# RD Sharma Solutions for Class 11 Maths Chapter 14 Quadratic Equations

## RD Sharma Solutions Class 11 Maths Chapter 14 â€“ Download Free PDF Updated Session 2021 – 22

RD Sharma Solutions for Class 11 Maths Chapter 14 – Quadratic EquationsÂ are provided here to help students study and score good marks in the board exams. 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 number system.

For a better understanding of the concepts, students can solve the exercise wise problems using the RD Sharma Solutions, which are developed by our expert faculty team at BYJUâ€™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

### 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: x2Â + 1 = 0

We know, i2Â = â€“1Â â‡’Â 1 = â€“i2

By substituting 1 = â€“i2Â in the above equation, we get

x2Â â€“ i2Â = 0

[By using the formula,Â a2Â â€“ b2Â = (a + b) (a â€“ b)]

(x + i) (x â€“ i) = 0

x + i = 0 or x â€“ i = 0

x = â€“i or x = i

âˆ´ The roots of the given equation are i, -i

2. 9x2Â + 4 = 0

Solution:

Given: 9x2Â + 4 = 0

9x2Â + 4 Ã— 1 = 0

We know, i2Â = â€“1Â â‡’Â 1 = â€“i2

By substituting 1 = â€“i2Â in the above equation, we get

So,

9x2Â + 4(â€“i2) = 0

9x2Â â€“ 4i2Â = 0

(3x)2Â â€“ (2i)2Â = 0

[By using the formula,Â a2Â â€“ b2Â = (a + b) (a â€“ b)]

(3x + 2i) (3x â€“ 2i) = 0

3x + 2i = 0 or 3x â€“ 2i = 0

3x = â€“2i or 3x = 2i

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

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

3. x2Â + 2x + 5 = 0

Solution:

Given: x2Â + 2x + 5 = 0

x2Â + 2x + 1 + 4 = 0

x2Â + 2(x) (1) + 12Â + 4 = 0

(x + 1)2Â + 4 = 0 [since,Â (a + b)2Â = a2Â + 2ab + b2]

(x + 1)2Â + 4 Ã— 1 = 0

We know, i2Â = â€“1Â â‡’Â 1 = â€“i2

By substituting 1 = â€“i2Â in the above equation, we get

(x + 1)2Â + 4(â€“i2) = 0

(x + 1)2Â â€“ 4i2Â = 0

(x + 1)2Â â€“ (2i)2Â = 0

[By using the formula,Â a2Â â€“ b2Â = (a + b) (a â€“ b)]

(x + 1 + 2i)(x + 1 â€“ 2i) = 0

x + 1 + 2i = 0 or x + 1 â€“ 2i = 0

x = â€“1 â€“ 2i or x = â€“1 + 2i

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

4. 4x2Â â€“ 12x + 25 = 0

Solution:

Given: 4x2Â â€“ 12x + 25 = 0

4x2Â â€“ 12x + 9 + 16 = 0

(2x)2Â â€“ 2(2x)(3) + 32Â + 16 = 0

(2x â€“ 3)2Â + 16 = 0 [Since,Â (a + b)2Â = a2Â + 2ab + b2]

(2x â€“ 3)2Â + 16 Ã— 1 = 0

We know, i2Â = â€“1Â â‡’Â 1 = â€“i2

By substituting 1 = â€“i2Â in the above equation, we get

(2x â€“ 3)2Â + 16(â€“i2) = 0

(2x â€“ 3)2Â â€“ 16i2Â = 0

(2x â€“ 3)2Â â€“ (4i)2Â = 0

[By using the formula, a2Â â€“ b2Â = (a + b) (a â€“ b)]

(2x â€“ 3 + 4i) (2x â€“ 3 â€“ 4i) = 0

2x â€“ 3 + 4i = 0 or 2x â€“ 3 â€“ 4i = 0

2x = 3 â€“ 4i or 2x = 3 + 4i

x = 3/2 â€“ 2i or x = 3/2 + 2i

âˆ´ The roots of the given equation are 3/2 + 2i, 3/2 â€“ 2i

5. x2Â + x + 1 = 0

Solution:

Given: x2Â + x + 1 = 0

x2 + x + Â¼ + Â¾ = 0

x2 + 2 (x) (1/2) + (1/2)2 + Â¾ = 0

(x + 1/2)2 + Â¾ = 0 [Since,Â (a + b)2Â = a2Â + 2ab + b2]

