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.

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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,

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

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,

We have i² = –1

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

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,

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

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,

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

∴ 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,

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

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 – (3√2 – 2i)x – √2i = 0

Given: x² – (3√2 – 2i)x – √2i = 0

We shall apply discriminant rule,

Where, x = (-b ±√(b² – 4ac))/2a

Here, a = 1, b = -(3√2 – 2i), c = –√2i

So,

(xi) x² – (√2 + i)x + √2i = 0

Given: x² – (√2 + i)x + √2i = 0

x² – (√2x + ix) + √2i = 0

x² – √2x – 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,

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