RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations

RD Sharma Solutions for Class 10 Maths Chapter 8 – Quadratic Equations are available here. As you prepare for the Class 10 Mathematics examination, understanding the key concepts and solving problems efficiently is essential for scoring full marks. To assist you, our team of subject experts has created the RD Sharma Solutions, offering step-by-step explanations for all exercise problems in line with the latest CBSE guidelines. If you aim for top marks, you're in the right place.

Chapter 8 of RD Sharma Class 10 focuses on Quadratic Equations, and it includes thirteen exercises. This chapter is crucial for Class 10 Mathematics, and mastering it is important. To help you understand and solve the problems effectively, our experts have provided detailed solutions to all exercises, available here at RD Sharma Solutions for Class 10.

The key topics covered in this chapter include:

• Determination of quadratic equations
• Formulation of quadratic equations
• Various methods to find the zeros or roots of quadratic equations, including factorization, completing the square, and using the quadratic formula
• Practical applications of quadratic equations in real-life scenarios

Download RD Sharma Class 10 Solutions Chapter 8 Quadratic Equation PDF

Access the RD Sharma Solutions for Class 10 Maths Chapter 8 – Quadratic Equations

1. Solve: x² – 5x + 6 = 0

Solution: We factor the quadratic: x² – 5x + 6 = 0

= (x – 2)(x – 3) = 0

Thus, x = 2 or x = 3.

2. Solve: 2x² + 7x + 3 = 0

Solution: Using the factorization method:

2x² + 6x + x + 3 = 0

(2x + 3)(x + 1) = 0

Thus, x = -3/2 or x = -1.

3. Solve: x² + 2√2x – 6 = 0

Solution: Using the quadratic formula:

x = [-2√2 ± √((2√2)² - 4×1×(-6))]/(2×1)

= [-2√2 ± √(8 + 24)]/2

= [-2√2 ± √32]/2

= [-2√2 ± 4√2]/2

Thus, x = (2√2)/2 = √2 or x = (-6√2)/2 = -3√2.

4. Solve: 3x² – 5x + 2 = 0

Solution: Factorizing:

3x² – 3x – 2x + 2 = 0

(3x – 2)(x – 1) = 0

Thus, x = 2/3 or x = 1.

5. Find the roots: x² – 4√3x + 12 = 0

Solution: Using the quadratic formula:

x = [-(-4√3) ± √((-4√3)² - 4×1×12)]/2

= [4√3 ± √(48 - 48)]/2

= (4√3 ± 0)/2

Thus, x = (4√3)/2 = 2√3.

(Repeated root.)

6. Solve: 2x² – 3x – 2 = 0

Solution: Using the quadratic formula:

x = [-(-3) ± √((-3)² - 4×2×(-2))]/(2×2)

= [3 ± √(9 + 16)]/4

= [3 ± √25]/4

Thus, x = (3 + 5)/4 = 2 or x = (3 – 5)/4 = -1/2.

7. Solve: x² – 7x + 10 = 0

Solution: Factorizing:

x² – 5x – 2x + 10 = 0

 (x – 5)(x – 2) = 0

Thus, x = 5 or x = 2.

8. Solve: 4x² – 12x + 9 = 0

Solution: Recognize it as a perfect square

(2x – 3)² = 0

Thus, 2x – 3 = 0 

x = 3/2.

9. Find the roots: x² + 5x + 6 = 0

Solution: Factorizing:

x² + 3x + 2x + 6 = 0

(x + 3)(x + 2) = 0

Thus, x = -3 or x = -2.

10. Solve: x² – 2x – 8 = 0

Solution: Factorizing:

x² – 4x + 2x – 8 = 0

(x – 4)(x + 2) = 0

Thus, x = 4 or x = -2.

11. Solve: 6x² – x – 2 = 0

Solution: Using factorization:

6x² – 4x + 3x – 2 = 0

 (3x – 2)(2x + 1) = 0

Thus,

x = 2/3 or x = -1/2.

12. Solve: 7x² + 14x + 7 = 0

Solution: Recognizing perfect square:

7(x² + 2x + 1) = 0

= 7(x + 1)² = 0

Thus,

x + 1 = 0 ⇒ x = -1.