(x + 1/2)2 + Â¾ Ã— 1 = 0

We know, i2Â = â€“1Â â‡’Â 1 = â€“i2

By substituting 1 = â€“i2Â in the above equation, we get

(x + Â½)2 + Â¾ (-1)2 = 0

(x + Â½)2 + Â¾ i2 = 0

(x + Â½)2 â€“ (âˆš3i/2)2 = 0

[By using the formula, a2Â â€“ b2Â = (a + b) (a â€“ b)]

(x + Â½ + âˆš3i/2) (x + Â½ – âˆš3i/2) = 0

(x + Â½ + âˆš3i/2) = 0 or (x + Â½ – âˆš3i/2) = 0

x = -1/2 – âˆš3i/2 or x = -1/2 + âˆš3i/2

âˆ´ The roots of the given equation are -1/2 + âˆš3i/2, -1/2 – âˆš3i/2

6. 4x2Â + 1 = 0

Solution:

Given: 4x2Â + 1 = 0

We know, i2Â = â€“1Â â‡’Â 1 = â€“i2

By substituting 1 = â€“i2Â in the above equation, we get

4x2Â â€“ i2Â = 0

(2x)2Â â€“ i2Â = 0

[By using the formula, a2Â â€“ b2Â = (a + b) (a â€“ b)]

(2x + i) (2x â€“ i) = 0

2x + i = 0 or 2x â€“ i = 0

2x = â€“i or 2x = i

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

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

7. x2Â â€“ 4x + 7 = 0

Solution:

Given: x2Â â€“ 4x + 7 = 0

x2Â â€“ 4x + 4 + 3 = 0

x2Â â€“ 2(x) (2) + 22Â + 3 = 0

(x â€“ 2)2Â + 3 = 0 [Since,Â (a â€“ b)2Â = a2Â â€“ 2ab + b2]

(x â€“ 2)2Â + 3 Ã— 1 = 0

We know, i2Â = â€“1Â â‡’Â 1 = â€“i2

By substituting 1 = â€“i2Â in the above equation, we get

(x â€“ 2)2Â + 3(â€“i2) = 0

(x â€“ 2)2Â â€“ 3i2Â = 0

(x â€“ 2)2Â â€“ (âˆš3i)2 = 0

[By using the formula, a2Â â€“ b2Â = (a + b) (a â€“ b)]

(x â€“ 2 + âˆš3i) (x â€“ 2 – âˆš3i) = 0

(x â€“ 2 + âˆš3i) = 0 or (x â€“ 2 – âˆš3i) = 0

x = 2 – âˆš3i or x = 2 + âˆš3i

x = 2 Â± âˆš3i

âˆ´ The roots of the given equation are 2 Â± âˆš3i

8. x2Â + 2x + 2 = 0

Solution:

Given: x2Â + 2x + 2 = 0

x2Â + 2x + 1 + 1 = 0

x2Â + 2(x)(1) + 12Â + 1 = 0

(x + 1)2Â + 1 = 0 [âˆµÂ (a + b)2Â = a2Â + 2ab + b2]

We know, i2Â = â€“1Â â‡’Â 1 = â€“i2

By substituting 1 = â€“i2Â in the above equation, we get

(x + 1)2Â + (â€“i2) = 0

(x + 1)2Â â€“ i2Â = 0

(x + 1)2Â â€“ (i)2Â = 0

[By using the formula, a2Â â€“ b2Â = (a + b) (a â€“ b)]

(x + 1 + i) (x + 1 â€“ i) = 0

x + 1 + i = 0 or x + 1 â€“ i = 0

x = â€“1 â€“ i or x = â€“1 + i

x = -1 Â± i

âˆ´ The roots of the given equation are -1 Â± i

9. 5x2Â â€“ 6x + 2 = 0

Solution:

Given: 5x2Â â€“ 6x + 2 = 0

We shall apply discriminant rule,

Where, x = (-b Â±âˆš(b2 â€“ 4ac))/2a

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

So,

x = (-(-6) Â±âˆš(-62 â€“ 4 (5)(2)))/ 2(5)

= (6 Â± âˆš(36-40))/10

= (6 Â± âˆš(-4))/10

= (6 Â± âˆš4(-1))/10

We have i2Â = â€“1

By substituting â€“1 = i2Â in the above equation, we get

x = (6 Â± âˆš4i2)/10

= (6 Â± 2i)/10

= 2(3Â±i)/10

= (3Â±i)/5

x = 3/5 Â± i/5

âˆ´ The roots of the given equation are 3/5 Â± i/5

10. 21x2Â + 9x + 1 = 0

Solution:

Given: 21x2Â + 9x + 1 = 0

We shall apply discriminant rule,

Where, x = (-b Â±âˆš(b2 â€“ 4ac))/2a

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

So,

x = (-9 Â±âˆš(92 â€“ 4 (21)(1)))/ 2(21)

= (-9 Â± âˆš(81-84))/42

= (-9 Â± âˆš(-3))/42

= (-9 Â± âˆš3(-1))/42

We have i2Â = â€“1

By substituting â€“1 = i2Â in the above equation, we get

x = (-9 Â± âˆš3i2)/42

= (-9 Â± âˆš(âˆš3i)2/42

= (-9 Â± âˆš3i)/42

= -9/42 Â± âˆš3i/42

= -3/14 Â± âˆš3i/42

âˆ´ The roots of the given equation are -3/14 Â± âˆš3i/42

11. x2Â â€“ x + 1 = 0

Solution:

Given: x2Â â€“ x + 1 = 0

x2 – x + Â¼ + Â¾ = 0

x2 – 2 (x) (1/2) + (1/2)2 + Â¾ = 0

(x – 1/2)2 + Â¾ = 0 [Since,Â (a + b)2Â = a2Â + 2ab + b2]

(x – 1/2)2 + Â¾ Ã— 1 = 0

We know, i2Â = â€“1Â â‡’Â 1 = â€“i2

By substituting 1 = â€“i2Â in the above equation, we get

(x – Â½)2 + Â¾ (-1)2 = 0

(x – Â½)2 + Â¾ (-i)2 = 0

(x – Â½)2 â€“ (âˆš3i/2)2 = 0

[By using the formula, a2Â â€“ b2Â = (a + b) (a â€“ b)]

(x – Â½ + âˆš3i/2) (x – Â½ – âˆš3i/2) = 0

(x – Â½ + âˆš3i/2) = 0 or (x – Â½ – âˆš3i/2) = 0

x = 1/2 – âˆš3i/2 or x = 1/2 + âˆš3i/2

âˆ´ The roots of the given equation are 1/2 + âˆš3i/2, 1/2 – âˆš3i/2

12. x2Â + x + 1 = 0

Solution:

Given: x2Â + x + 1 = 0

x2 + x + Â¼ + Â¾ = 0

x2 + 2 (x) (1/2) + (1/2)2 + Â¾ = 0

(x + 1/2)2 + Â¾ = 0 [Since,Â (a + b)2Â = a2Â + 2ab + b2]

(x + 1/2)2 + Â¾ Ã— 1 = 0

We know, i2Â = â€“1Â â‡’Â 1 = â€“i2

By substituting 1 = â€“i2Â in the above equation, we get

(x + Â½)2 + Â¾ (-1)2 = 0

(x + Â½)2 + Â¾ i2 = 0

(x + Â½)2 â€“ (âˆš3i/2)2 = 0

[By using the formula, a2Â â€“ b2Â = (a + b) (a â€“ b)]

(x + Â½ + âˆš3i/2) (x + Â½ – âˆš3i/2) = 0

(x + Â½ + âˆš3i/2) = 0 or (x + Â½ – âˆš3i/2) = 0

x = -1/2 – âˆš3i/2 or x = -1/2 + âˆš3i/2

âˆ´ The roots of the given equation are -1/2 + âˆš3i/2, -1/2 – âˆš3i/2

13. 17x2Â â€“ 8x + 1 = 0

Solution:

Given: 17x2Â â€“ 8x + 1 = 0

We shall apply discriminant rule,

Where, x = (-b Â±âˆš(b2 â€“ 4ac))/2a

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

So,

x = (-(-8) Â±âˆš(-82 â€“ 4 (17)(1)))/ 2(17)

= (8 Â± âˆš(64-68))/34

= (8 Â± âˆš(-4))/34

= (8 Â± âˆš4(-1))/34

We have i2Â = â€“1

By substituting â€“1 = i2Â in the above equation, we get

x = (8 Â± âˆš(2i)2)/34

= (8 Â± 2i)/34

= 2(4Â±i)/34

= (4Â±i)/17

x = 4/17 Â± i/17

âˆ´ The roots of the given equation are 4/17 Â± i/17

EXERCISE 14.2 PAGE NO: 14.13

1. Solving the following quadratic equations by factorization method:

(i) x2Â + 10ix â€“ 21 = 0

(ii) x2Â + (1 â€“ 2i)x â€“ 2i = 0

(iii) x2 â€“ (2âˆš3 + 3i) x + 6âˆš3i = 0

(iv) 6x2Â â€“ 17ix â€“ 12 = 0

Solution:

(i) x2Â + 10ix â€“ 21 = 0

Given: x2Â + 10ix â€“ 21 = 0

x2Â + 10ix â€“ 21 Ã— 1 = 0

We know, i2Â = â€“1Â â‡’Â 1 = â€“i2

By substituting 1 = â€“i2Â in the above equation, we get

x2Â + 10ix â€“ 21(â€“i2) = 0

x2Â + 10ix + 21i2Â = 0

x2Â + 3ix + 7ix + 21i2Â = 0

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

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

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

x = â€“3i or â€“7i

âˆ´ The roots of the given equation are â€“3i, â€“7i

(ii) x2Â + (1 â€“ 2i)x â€“ 2i = 0

Given: x2Â + (1 â€“ 2i)x â€“ 2i = 0

x2Â + x â€“ 2ix â€“ 2i = 0

x(x + 1) â€“ 2i(x + 1) = 0

(x + 1) (x â€“ 2i) = 0

x + 1 = 0 or x â€“ 2i = 0

x = â€“1 or 2i

âˆ´ The roots of the given equation are â€“1, 2i

(iii) x2 â€“ (2âˆš3 + 3i) x + 6âˆš3i = 0

Given: x2 â€“ (2âˆš3 + 3i) x + 6âˆš3i = 0

x2 â€“ (2âˆš3x + 3ix) + 6âˆš3i = 0

x2 â€“ 2âˆš3x – 3ix + 6âˆš3i = 0

x(x – 2âˆš3) â€“ 3i(x – 2âˆš3) = 0

(x – 2âˆš3) (x â€“ 3i) = 0

(x – 2âˆš3) = 0 or (x â€“ 3i) = 0

x = 2âˆš3 or x = 3i

âˆ´ The roots of the given equation are 2âˆš3, 3i

(iv) 6x2Â â€“ 17ix â€“ 12 = 0

Given: 6x2Â â€“ 17ix â€“ 12 = 0

6x2Â â€“ 17ix â€“ 12 Ã— 1 = 0

We know, i2Â = â€“1Â â‡’Â 1 = â€“i2

By substituting 1 = â€“i2Â in the above equation, we get

6x2Â â€“ 17ix â€“ 12(â€“i2) = 0

6x2Â â€“ 17ix + 12i2Â = 0

6x2Â â€“ 9ix â€“ 8ix + 12i2Â = 0

3x(2x â€“ 3i) â€“ 4i(2x â€“ 3i) = 0

(2x â€“ 3i) (3x â€“ 4i) = 0

2x â€“ 3i = 0 or 3x â€“ 4i = 0

2x = 3i or 3x = 4i

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

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

2. Solve the following quadratic equations:

(i) x2 â€“ (3âˆš2 + 2i) x + 6âˆš2i = 0

(ii) x2 â€“ (5 – i) x + (18 + i) = 0

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

(iv) x2 â€“ (2 + i)x â€“ (1 â€“ 7i) = 0

(v) ix2Â â€“ 4x â€“ 4i = 0

(vi) x2Â + 4ix â€“ 4 = 0

(vii) 2x2 + âˆš15ix â€“ i = 0Â

(viii) x2Â â€“ x + (1 + i) = 0

(ix) ix2Â â€“ x + 12i = 0

(x) x2 â€“ (3âˆš2 â€“ 2i)x – âˆš2i = 0

(xi) x2 â€“ (âˆš2 + i)x + âˆš2i = 0

(xii) 2x2 â€“ (3 + 7i)x + (9i – 3) = 0

Solution:

(i) x2 â€“ (3âˆš2 + 2i) x + 6âˆš2i = 0

Given: x2 â€“ (3âˆš2 + 2i) x + 6âˆš2i = 0

x2 â€“ (3âˆš2x + 2ix) + 6âˆš2i = 0

x2 â€“ 3âˆš2x – 2ix + 6âˆš2i = 0

x(x – 3âˆš2) â€“ 2i(x – 3âˆš2) = 0

(x – 3âˆš2) (x â€“ 2i) = 0

(x – 3âˆš2) = 0 or (x â€“ 2i) = 0

x = 3âˆš2 or x = 2i

âˆ´ The roots of the given equation are 3âˆš2, 2i

(ii) x2 â€“ (5 – i) x + (18 + i) = 0

Given: x2 â€“ (5 – i) x + (18 + i) = 0

We shall apply discriminant rule,

Where, x = (-b Â±âˆš(b2 â€“ 4ac))/2a

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

So,

We can write 48 + 14i = 49 â€“ 1 + 14i

So,

48 + 14i = 49 + i2Â + 14i [âˆµÂ i2Â = â€“1]

