RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.3 Quadratic Equations
RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.3 Quadratic Equations has been provided here. Students can refer to these questions before their examinations for better preparation.
Neha Tanna19 Nov, 2024
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RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.3: Chapter 8, Exercise 8.3 of RD Sharma's Class 10 Maths focuses on solving quadratic equations using the quadratic formula. This exercise emphasizes the formula , which is derived from the standard quadratic equation.
Students learn to calculate the discriminant (D), and its role in determining the nature of roots—real and distinct ( D>0 ), real and equal ( D=0 ), or imaginary ( D<0 ). The problems enhance conceptual understanding and application skills for solving quadratic equations systematically.RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.3 Overview
Chapter 8 of RD Sharma's Class 10 Maths book focuses on Quadratic Equations, a foundational topic in algebra. Exercise 8.3 emphasizes solving quadratic equations using methods like factorization and the quadratic formula, offering practical applications in real-life problem-solving scenarios. Mastery of these concepts is essential for understanding advanced mathematical topics in higher studies and for competitive exams. This exercise helps students develop analytical thinking and problem-solving skills, which are crucial for physics and engineering. By providing step-by-step solutions, RD Sharma ensures a deeper understanding of the topic, helping students build confidence in tackling quadratic equations systematically.RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.3 PDF
Below, we have provided the RD Sharma Solutions for Class 10 Maths Chapter 8, Exercise 8.3 on Quadratic Equations in a downloadable PDF format. These solutions are designed to help students understand and solve quadratic equations effectively. They include step-by-step explanations to make learning easy and efficient. Download the PDF now to simplify your preparation and excel in your exams!RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.3 PDF
RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.3 Quadratic Equations
Below is the RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.3 Quadratic Equations -Solve the following quadratic equation by factorization:
1. (x – 4)(x + 2) = 0
Solution:
Given, (x – 4) (x + 2) = 0 So, either x – 4 = 0 ⇒ x = 4 Or, x + 2 = 0, ⇒ x = – 2 Thus, the roots of the given quadratic equation are 4 and -2, respectively.2. (2x + 3) (3x – 7) = 0
Solution:
Given, (2x + 3) (3x – 7) = 0. So, either 2x + 3 = 0, ⇒ x = – 3/2 Or, 3x -7 = 0, ⇒ x = 7/3 Thus, the roots of the given quadratic equation are x = -3/2 and x = 7/3, respectively.3. 3x 2 – 14x – 5 = 0
Solution:
Given. 3x 2 – 14x – 5 = 0 ⇒ 3x 2 – 14x – 5 = 0 ⇒ 3x 2 – 15x + x – 5 = 0 ⇒ 3x(x – 5) + 1(x – 5) = 0 ⇒ (3x + 1)(x – 5) = 0 Now, either 3x + 1 = 0 ⇒ x = -1/3 Or, x – 5 = 0 ⇒ x = 5 Thus, the roots of the given quadratic equation are 5 and x = – 1/3, respectively.4. Find the roots of the equation 9x 2 – 3x – 2 = 0.
Solution:
Given, 9x 2 – 3x – 2 = 0. ⇒ 9x 2 – 3x – 2 = 0. ⇒ 9x 2 – 6x + 3x – 2 = 0 ⇒ 3x (3x – 2) + 1(3x – 2) = 0 ⇒ (3x – 2)(3x + 1) = 0 Now, either 3x – 2 = 0 ⇒ x = 2/3 Or, 3x + 1= 0 ⇒ x = -1/3 Thus, the roots of the given quadratic equation are x = 2/3 and x = -1/3, respectively.
5.
