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 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. x ^{2} + 1 = 0**

**Solution:**

Given: x^{2}Â + 1 = 0

We know, i^{2}Â = â€“1Â â‡’Â 1 = â€“i^{2}

By substituting 1 = â€“i^{2}Â in the above equation, we get

x^{2}Â â€“ i^{2}Â = 0

^{2}Â â€“ b

^{2}Â = (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. 9x ^{2}Â + 4 = 0**

**Solution:**

Given: 9x^{2}Â + 4 = 0

9x^{2}Â + 4 Ã— 1 = 0

We know, i^{2}Â = â€“1Â â‡’Â 1 = â€“i^{2}

By substituting 1 = â€“i^{2}Â in the above equation, we get

So,

9x^{2}Â + 4(â€“i^{2}) = 0

9x^{2}Â â€“ 4i^{2}Â = 0

(3x)^{2}Â â€“ (2i)^{2}Â = 0

^{2}Â â€“ b

^{2}Â = (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. x ^{2}Â + 2x + 5 = 0**

**Solution:**

Given: x^{2}Â + 2x + 5 = 0

x^{2}Â + 2x + 1 + 4 = 0

x^{2}Â + 2(x) (1) + 1^{2}Â + 4 = 0

(x + 1)^{2}Â + 4 = 0 [since,Â (a + b)^{2}Â = a^{2}Â + 2ab + b^{2}]

(x + 1)^{2}Â + 4 Ã— 1 = 0

We know, i^{2}Â = â€“1Â â‡’Â 1 = â€“i^{2}

By substituting 1 = â€“i^{2}Â in the above equation, we get

(x + 1)^{2}Â + 4(â€“i^{2}) = 0

(x + 1)^{2}Â â€“ 4i^{2}Â = 0

(x + 1)^{2}Â â€“ (2i)^{2}Â = 0

^{2}Â â€“ b

^{2}Â = (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. 4x ^{2}Â â€“ 12x + 25 = 0**

**Solution:**

Given: 4x^{2}Â â€“ 12x + 25 = 0

4x^{2}Â â€“ 12x + 9 + 16 = 0

(2x)^{2}Â â€“ 2(2x)(3) + 3^{2}Â + 16 = 0

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

(2x â€“ 3)^{2}Â + 16 Ã— 1 = 0

We know, i^{2}Â = â€“1Â â‡’Â 1 = â€“i^{2}

By substituting 1 = â€“i^{2}Â in the above equation, we get

(2x â€“ 3)^{2}Â + 16(â€“i^{2}) = 0

(2x â€“ 3)^{2}Â â€“ 16i^{2}Â = 0

(2x â€“ 3)^{2}Â â€“ (4i)^{2}Â = 0

^{2}Â â€“ b

^{2}Â = (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. x ^{2}Â + x + 1 = 0**

**Solution:**

Given: x^{2}Â + x + 1 = 0

x^{2} + x + Â¼ + Â¾ = 0

x^{2} + 2 (x) (1/2) + (1/2)^{2} + Â¾ = 0

(x + 1/2)^{2} + Â¾ = 0 [Since,Â (a + b)^{2}Â = a^{2}Â + 2ab + b^{2}]

(x + 1/2)^{2} + Â¾ Ã— 1 = 0

We know, i^{2}Â = â€“1Â â‡’Â 1 = â€“i^{2}

By substituting 1 = â€“i^{2}Â in the above equation, we get

(x + Â½)^{2} + Â¾ (-1)^{2} = 0

(x + Â½)^{2} + Â¾ i^{2} = 0

(x + Â½)^{2} â€“ (**âˆš**3i/2)^{2} = 0

^{2}Â â€“ b

^{2}Â = (a + b) (a â€“ b)]

(x + Â½ + **âˆš**3i/2) (x + Â½ – **âˆš**3i/2) = 0

(x + Â½ + **âˆš**3i/2) = 0 or (x + Â½ – **âˆš**3i/2) = 0

x = -1/2 – **âˆš**3i/2 or x = -1/2 + **âˆš**3i/2

âˆ´ The roots of the given equation are -1/2 + **âˆš**3i/2, -1/2 – **âˆš**3i/2

**6. 4x ^{2}Â + 1 = 0**

**Solution:**

Given: 4x^{2}Â + 1 = 0

We know, i^{2}Â = â€“1Â â‡’Â 1 = â€“i^{2}

By substituting 1 = â€“i^{2}Â in the above equation, we get

4x^{2}Â â€“ i^{2}Â = 0

(2x)^{2}Â â€“ i^{2}Â = 0

^{2}Â â€“ b

^{2}Â = (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. x ^{2}Â â€“ 4x + 7 = 0**