= 72Â + i2Â + 2(7)(i)

= (7 + i)2Â [Since,Â (a + b)2Â = a2Â + b2Â + 2ab]

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

x = 2 + 3i or 3 â€“ 4i

âˆ´ The roots of the given equation are 3 â€“ 4i, 2 + 3i

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

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

We shall apply discriminant rule,

Where, x = (-b Â±âˆš(b2 â€“ 4ac))/2a

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

So,

We have i2Â = â€“1

By substituting â€“1 = i2Â in the above equation, we get

x = (1 â€“ i) or 4/5 â€“ 2i/5

âˆ´ The roots of the given equation are (1 â€“ i), 4/5 â€“ 2i/5

(iv) x2 â€“ (2 + i)x â€“ (1 â€“ 7i) = 0

Given: x2 â€“ (2 + i)x â€“ (1 â€“ 7i) = 0

We shall apply discriminant rule,

Where, x = (-b Â±âˆš(b2 â€“ 4ac))/2a

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

So,

We can write 7 â€“ 24i = 16 â€“ 9 â€“ 24i

7 â€“ 24i = 16 + 9(â€“1) â€“ 24i

= 16 + 9i2Â â€“ 24i [âˆµÂ i2Â = â€“1]

= 42Â + (3i)2Â â€“ 2(4) (3i)

= (4 â€“ 3i)2Â [âˆµÂ (a â€“ b)2Â = a2Â â€“ b2Â + 2ab]

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

x = 3 â€“ i or -1 + 2i

âˆ´ The roots of the given equation are (-1 + 2i), (3 â€“ i)

(v) ix2Â â€“ 4x â€“ 4i = 0

Given: ix2Â â€“ 4x â€“ 4i = 0

ix2Â + 4x(â€“1) â€“ 4i = 0 [We know, i2Â = â€“1]

So by substituting â€“1 = i2Â in the above equation, we get

ix2Â + 4xi2Â â€“ 4i = 0

i(x2Â + 4ix â€“ 4) = 0

x2Â + 4ix â€“ 4 = 0

x2Â + 4ix + 4(â€“1) = 0

x2Â + 4ix + 4i2Â = 0Â [Since,Â i2Â = â€“1]

x2Â + 2ix + 2ix + 4i2Â = 0

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

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

(x + 2i)2Â = 0

x + 2i = 0

x = â€“2i, -2i

âˆ´ The roots of the given equation are â€“2i, â€“2i

(vi) x2Â + 4ix â€“ 4 = 0

Given: x2Â + 4ix â€“ 4 = 0

x2Â + 4ix + 4(â€“1) = 0 [We know, i2Â = â€“1]

So by substituting â€“1 = i2Â in the above equation, we get

x2Â + 4ix + 4i2Â = 0

x2Â + 2ix + 2ix + 4i2Â = 0

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

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

(x + 2i)2Â = 0

x + 2i = 0

x = â€“2i, -2i

âˆ´ The roots of the given equation are â€“2i, â€“2i

(vii) 2x2 + âˆš15ix â€“ i = 0Â

Given: 2x2 + âˆš15ix â€“ i = 0Â

We shall apply discriminant rule,

Where, x = (-b Â±âˆš(b2 â€“ 4ac))/2a

Here, a = 2, b = âˆš15i, c = -i

So,

We can write 15 â€“ 8i = 16 â€“ 1 â€“ 8i

15 â€“ 8i = 16 + (â€“1) â€“ 8i

= 16 + i2Â â€“ 8i [âˆµÂ i2Â = â€“1]

= 42Â + (i)2Â â€“ 2(4)(i)

= (4 â€“ i)2Â [Since,Â (a â€“ b)2Â = a2Â â€“ b2Â + 2ab]

By using the result 15 â€“ 8i = (4 â€“ i)2, we get

âˆ´ The roots of the given equation are [1+ (4 – âˆš15)i/4] , [-1 -(4 + âˆš15)i/4]

(viii) x2Â â€“ x + (1 + i) = 0

Given: x2Â â€“ x + (1 + i) = 0

We shall apply discriminant rule,

Where, x = (-b Â±âˆš(b2 â€“ 4ac))/2a

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

So,

We can write 3 + 4i = 4 â€“ 1 + 4i

3 + 4i = 4 + i2Â + 4i [âˆµÂ i2Â = â€“1]

