RD Sharma Solutions Class 11 Maths Chapter 14 – Quadratic Equations (Updated For 2024)

RD Sharma Solutions Class 11 Maths Chapter 14 – Quadratic Equations: The topic quadratic equation might seem familiar right? Here, RD Sharma Solutions Class 11 Maths Chapter 14 Quadratic Equations will solve your queries related to real coefficients and real roots. So without any second question grab your RD Sharma Solutions Class 11 Maths Chapter 14 and begin your test preparation.

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RD Sharma Solutions Class 11 Maths Chapter 14

Exercise-wise: RD Sharma Solutions Class 11 Maths Chapter 14 Quadratic Equations

RD Sharma Solutions Exercise 14.1

RD Sharma Solutions Exercise 14.2

Access RD Sharma Solutions Class 11 Maths Chapter 14

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 the 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 the 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 the 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

Important Topics This Chapter Includes-

Going through the important topics before you actually begin your exam preparation is one intelligent move. Thus to simplify the situation we compiled the important topics to learn from the chapter-

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

Access Important Chapters of RD Sharma Solutions Class 11 Maths 

• Chapter 1 – Sets
• Chapter 2 – Relations
• Chapter 3 – Functions
• Chapter 4 – Measurement of Angles
• Chapter 5 – Trigonometric Functions
• Chapter 6 – Graphs of Trigonometric Functions
• Chapter 7 – Trigonometric Ratios of Compound Angles
• Chapter 8 – Transformation Formulae
• Chapter 9 – Trigonometric Ratios of Multiple and Sub-Multiple Angles
• Chapter 10 – Sine and Cosine Formulae and Their Applications
• Chapter 11 – Trigonometric Equations
• Chapter 12 – Mathematical Induction
• Chapter 13 – Complex Numbers

Well, this is everything that you need from the RD Sharma Solutions Class 11 Maths Chapter 14 Quadratic Equations. If you find any issue concerning the same make sure to comment down. Best for your CBSE class 11 maths exam. 

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