Solution:
Given,
6. 6x 2 + 11x + 3 = 0
Solution:
Given equation is 6x 2 + 11x + 3 = 0. ⇒ 6x 2 + 9x + 2x + 3 = 0 ⇒ 3x (2x + 3) + 1(2x + 3) = 0 ⇒ (2x +3) (3x + 1) = 0 Now, either 2x + 3 = 0 ⇒ x = -3/2 Or, 3x + 1= 0 ⇒ x = -1/3 Thus, the roots of the given quadratic equation are x = -3/2 and x = -1/3, respectively.7. 5x 2 – 3x – 2 = 0
Solution:
Given equation is 5x 2 – 3x – 2 = 0. ⇒ 5x 2 – 3x – 2 = 0. ⇒ 5x 2 – 5x + 2x – 2 = 0 ⇒ 5x(x – 1) + 2(x – 1) = 0 ⇒ (5x + 2)(x – 1) = 0 Now, either 5x + 2 = 0 ⇒x = -2/5 Or, x -1= 0 ⇒x = 1 Thus, the roots of the given quadratic equation are 1 and x = -2/5, respectively.8. 48x 2 – 13x – 1 = 0
Solution:
Given equation is 48x 2 – 13x – 1 = 0. ⇒ 48x 2 – 13x – 1 = 0. ⇒ 48x 2 – 16x + 3x – 1 = 0. ⇒ 16x(3x – 1) + 1(3x – 1) = 0 ⇒ (16x + 1)(3x – 1) = 0 Either 16x + 1 = 0 ⇒ x = -1/16 Or, 3x – 1=0 ⇒ x = 1/3 Thus, the roots of the given quadratic equation are x = -1/16 and x = 1/3, respectively.9. 3x 2 = -11x – 10
Solution:
Given equation is 3x 2 = -11x – 10 ⇒ 3x 2 + 11x + 10 = 0 ⇒ 3x 2 + 6x + 5x + 10 = 0 ⇒ 3x(x + 2) + 5(x + 2) = 0 ⇒ (3x + 5)(x + 2) = 0 Now, either 3x + 5 = 0 ⇒ x = -5/3 Or, x + 2 = 0 ⇒ x = -2 Thus, the roots of the given quadratic equation are x = -5/3 and -2, respectively.10. 25x(x + 1) = – 4
Solution:
Given equation is 25x(x + 1) = -4 25x(x + 1) = -4 ⇒ 25x 2 + 25x + 4 = 0 ⇒ 25x 2 + 20x + 5x + 4 = 0 ⇒ 5x (5x + 4) + 1(5x + 4) = 0 ⇒ (5x + 4)(5x + 1) = 0 Now, either 5x + 4 = 0 therefore x = – 4/5 Or, 5x + 1 = 0 therefore x = -1 /5 Thus, the roots of the given quadratic equation are x = – 4/5 and x = -1/5, respectively.11. 16x – 10/x = 27
Solution:
Given, 16x – 10/x = 27 On multiplying x on both sides we have, ⇒ 16x 2 – 10 = 27x ⇒ 16x 2 – 27x – 10 = 0 ⇒ 16x 2 – 32x + 5x – 10 = 0 ⇒ 16x(x – 2) +5(x – 2) = 0 ⇒ (16x + 5) (x – 2) = 0 Now, either 16x + 5 = 0 ⇒ x = -5/16 Or, x – 2 = 0 ⇒ x = 2 Thus, the roots of the given quadratic equation are x = – 5/16 and x = 2, respectively.
12.
Solution:
Given equation is,
On cross multiplying on both the sides we get,
2 = 3x(x – 2)
2 = 3x
2
– 6x
3x
2
– 6x – 2 = 0
⇒ 3x
2
– 3x – 3x – 2 = 0
Now, either
are the solutions of the given quadratic equations.
13. x – 1/x = 3, x ≠ 0
Solution:
Given, x – 1/x = 3 On multiplying x on both sides, we have, ⇒ x 2 – 1 = 3x ⇒ x 2 – 3x – 1 = 0
14.
Solution:
Given,
15.
Solution:
Given,
On cross multiplying we get,
⇒ x(3x – 8) = 8(x – 3)(x – 2)
⇒ 3x
2
– 8x = 8(x
2
– 5x + 6)
⇒ 8x
2
– 40x + 48 – (3x
2
– 8x) = 0
⇒ 5x
2
– 32x + 48 = 0
⇒ 5x
2
– 20x – 12x + 48 = 0
⇒ 5x(x – 4) – 12(x – 4) = 0
⇒ (x – 4)(5x – 12) = 0
Now, either x – 4 = 0 ⇒ x = 4
Or, 5x – 12 = 0 ⇒ x = 12/5
Thus, the roots of the given quadratic equation are 12/5 and 4, respectively.
16. a 2 x 2 – 3abx + 2b 2 = 0
Solution:
Given equation is a 2 x 2 – 3abx + 2b 2 = 0 ⇒ a 2 x 2 – abx – 2abx + 2b 2 = 0 ⇒ ax(ax – b) – 2b(ax – b) = 0 ⇒ (ax – b)(ax – 2b) = 0 Now, either ax – b = 0 ⇒ x = b/a Or, ax – 2b = 0 ⇒ x = 2b/a Thus, the roots of the quadratic equation are x = 2b/a and x = b/a, respectively.17. 9x 2 – 6b 2 x – (a 4 – b 4 ) = 0
Solution:
Given,
Thus, the roots of the quadratic equation are x = (b
2
– a
2
)/3 and x = (a
2
+ b
2
)/3, respectively.