= 22Â + i2Â + 2(2) (i)

= (2 + i)2Â [Since,Â (a + b)2Â = a2Â + b2Â + 2ab]

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

x = 2i/2 or (2 â€“ 2i)/2

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

x = i or (1 – i)

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

(ix) ix2Â â€“ x + 12i = 0

Given: ix2Â â€“ x + 12i = 0

ix2Â + x(â€“1) + 12i = 0 [We know, i2Â = â€“1]

so by substituting â€“1 = i2Â in the above equation, we get

ix2Â + xi2Â + 12i = 0

i(x2Â + ix + 12) = 0

x2Â + ix + 12 = 0

x2Â + ix â€“ 12(â€“1) = 0

x2Â + ix â€“ 12i2Â = 0Â [Since,Â i2Â = â€“1]

x2Â â€“ 3ix + 4ix â€“ 12i2Â = 0

x(x â€“ 3i) + 4i(x â€“ 3i) = 0

(x â€“ 3i) (x + 4i) = 0

x â€“ 3i = 0 or x + 4i = 0

x = 3i or â€“4i

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

(x) x2 â€“ (3âˆš2 â€“ 2i)x – âˆš2i = 0

Given: x2 â€“ (3âˆš2 â€“ 2i)x – âˆš2i = 0

We shall apply discriminant rule,

Where, x = (-b Â±âˆš(b2 â€“ 4ac))/2a

Here, a = 1, b = -(3âˆš2 â€“ 2i), c = –âˆš2i

So,

(xi) x2 â€“ (âˆš2 + i)x + âˆš2i = 0

Given: x2 â€“ (âˆš2 + i)x + âˆš2i = 0

x2 â€“ (âˆš2x + ix) + âˆš2i = 0

x2 â€“ âˆš2x – ix + âˆš2i = 0

x(x – âˆš2) â€“ i(x – âˆš2) = 0

(x – âˆš2) (x – i) = 0

(x – âˆš2) = 0 or (x – i) = 0

x = âˆš2 or x = i

âˆ´ The roots of the given equation are i, âˆš2

(xii) 2x2 â€“ (3 + 7i)x + (9i – 3) = 0

Given: 2x2 â€“ (3 + 7i)x + (9i – 3) = 0

We shall apply discriminant rule,

Where, x = (-b Â±âˆš(b2 â€“ 4ac))/2a

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

So,

We can write 16 + 30i = 25 â€“ 9 + 30i

16 + 30i = 25 + 9(â€“1) + 30i

= 25 + 9i2Â + 30i [âˆµÂ i2Â = â€“1]

= 52Â + (3i)2Â + 2(5)(3i)

= (5 + 3i)2Â [âˆµÂ (a + b)2Â = a2Â + b2Â + 2ab]

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

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

Exercise 14.1 Solutions

Exercise 14.2 Solutions

## Frequently Asked Questions on RD Sharma Solutions for Class 11 Maths Chapter 14

### What are the ways to learn the topics from RD Sharma Solutions for Class 11 Maths Chapter 14 faster?

Students can quickly comprehend the key topics by referring to the RD Sharma Solutions for Class 11 Maths Chapter 14. It comprises solutions to textbook questions written in an elaborate manner highlighting the important points. Further, all the solutions are formulated by subject experts as per the latest CBSE syllabus and guidelines.

### How should I prepare for the board exam using the RD Sharma Solutions for Class 11 Maths Chapter 14?

Students can refer to CBSE syllabus and the correct study materials like the RD Sharma Solutions provided at BYJUâ€™S should be chosen to score good marks in the exam. Students need to learn the chapter and revise them on a regular basis to get a grip on the concepts. Students should refer to the solutions while learning a new chapter to understand the method of answering questions as per the latest CBSE guidelines. The syllabus for the academic year should be understood before the exam to know the marks weightage of each concept covered in the chapter.

### Can the RD Sharma Solutions for Class 11 Maths Chapter 14 really help students with their board exam preparations?

The Class 11 board exams are crucial in every studentâ€™s academic life as it lays a foundation for all career goals. The resource primarily increases the logical and analytical thinking skills of students which are vital for taking an exam. Before kickstarting preparations for the exam, students must have an overview of the syllabus and mark weightage for the concepts as per the CBSE guidelines. The solutions are prepared by subject matter experts at BYJUâ€™S with an aim to clarify doubts of students while learning the solutions to the textbook questions.