**Solution:**

Given: x^{2}Â â€“ 4x + 7 = 0

x^{2}Â â€“ 4x + 4 + 3 = 0

x^{2}Â â€“ 2(x) (2) + 2^{2}Â + 3 = 0

(x â€“ 2)^{2}Â + 3 = 0 [Since,Â (a â€“ b)^{2}Â = a^{2}Â â€“ 2ab + b^{2}]

(x â€“ 2)^{2}Â + 3 Ã— 1 = 0

We know, i^{2}Â = â€“1Â â‡’Â 1 = â€“i^{2}

By substituting 1 = â€“i^{2}Â in the above equation, we get

(x â€“ 2)^{2}Â + 3(â€“i^{2}) = 0

(x â€“ 2)^{2}Â â€“ 3i^{2}Â = 0

(x â€“ 2)^{2}Â â€“ (**âˆš**3i)^{2} = 0

^{2}Â â€“ b

^{2}Â = (a + b) (a â€“ b)]

(x â€“ 2 + **âˆš**3i) (x â€“ 2 – **âˆš**3i) = 0

(x â€“ 2 + **âˆš**3i) = 0 or (x â€“ 2 – **âˆš**3i) = 0

x = 2 – **âˆš**3i or x = 2 + **âˆš**3i

x = 2 **Â±** **âˆš**3i

âˆ´ The roots of the given equation are 2 **Â±** **âˆš**3i

**8. x ^{2}Â + 2x + 2 = 0**

**Solution:**

Given: x^{2}Â + 2x + 2 = 0

x^{2}Â + 2x + 1 + 1 = 0

x^{2}Â + 2(x)(1) + 1^{2}Â + 1 = 0

(x + 1)^{2}Â + 1 = 0 [âˆµÂ (a + b)^{2}Â = a^{2}Â + 2ab + b^{2}]

We know, i^{2}Â = â€“1Â â‡’Â 1 = â€“i^{2}

By substituting 1 = â€“i^{2}Â in the above equation, we get

(x + 1)^{2}Â + (â€“i^{2}) = 0

(x + 1)^{2}Â â€“ i^{2}Â = 0

(x + 1)^{2}Â â€“ (i)^{2}Â = 0

^{2}Â â€“ b

^{2}Â = (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. 5x ^{2}Â â€“ 6x + 2 = 0**

**Solution:**

Given: 5x^{2}Â â€“ 6x + 2 = 0

We shall apply discriminant rule,

Where, x = (-b **Â±âˆš**(b^{2} â€“ 4ac))/2a

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

So,

x = (-(-6)** Â±âˆš**(-6^{2} â€“ 4 (5)(2)))/ 2(5)

= (6 **Â± âˆš**(36-40))/10

= (6 **Â± âˆš**(-4))/10

= (6 **Â± âˆš**4(-1))/10

We have i^{2}Â = â€“1

By substituting â€“1 = i^{2}Â in the above equation, we get

x = (6 **Â± âˆš**4i^{2})/10

= (6 **Â± **2i)/10

= 2(3**Â±**i)/10

= (3**Â±**i)/5

x = 3/5 **Â±** i/5

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

**10. 21x ^{2}Â + 9x + 1 = 0**

**Solution:**

Given: 21x^{2}Â + 9x + 1 = 0

We shall apply discriminant rule,

Where, x = (-b **Â±âˆš**(b^{2} â€“ 4ac))/2a

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

So,

x = (-9** Â±âˆš**(9^{2} â€“ 4 (21)(1)))/ 2(21)

= (-9 **Â± âˆš**(81-84))/42

= (-9 **Â± âˆš**(-3))/42

= (-9 **Â± âˆš**3(-1))/42

We have i^{2}Â = â€“1

By substituting â€“1 = i^{2}Â in the above equation, we get

x = (-9 **Â± âˆš**3i^{2})/42

= (-9 **Â± âˆš**(**âˆš**3i)^{2}/42

= (-9 **Â± âˆš**3i)/42

= -9/42 **Â± âˆš**3i/42

= -3/14 **Â± âˆš**3i/42

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

**11. x ^{2}Â â€“ x + 1 = 0**

**Solution:**

Given: x^{2}Â â€“ x + 1 = 0

x^{2} – x + Â¼ + Â¾ = 0

x^{2} – 2 (x) (1/2) + (1/2)^{2} + Â¾ = 0

(x – 1/2)^{2} + Â¾ = 0 [Since,Â (a + b)^{2}Â = a^{2}Â + 2ab + b^{2}]

