x – 3i) (3x – 4i) = 0

2x – 3i = 0 or 3x – 4i = 0

2x = 3i or 3x = 4i

x = 3i/2 or x = 4i/3

∴ The roots of the given equation are 3i/2, 4i/3

2. Solve the following quadratic equations:

(i) x2 – (3√2 + 2i) x + 6√2i = 0

(ii) x2 – (5 – i) x + (18 + i) = 0

(iii) (2 + i)x2 – (5- i)x + 2 (1 – i) = 0

(iv) x2 – (2 + i)x – (1 – 7i) = 0

(v) ix2 – 4x – 4i = 0

(vi) x2 + 4ix – 4 = 0

(vii) 2x2 + √15ix – i = 0 

(viii) x2 – x + (1 + i) = 0

(ix) ix2 – x + 12i = 0

(x) x2 – (3√2 – 2i)x – √2i = 0 

(xi) x2 – (√2 + i)x + √2i = 0

(xii) 2x2 – (3 + 7i)x + (9i – 3) = 0

Solution:

(i) x2 – (3√2 + 2i) x + 6√2i = 0

Given: x2 – (3√2 + 2i) x + 6√2i = 0

x2 – (3√2x + 2ix) + 6√2i = 0

x2 – 3√2x – 2ix + 6√2i = 0

x(x – 3√2) – 2i(x – 3√2) = 0

(x – 3√2) (x – 2i) = 0

(x – 3√2) = 0 or (x – 2i) = 0

x = 3√2 or x = 2i

∴ The roots of the given equation are 3√2, 2i

(ii) x2 – (5 – i) x + (18 + i) = 0

Given: x2 – (5 – i) x + (18 + i) = 0

We shall apply discriminant rule,

Where, x = (-b ±√(b2 – 4ac))/2a

Here, a = 1, b = -(5-i), c = (18+i)

So,

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 1

We can write 48 + 14i = 49 – 1 + 14i

So,

48 + 14i = 49 + i2 + 14i [∵ i2 = –1]

= 72 + i2 + 2(7)(i)

= (7 + i)2 [Since, (a + b)2 = a2 + b2 + 2ab]

By using the result 48 + 14i = (7 + i) 2, we get

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 2

 x = 2 + 3i or 3 – 4i

∴ The roots of the given equation are 3 – 4i, 2 + 3i

(iii) (2 + i)x2 – (5- i)x + 2 (1 – i) = 0

Given: (2 + i)x2 – (5- i)x + 2 (1 – i) = 0

We shall apply discriminant rule,

Where, x = (-b ±√(b2 – 4ac))/2a

Here, a = (2+i), b = -(5-i), c = 2(1-i)

So,

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 3

We have i2 = –1

By substituting –1 = i2 in the above equation, we get

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 4

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 5

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 6

x = (1 – i) or 4/5 – 2i/5

∴ The roots of the given equation are (1 – i), 4/5 – 2i/5

(iv) x2 – (2 + i)x – (1 – 7i) = 0

Given: x2 – (2 + i)x – (1 – 7i) = 0

We shall apply discriminant rule,

Where, x = (-b ±√(b2 – 4ac))/2a

Here, a = 1, b = -(2+i), c = -(1-7i)

So,

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 7

We can write 7 – 24i = 16 – 9 – 24i

7 – 24i = 16 + 9(–1) – 24i

= 16 + 9i2 – 24i [∵ i2 = –1]

= 42 + (3i)2 – 2(4) (3i)

= (4 – 3i)2 [∵ (a – b)2 = a2 – b2 + 2ab]

By using the result 7 – 24i = (4 – 3i)2, we get

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 8

x = 3 – i or -1 + 2i

∴ The roots of the given equation are (-1 + 2i), (3 – i)

(v) ix2 – 4x – 4i = 0

Given: ix2 – 4x – 4i = 0

ix2 + 4x(–1) – 4i = 0 [We know, i2 = –1]

So by substituting –1 = i2 in the above equation, we get

ix2 + 4xi2 – 4i = 0

i(x2 + 4ix – 4) = 0

x2 + 4ix – 4 = 0

x2 + 4ix + 4(–1) = 0

x2 + 4ix + 4i2 = 0 [Since, i2 = –1]

x2 + 2ix + 2ix + 4i2 = 0

x(x + 2i) + 2i(x + 2i) = 0

(x + 2i) (x + 2i) = 0

(x + 2i)2 = 0

x + 2i = 0

x = –2i, -2i

∴ The roots of the given equation are –2i, –2i

(vi) x2 + 4ix – 4 = 0

Given: x2 + 4ix – 4 = 0

x2 + 4ix + 4(–1) = 0 [We know, i2 = –1]