18. 4x 2 + 4bx – (a 2 – b 2 ) = 0
Solution:
Given, 4x 2 + 4bx – (a 2 – b 2 ) = 0 For factorizing, 4(a 2 – b 2 ) = -4(a – b) (a + b) = [-2(a-b)] [2(a + b)] ⇒ 2(b – a)*2(b + a) ⇒ 4x 2 + (2(b – a) + 2(b + a)) – (a – b)(a + b) = 0 So, now 4x 2 + 2(b – a)x++ 2(b + a)x + (b – a)(a + b) = 0 ⇒ 2x(2x + (b – a)) +(a + b)(2x + (b – a)) = 0 ⇒ (2x + (b – a))(2x + b + a) = 0 Now, either (2x + (b – a)) = 0 ⇒x = (a – b)/2 Or, (2x + b + a) = 0 ⇒ x = -(a + b)/2 Thus, the roots of the given quadratic equation are x = -(a + b)/2 and x = (a – b)/2, respectively.19. ax 2 + (4a 2 – 3b)x – 12ab = 0
Solution:
Given equation is ax 2 + (4a 2 – 3b)x – 12ab = 0 ⇒ ax 2 + 4a 2 x – 3bx – 12ab = 0 ⇒ ax(x + 4a) – 3b(x + 4a) = 0 ⇒ (x + 4a)(ax – 3b) = 0 Now, either x + 4a = 0 ⇒ x = -4a Or, ax – 3b = 0 ⇒ x = 3b/a Thus, the roots of the given quadratic equation are x = 3b/a and -4a, respectively.20. 2x 2 + ax – a 2 = 0
Solution:
Given,
21. 16/x – 1 = 15/(x + 1), x ≠ 0, -1
Solution:
Given,
22. x ≠ -2, 3/2
Solution:
Given,
On cross-multiplying we get,
(x + 3)(2x – 3) = (x + 2)(3x – 7)
⇒ 2x
2
– 3x + 6x – 9 = 3x
2
– x – 14
⇒ 2x
2
+ 3x – 9 = 3x
2
– x – 14
⇒ x
2
– 3x – x – 14 + 9 = 0
⇒ x
2
– 5x + x – 5 = 0
⇒ x(x – 5) + 1(x – 5) = 0
⇒ (x – 5)(x + l) – 0
Now, either x – 5 = 0 or x + 1 = 0
⇒ x = 5 and x = -1
Thus, the roots of the given quadratic equation are 5 and -1, respectively.
23. x ≠ 3, 4
Solution:
The given equation is
On cross-multiplying, we have
3(4x
2
– 19x + 20) = 25(x
2
– 7x + 12)
⇒ 12x
2
– 57x + 60 = 25x
2
– 175x + 300
⇒13x
2
– 78x – 40x + 240 = 0
⇒13x
2
– 118x + 240 = 0
⇒13x
2
– 78x – 40x + 240 = 0
⇒13x(x – 6) – 40(x – 6) = 0
⇒ (x – 6)(13x – 40) = 0
Now, either x – 6 = 0 ⇒x = 6
Or, 13x – 40 = 0 ⇒x = 40/13
Thus, the roots of the given quadratic equation are 6 and 40/13, respectively.
24. x ≠ 0, 2
Solution:
Given equation is,
On cross-multiplying, we get
4(2x
2
+ 2) = 17(x
2
– 2x)
⇒ 8x
2
+ 8 = 17x
2
– 34x
⇒ 9x
2
– 34x – 8 = 0
⇒ 9x
2
– 36x + 2x – 8 = 0
⇒ 9x(x – 4) + 2(x – 4) = 0
⇒ 9x + 2)(x – 4) = 0
Now, either 9x + 2 = 0 ⇒x = -2/9
Or, x – 4 = 0 ⇒ x = 4
Thus, the roots of the given quadratic equation are x = -2/9 and 4, respectively.
25.
Solution:
Given equation is,
26.