(x – 1/2)^{2} + Â¾ Ã— 1 = 0

We know, i^{2}Â = â€“1Â â‡’Â 1 = â€“i^{2}

By substituting 1 = â€“i^{2}Â in the above equation, we get

(x – Â½)^{2} + Â¾ (-1)^{2} = 0

(x – Â½)^{2} + Â¾ (-i)^{2} = 0

(x – Â½)^{2} â€“ (**âˆš**3i/2)^{2} = 0

^{2}Â â€“ b

^{2}Â = (a + b) (a â€“ b)]

(x – Â½ + **âˆš**3i/2) (x – Â½ – **âˆš**3i/2) = 0

(x – Â½ + **âˆš**3i/2) = 0 or (x – Â½ – **âˆš**3i/2) = 0

x = 1/2 – **âˆš**3i/2 or x = 1/2 + **âˆš**3i/2

âˆ´ The roots of the given equation are 1/2 + **âˆš**3i/2, 1/2 – **âˆš**3i/2

**12. x ^{2}Â + x + 1 = 0**

**Solution:**

Given: x^{2}Â + x + 1 = 0

x^{2} + x + Â¼ + Â¾ = 0

x^{2} + 2 (x) (1/2) + (1/2)^{2} + Â¾ = 0

(x + 1/2)^{2} + Â¾ = 0 [Since,Â (a + b)^{2}Â = a^{2}Â + 2ab + b^{2}]

(x + 1/2)^{2} + Â¾ Ã— 1 = 0

We know, i^{2}Â = â€“1Â â‡’Â 1 = â€“i^{2}

By substituting 1 = â€“i^{2}Â in the above equation, we get

(x + Â½)^{2} + Â¾ (-1)^{2} = 0

(x + Â½)^{2} + Â¾ i^{2} = 0

(x + Â½)^{2} â€“ (**âˆš**3i/2)^{2} = 0

^{2}Â â€“ b

^{2}Â = (a + b) (a â€“ b)]

(x + Â½ + **âˆš**3i/2) (x + Â½ – **âˆš**3i/2) = 0

(x + Â½ + **âˆš**3i/2) = 0 or (x + Â½ – **âˆš**3i/2) = 0

x = -1/2 – **âˆš**3i/2 or x = -1/2 + **âˆš**3i/2

âˆ´ The roots of the given equation are -1/2 + **âˆš**3i/2, -1/2 – **âˆš**3i/2

**13. 17x ^{2}Â â€“ 8x + 1 = 0**

**Solution:**

Given: 17x^{2}Â â€“ 8x + 1 = 0

We shall apply discriminant rule,

Where, x = (-b **Â±âˆš**(b^{2} â€“ 4ac))/2a

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

So,

x = (-(-8)** Â±âˆš**(-8^{2} â€“ 4 (17)(1)))/ 2(17)

= (8 **Â± âˆš**(64-68))/34

= (8 **Â± âˆš**(-4))/34

= (8 **Â± âˆš**4(-1))/34

We have i^{2}Â = â€“1

By substituting â€“1 = i^{2}Â in the above equation, we get

x = (8 **Â± âˆš**(2i)^{2})/34

= (8 **Â± **2i)/34

= 2(4**Â±**i)/34

= (4**Â±**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) x ^{2}Â + 10ix â€“ 21 = 0**

**(ii) x ^{2}Â + (1 â€“ 2i)x â€“ 2i = 0**

**(iii) x ^{2} â€“ (2âˆš3 + 3i) x + 6âˆš3i = 0**

**(iv) 6x ^{2}Â â€“ 17ix â€“ 12 = 0**

**Solution:**

**(i) **x^{2}Â + 10ix â€“ 21 = 0

Given: x^{2}Â + 10ix â€“ 21 = 0

x^{2}Â + 10ix â€“ 21 Ã— 1 = 0

We know, i^{2}Â = â€“1Â â‡’Â 1 = â€“i^{2}

By substituting 1 = â€“i^{2}Â in the above equation, we get

x^{2}Â + 10ix â€“ 21(â€“i^{2}) = 0

x^{2}Â + 10ix + 21i^{2}Â = 0

x^{2}Â + 3ix + 7ix + 21i^{2}Â = 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) **x^{2}Â + (1 â€“ 2i)x â€“ 2i = 0