So by substituting –1 = i2 in the above equation, we get

x2 + 4ix + 4i2 = 0

x2 + 2ix + 2ix + 4i2 = 0

x(x + 2i) + 2i(x + 2i) = 0

(x + 2i) (x + 2i) = 0

(x + 2i)2 = 0

x + 2i = 0

x = –2i, -2i

∴ The roots of the given equation are –2i, –2i

(vii) 2x2 + √15ix – i = 0 

Given: 2x2 + √15ix – i = 0 

We shall apply discriminant rule,

Where, x = (-b ±√(b2 – 4ac))/2a

Here, a = 2, b = √15i, c = -i

So,

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 9

We can write 15 – 8i = 16 – 1 – 8i

15 – 8i = 16 + (–1) – 8i

= 16 + i2 – 8i [∵ i2 = –1]

= 42 + (i)2 – 2(4)(i)

= (4 – i)2 [Since, (a – b)2 = a2 – b2 + 2ab]

By using the result 15 – 8i = (4 – i)2, we get

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 10

∴ The roots of the given equation are [1+ (4 – √15)i/4] , [-1 -(4 + √15)i/4]

(viii) x2 – x + (1 + i) = 0

Given: x2 – x + (1 + i) = 0

We shall apply discriminant rule,

Where, x = (-b ±√(b2 – 4ac))/2a

Here, a = 1, b = -1, c = (1+i)

So,

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 11
RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 12

We can write 3 + 4i = 4 – 1 + 4i

3 + 4i = 4 + i2 + 4i [∵ i2 = –1]

= 22 + i2 + 2(2) (i)

= (2 + i)2 [Since, (a + b)2 = a2 + b2 + 2ab]

By using the result 3 + 4i = (2 + i)2, we get

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 13

x = 2i/2 or (2 – 2i)/2

x = i or 2(1-i)/2

x = i or (1 – i)

∴ The roots of the given equation are (1-i), i

(ix) ix2 – x + 12i = 0

Given: ix2 – x + 12i = 0

ix2 + x(–1) + 12i = 0 [We know, i2 = –1]

so by substituting –1 = i2 in the above equation, we get

ix2 + xi2 + 12i = 0

i(x2 + ix + 12) = 0

x2 + ix + 12 = 0

x2 + ix – 12(–1) = 0

x2 + ix – 12i2 = 0 [Since, i2 = –1]

x2 – 3ix + 4ix – 12i2 = 0

x(x – 3i) + 4i(x – 3i) = 0

(x – 3i) (x + 4i) = 0

x – 3i = 0 or x + 4i = 0

x = 3i or –4i

∴ The roots of the given equation are -4i, 3i

(x) x2 – (32 – 2i)x – 2i = 0 

Given: x2 – (32 – 2i)x – 2i = 0 

We shall apply discriminant rule,

Where, x = (-b ±√(b2 – 4ac))/2a

Here, a = 1, b = -(32 – 2i), c = –2i

So,

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 14

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 15

(xi) x2 – (2 + i)x + 2i = 0

Given: x2 – (2 + i)x + 2i = 0

x2 – (2x + ix) + 2i = 0

x22x – ix + 2i = 0

x(x – 2) – i(x – 2) = 0

(x – 2) (x – i) = 0

(x – 2) = 0 or (x – i) = 0

x = 2 or x = i

∴ The roots of the given equation are i, 2

(xii) 2x2 – (3 + 7i)x + (9i – 3) = 0

Given: 2x2 – (3 + 7i)x + (9i – 3) = 0

We shall apply discriminant rule,

Where, x = (-b ±√(b2 – 4ac))/2a

Here, a = 2, b = -(3 + 7i), c = (9i – 3)

So,

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 16

We can write 16 + 30i = 25 – 9 + 30i

16 + 30i = 25 + 9(–1) + 30i

= 25 + 9i2 + 30i [∵ i2 = –1]

= 52 + (3i)2 + 2(5)(3i)

= (5 + 3i)2 [∵ (a + b)2 = a2 + b2 + 2ab]

By using the result 16 + 30i = (5 + 3i)2, we get

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 17

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic Equations image - 18

Leave a Comment

Your email address will not be published. Required fields are marked *