RD Sharma Solutions for Class 11 Chapter 14 - Quadratic Equations

In earlier classes, we have studied quadratic equations with real coefficients and real roots only. In this chapter, we shall study quadratic equations with real coefficients and complex roots. We shall also discuss quadratic equations with complex coefficients and their solutions in the complex numbers system. For a better understanding of the concepts, students can solve the exercise wise problems using the solutions, which are developed by our expert faculty team at BYJU’S. Students aspiring to secure high marks in their examination are advised to practice the solutions on a regular basis. RD Sharma Class 11 Maths Solutions pdf can be downloaded from the links provided below.

Chapter 14 – Quadratic Equations contains two exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

  • Some useful definitions and results.
  • Quadratic equations with real coefficients.
  • Quadratic equations with complex coefficients.

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

 

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

EXERCISE 14.1 PAGE NO: 14.5

Solve the following quadratic equations by factorization method only:

1. x2 + 1 = 0

Solution:

Given: x2 + 1 = 0

We know, i2 = –1 ⇒ 1 = –i2

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

x2 – i2 = 0

[By using the formula, a2 – b2 = (a + b) (a – b)]

(x + i) (x – i) = 0

x + i = 0 or x – i = 0

x = –i or x = i

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

2. 9x2 + 4 = 0

Solution:

Given: 9x2 + 4 = 0

9x2 + 4 × 1 = 0

We know, i2 = –1 ⇒ 1 = –i2

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

So,

9x2 + 4(–i2) = 0

9x2 – 4i2 = 0

(3x)2 – (2i)2 = 0

[By using the formula, a2 – b2 = (a + b) (a – b)]

(3x + 2i) (3x – 2i) = 0

3x + 2i = 0 or 3x – 2i = 0

3x = –2i or 3x = 2i

x = -2i/3 or x = 2i/3

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

3. x2 + 2x + 5 = 0

Solution:

Given: x2 + 2x + 5 = 0

x2 + 2x + 1 + 4 = 0

x2 + 2(x) (1) + 12 + 4 = 0

(x + 1)2 + 4 = 0 [since, (a + b)2 = a2 + 2ab + b2]

(x + 1)2 + 4 × 1 = 0

We know, i2 = –1 ⇒ 1 = –i2

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

(x + 1)2 + 4(–i2) = 0

(x + 1)2 – 4i2 = 0

(x + 1)2 – (2i)2 = 0

[By using the formula, a2 – b2 = (a + b) (a – b)]

(x + 1 + 2i)(x + 1 – 2i) = 0

x + 1 + 2i = 0 or x + 1 – 2i = 0

x = –1 – 2i or x = –1 + 2i

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

4. 4x2 – 12x + 25 = 0

Solution:

Given: 4x2 – 12x + 25 = 0

4x2 – 12x + 9 + 16 = 0

(2x)2 – 2(2x)(3) + 32 + 16 = 0

(2x – 3)2 + 16 = 0 [Since, (a + b)2 = a2 + 2ab + b2]

(2x – 3)2 + 16 × 1 = 0

We know, i2 = –1 ⇒ 1 = –i2

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

(2x – 3)2 + 16(–i2) = 0

(2x – 3)2 – 16i2 = 0

(2x – 3)2 – (4i)2 = 0

[By using the formula, a2 – b2 = (a + b) (a – b)]

(2x – 3 + 4i) (2x – 3 – 4i) = 0

2x – 3 + 4i = 0 or 2x – 3 – 4i = 0

2x = 3 – 4i or 2x = 3 + 4i

x = 3/2 – 2i or x = 3/2 + 2i

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

5. x2 + x + 1 = 0

Solution:

Given: x2 + x + 1 = 0

x2 + x + ¼ + ¾ = 0

x2 + 2 (x) (1/2) + (1/2)2 + ¾ = 0

(x + 1/2)2 + ¾ = 0 [Since, (a + b)2 = a2 + 2ab + b2]

(x + 1/2)2 + ¾ × 1 = 0

We know, i2 = –1 ⇒ 1 = –i2

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

(x + ½)2 + ¾ (-1)2 = 0

(x + ½)2 + ¾ i2 = 0

(x + ½)2 – (3i/2)2 = 0

[By using the formula, a2 – b2 = (a + b) (a – b)]

(x + ½ + 3i/2) (x + ½ – 3i/2) = 0

(x + ½ + 3i/2) = 0 or (x + ½ – 3i/2) = 0

x = -1/2 – 3i/2 or x = -1/2 + 3i/2

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

6. 4x2 + 1 = 0

Solution:

Given: 4x2 + 1 = 0

We know, i2 = –1 ⇒ 1 = –i2

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

4x2 – i2 = 0

(2x)2 – i2 = 0

[By using the formula, a2 – b2 = (a + b) (a – b)]

(2x + i) (2x – i) = 0

2x + i = 0 or 2x – i = 0

2x = –i or 2x = i

x = -i/2 or x = i/2

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

7. x2 – 4x + 7 = 0

Solution:

Given: x2 – 4x + 7 = 0

x2 – 4x + 4 + 3 = 0

x2 – 2(x) (2) + 22 + 3 = 0

(x – 2)2 + 3 = 0 [Since, (a – b)2 = a2 – 2ab + b2]

(x – 2)2 + 3 × 1 = 0

We know, i2 = –1 ⇒ 1 = –i2

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

(x – 2)2 + 3(–i2) = 0

(x – 2)2 – 3i2 = 0

(x – 2)2 – (3i)2 = 0

[By using the formula, a2 – b2 = (a + b) (a – b)]