Given: x^{2}Â + (1 â€“ 2i)x â€“ 2i = 0

x^{2}Â + 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) **x^{2} â€“ (2âˆš3 + 3i) x + 6âˆš3i = 0

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

x^{2} â€“ (2âˆš3x + 3ix) + 6âˆš3i = 0

x^{2} 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) **6x^{2}Â â€“ 17ix â€“ 12 = 0

Given: 6x^{2}Â â€“ 17ix â€“ 12 = 0

6x^{2}Â â€“ 17ix â€“ 12 Ã— 1 = 0

We know, i^{2}Â = â€“1Â â‡’Â 1 = â€“i^{2}

By substituting 1 = â€“i^{2}Â in the above equation, we get

6x^{2}Â â€“ 17ix â€“ 12(â€“i^{2}) = 0

6x^{2}Â â€“ 17ix + 12i^{2}Â = 0

6x^{2}Â â€“ 9ix â€“ 8ix + 12i^{2}Â = 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) x ^{2} â€“ (3âˆš2 + 2i) x + 6âˆš2i = 0**

**(ii) x ^{2} â€“ (5 – i) x + (18 + i) = 0**

**(iii) (2 + i)x ^{2} â€“ (5- i)x + 2 (1 – i) = 0**

**(iv) x ^{2} â€“ (2 + i)x â€“ (1 â€“ 7i) = 0**

**(v) ix ^{2}Â â€“ 4x â€“ 4i = 0**

**(vi) x ^{2}Â + 4ix â€“ 4 = 0**

**(vii) 2x ^{2} + âˆš15ix â€“ i = 0Â **

**(viii) x ^{2}Â â€“ x + (1 + i) = 0**

**(ix) ix ^{2}Â â€“ x + 12i = 0**

**(x) x ^{2} â€“ (3âˆš2 â€“ 2i)x – âˆš2i = 0**

**(xi) x ^{2} â€“ (âˆš2 + i)x + âˆš2i = 0**

**(xii) 2x ^{2} â€“ (3 + 7i)x + (9i – 3) = 0**

**Solution:**

**(i) **x^{2} â€“ (3âˆš2 + 2i) x + 6âˆš2i = 0

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

x^{2} â€“ (3âˆš2x + 2ix) + 6âˆš2i = 0

x^{2} 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) **x^{2} â€“ (5 – i) x + (18 + i) = 0

Given: x^{2} â€“ (5 – i) x + (18 + i) = 0

We shall apply discriminant rule,

Where, x = (-b **Â±âˆš**(b^{2} 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 + i^{2}Â + 14i [âˆµÂ i^{2}Â = â€“1]

= 7^{2}Â + i^{2}Â + 2(7)(i)

= (7 + i)^{2}Â [Since,Â (a + b)^{2}Â = a^{2}Â + b^{2}Â + 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)x^{2} â€“ (5- i)x + 2 (1 – i) = 0

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

We shall apply discriminant rule,

Where, x = (-b **Â±âˆš**(b^{2} â€“ 4ac))/2a

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

So,

We have i^{2}Â = â€“1

By substituting â€“1 = i^{2}Â 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) **x^{2} â€“ (2 + i)x â€“ (1 â€“ 7i) = 0

Given: x^{2} â€“ (2 + i)x â€“ (1 â€“ 7i) = 0

We shall apply discriminant rule,

Where, x = (-b **Â±âˆš**(b^{2} 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 + 9i^{2}Â â€“ 24i [âˆµÂ i^{2}Â = â€“1]

= 4^{2}Â + (3i)^{2}Â â€“ 2(4) (3i)

= (4 â€“ 3i)^{2}Â [âˆµÂ (a â€“ b)^{2}Â = a^{2}Â â€“ b^{2}Â + 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)** ix^{2}Â â€“ 4x â€“ 4i = 0

Given: ix^{2}Â â€“ 4x â€“ 4i = 0

ix^{2}Â + 4x(â€“1) â€“ 4i = 0 [We know, i^{2}Â = â€“1]