, x ≠ 0Solution:
Given equation is,
On cross multiplying, we have
x(3x – 5) = 6(x
2
– 3x + 2)
⇒ 3x
2
– 5x = 6x
2
– 18x + 12
⇒ 3x
2
– 13x + 12 = 0
⇒ 3x
2
– 9x – 4x + 12 = 0
⇒ 3x(x – 3) – 4(x – 3) = 0
⇒ (x – 3)(3x – 4) = 0
Now, either x – 3 = 0 ⇒ x = 3
Or, 3x – 4 = 0 ⇒ 4/3.
Thus, the roots of the given quadratic equation are 3 and 4/3, respectively.
27.
, x ≠ 1, -1
Solution:
The given equation is,
On cross – multiplying we have,
⇒ 6(4x) = 5(x
2
– 1) = 24x
⇒ 5x
2
– 5 = 5x
2
– 24x – 5 =0
⇒ 5x
2
– 25x + x – 5 = 0
⇒ 5x(x – 5) + 1(x – 5) = 0
⇒ (5x + 1)(x – 5) = 0
Now, either x – 5 = 0 ⇒ x = 5
Or, 5x + 1 = 0 ⇒ x = −1/5
Thus, the roots of the given quadratic equation are x = −1/5 and 5, respectively.
28.
, x ≠ 1, -1/2
Solution:
The given equation is,
On cross – multiplying we have,
⇒ 2(5x
2
+ 2x + 2) = 5(2x
2
– x – 1)
⇒ 10x
2
+ 4x + 4 = 10x
2
– 5x – 5
⇒ 4x + 5x + 4 + 5 = 0
⇒ 9x + 9 = 0
⇒ 9x = -9
Thus, x = -1 is the only root of the given equation.
29.
Solution:
Given equation is,
Thus, the roots of the given quadratic equation are x = 1 and x = -2, respectively.
30.
Solution:
Given equation is,
On cross-multiplying, we have,
3(2x
2
– 22x + 58) = 10(x
2
– 12x + 35)
⇒ 6x
2
– 66x + 174 = 10x
2
– 120x + 350
⇒ 4x
2
– 54x + 176 = 0
⇒ 2x
2
– 27x + 88 = 0
⇒ 2x
2
– 16x – 11x + 88 = 0
⇒ 2x(x – 8) – 11(x + 8) = 0
⇒ (x – 8)(2x – 11) = 0
Now, either x – 8 = 0 ⇒ x = 8
Or, 2x – 11 = 0 ⇒ x = 11/2
Thus, the roots of the given quadratic equation are x = 11/2 and 8, respectively.
Benefits of Solving RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.3
Conceptual Clarity
Exercise 8.3 focuses on methods to solve quadratic equations, including factorization, completing the square, and using the quadratic formula. Practicing these helps in mastering the fundamental concepts and ensures clarity on the topic.Strengthening Problem-Solving Skills
This exercise offers a variety of problems, enabling students to apply different approaches to solving quadratic equations. Regular practice builds critical thinking and problem-solving abilities.Preparation for Board Exams
Quadratic equations form an important part of the Class 10 Maths syllabus. Solving RD Sharma's problems ensures thorough preparation for board exams as the book aligns well with the NCERT syllabus and exam pattern.Confidence Boost
RD Sharma solutions provide step-by-step explanations for each problem. Following these solutions helps students gain confidence in solving complex problems independently.Enhances Speed and Accuracy
Solving numerous problems improves calculation speed and enhances accuracy, which is crucial for board exams and other competitive exams.Application in Real-Life Problems
Quadratic equations have practical applications in physics, engineering, and other sciences. Understanding how to solve them equips students to handle real-life problems requiring mathematical solutions.Foundation for Higher Studies
Quadratic equations are essential for advanced mathematics, especially in Algebra and Calculus. Solving RD Sharma exercises strengthens the foundation required for higher classes and competitive exams.RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.3 FAQs
Who uses quadratic equations in real life?
Quadratic Equations are used in physics, engineering, economics, and computer science for modeling various phenomena and designing structures.
What makes a problem quadratic?
A quadratic equation is the equation of the 2nd degree. This means that it comprises at least one (1) term that is squared. One of the standard formulas for solving quadratic equations is 'ax² + bx + c = 0' here a, b, and c are constants or numerical coefficients.
Can quadratic formula solve everything?
Every quadratic equation has solutions if you allow for complex numbers.
What is the quadratic formula technique?
Step 1: Identify a, b, and c in the quadratic equation a x 2 + b x + c = 0 . Step 2: Substitute the values from step 1 into the quadratic formula x = − b ± b 2 − 4 a c 2 a . Step 3: Simplify, making sure to follow the order of operations.
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