(x – 2 + 3i) (x – 2 – 3i) = 0

(x – 2 + 3i) = 0 or (x – 2 – 3i) = 0

x = 2 – 3i or x = 2 + 3i

x = 2 ± 3i

∴ The roots of the given equation are 2 ± 3i

8. x2 + 2x + 2 = 0

Solution:

Given: x2 + 2x + 2 = 0

x2 + 2x + 1 + 1 = 0

x2 + 2(x)(1) + 12 + 1 = 0

(x + 1)2 + 1 = 0 [∵ (a + b)2 = a2 + 2ab + b2]

We know, i2 = –1 ⇒ 1 = –i2

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

(x + 1)2 + (–i2) = 0

(x + 1)2 – i2 = 0

(x + 1)2 – (i)2 = 0

[By using the formula, a2 – b2 = (a + b) (a – b)]

(x + 1 + i) (x + 1 – i) = 0

x + 1 + i = 0 or x + 1 – i = 0

x = –1 – i or x = –1 + i

x = -1 ± i

∴ The roots of the given equation are -1 ± i

9. 5x2 – 6x + 2 = 0

Solution:

Given: 5x2 – 6x + 2 = 0

We shall apply discriminant rule,

Where, x = (-b ±√(b2 – 4ac))/2a

Here, a = 5, b = -6, c = 2

So,

x = (-(-6) ±√(-62 – 4 (5)(2)))/ 2(5)

= (6 ± √(36-40))/10

= (6 ± √(-4))/10

= (6 ± √4(-1))/10

We have i2 = –1

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

x = (6 ± √4i2)/10

= (6 ± 2i)/10

= 2(3±i)/10

= (3±i)/5

x = 3/5 ± i/5

∴ The roots of the given equation are 3/5 ± i/5

10. 21x2 + 9x + 1 = 0

Solution:

Given: 21x2 + 9x + 1 = 0

We shall apply discriminant rule,

Where, x = (-b ±√(b2 – 4ac))/2a

Here, a = 21, b = 9, c = 1

So,

x = (-9 ±√(92 – 4 (21)(1)))/ 2(21)

= (-9 ± √(81-84))/42

= (-9 ± √(-3))/42

= (-9 ± √3(-1))/42

We have i2 = –1

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

x = (-9 ± √3i2)/42

= (-9 ± √(3i)2/42

= (-9 ± √3i)/42

= -9/42 ± √3i/42

= -3/14 ± √3i/42

∴ The roots of the given equation are -3/14 ± √3i/42

11. x2 – x + 1 = 0

Solution:

Given: x2 – x + 1 = 0

x2 – x + ¼ + ¾ = 0

x2 – 2 (x) (1/2) + (1/2)2 + ¾ = 0

(x – 1/2)2 + ¾ = 0 [Since, (a + b)2 = a2 + 2ab + b2]

(x – 1/2)2 + ¾ × 1 = 0

We know, i2 = –1 ⇒ 1 = –i2

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

(x – ½)2 + ¾ (-1)2 = 0

(x – ½)2 + ¾ (-i)2 = 0

(x – ½)2 – (3i/2)2 = 0

[By using the formula, a2 – b2 = (a + b) (a – b)]

(x – ½ + 3i/2) (x – ½ – 3i/2) = 0

(x – ½ + 3i/2) = 0 or (x – ½ – 3i/2) = 0

x = 1/2 – 3i/2 or x = 1/2 + 3i/2

∴ The roots of the given equation are 1/2 + 3i/2, 1/2 – 3i/2

12. x2 + x + 1 = 0

Solution:

Given: x2 + x + 1 = 0

x2 + x + ¼ + ¾ = 0

x2 + 2 (x) (1/2) + (1/2)2 + ¾ = 0

(x + 1/2)2 + ¾ = 0 [Since, (a + b)2 = a2 + 2ab + b2]

(x + 1/2)2 + ¾ × 1 = 0

We know, i2 = –1 ⇒ 1 = –i2

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

(x + ½)2 + ¾ (-1)2 = 0

(x + ½)2 + ¾ i2 = 0

(x + ½)2 – (3i/2)2 = 0

[By using the formula, a2 – b2 = (a + b) (a – b)]

(x + ½ + 3i/2) (x + ½ – 3i/2) = 0

(x + ½ + 3i/2) = 0 or (x + ½ – 3i/2) = 0

x = -1/2 – 3i/2 or x = -1/2 + 3i/2

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

13. 17x2 – 8x + 1 = 0

Solution:

Given: 17x2 – 8x + 1 = 0

We shall apply discriminant rule,

Where, x = (-b ±√(b2 – 4ac))/2a

Here, a = 17, b = -8, c = 1

So,

x = (-(-8) ±√(-82 – 4 (17)(1)))/ 2(17)

= (8 ± √(64-68))/34

= (8 ± √(-4))/34

= (8 ± √4(-1))/34

We have i2 = –1

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

x = (8 ± √(2i)2)/34

= (8 ± 2i)/34

= 2(4±i)/34

= (4±i)/17

x = 4/17 ± i/17

∴ The roots of the given equation are 4/17 ± i/17


EXERCISE 14.2 PAGE NO: 14.13

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


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

Exercise 14.1 Solutions

Exercise 14.2 Solutions

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