So by substituting â€“1 = i^{2}Â in the above equation, we get

ix^{2}Â + 4xi^{2}Â â€“ 4i = 0

i(x^{2}Â + 4ix â€“ 4) = 0

x^{2}Â + 4ix â€“ 4 = 0

x^{2}Â + 4ix + 4(â€“1) = 0

x^{2}Â + 4ix + 4i^{2}Â = 0Â [Since,Â i^{2}Â = â€“1]

x^{2}Â + 2ix + 2ix + 4i^{2}Â = 0

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

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

(x + 2i)^{2}Â = 0

x + 2i = 0

x = â€“2i, -2i

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

**(vi) **x^{2}Â + 4ix â€“ 4 = 0

Given: x^{2}Â + 4ix â€“ 4 = 0

x^{2}Â + 4ix + 4(â€“1) = 0 [We know, i^{2}Â = â€“1]

So by substituting â€“1 = i^{2}Â in the above equation, we get

x^{2}Â + 4ix + 4i^{2}Â = 0

x^{2}Â + 2ix + 2ix + 4i^{2}Â = 0

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

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

(x + 2i)^{2}Â = 0

x + 2i = 0

x = â€“2i, -2i

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

**(vii) **2x^{2} + âˆš15ix â€“ i = 0**Â **

Given: 2x^{2} + âˆš15ix â€“ i = 0**Â **

We shall apply discriminant rule,

Where, x = (-b **Â±âˆš**(b^{2} 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 + i^{2}Â â€“ 8i [âˆµÂ i^{2}Â = â€“1]

= 4^{2}Â + (i)^{2}Â â€“ 2(4)(i)

= (4 â€“ i)^{2}Â [Since,Â (a â€“ b)^{2}Â = a^{2}Â â€“ b^{2}Â + 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)** x^{2}Â â€“ x + (1 + i) = 0

Given: x^{2}Â â€“ x + (1 + i) = 0

We shall apply discriminant rule,

Where, x = (-b **Â±âˆš**(b^{2} â€“ 4ac))/2a

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

So,

We can write 3 + 4i = 4 â€“ 1 + 4i

3 + 4i = 4 + i^{2}Â + 4i [âˆµÂ i^{2}Â = â€“1]

= 2^{2}Â + i^{2}Â + 2(2) (i)

= (2 + i)^{2}Â [Since,Â (a + b)^{2}Â = a^{2}Â + b^{2}Â + 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) **ix^{2}Â â€“ x + 12i = 0

Given: ix^{2}Â â€“ x + 12i = 0

ix^{2}Â + x(â€“1) + 12i = 0 [We know, i^{2}Â = â€“1]

so by substituting â€“1 = i^{2}Â in the above equation, we get

ix^{2}Â + xi^{2}Â + 12i = 0

i(x^{2}Â + ix + 12) = 0

x^{2}Â + ix + 12 = 0

x^{2}Â + ix â€“ 12(â€“1) = 0

x^{2}Â + ix â€“ 12i^{2}Â = 0Â [Since,Â i^{2}Â = â€“1]

x^{2}Â â€“ 3ix + 4ix â€“ 12i^{2}Â = 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)** x^{2} â€“ (3**âˆš**2 â€“ 2i)x – **âˆš**2i = 0

Given: x^{2} â€“ (3**âˆš**2 â€“ 2i)x – **âˆš**2i = 0

We shall apply discriminant rule,

Where, x = (-b **Â±âˆš**(b^{2} â€“ 4ac))/2a

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

So,

**(xi)** x^{2} â€“ (**âˆš**2 + i)x + **âˆš**2i = 0

Given: x^{2} â€“ (**âˆš**2 + i)x + **âˆš**2i = 0

x^{2} â€“ (**âˆš**2x + ix) + **âˆš**2i = 0

x^{2} â€“ **âˆš**2x – ix + **âˆš**2i = 0

x(x – **âˆš**2) â€“ i(x – **âˆš**2) = 0

(x – **âˆš**2) (x – i) = 0

(x – **âˆš**2) = 0 or (x – i) = 0

x = **âˆš**2 or x = i

âˆ´ The roots of the given equation are i, **âˆš**2

**(xii) **2x^{2} â€“ (3 + 7i)x + (9i – 3) = 0

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

We shall apply discriminant rule,

Where, x = (-b **Â±âˆš**(b^{2} â€“ 4ac))/2a

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

So,

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

16 + 30i = 25 + 9(â€“1) + 30i

= 25 + 9i^{2}Â + 30i [âˆµÂ i^{2}Â = â€“1]

= 5^{2}Â + (3i)^{2}Â + 2(5)(3i)

= (5 + 3i)^{2}Â [âˆµÂ (a + b)^{2}Â = a^{2}Â + b^{2}Â + 2ab]

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