Class 10 RD Sharma Solutions Chapter 8 Quadratic Equations Exercise 8.3 | Set 2 (original) (raw)
Last Updated : 1 Oct, 2024
The Quadratic equations are fundamental to algebra and appear frequently in various mathematical contexts. They play a crucial role in understanding polynomial functions and solving real-world problems. In this article, we will delve into Exercise 8.3 | Set 2 from RD Sharmaβs Class 10 Mathematics book which focuses on solving quadratic equations.
This exercise helps students practice and master different techniques for solving these equations.
What is a Quadratic Equation?
A quadratic equation is a polynomial equation of degree 2 which can be written in the standard form:
ax2 + bx + c = 0
where a, b, and c are constants, and πβ 0. The solutions to a quadratic equation can be found using various methods including factoring completing the square and using the quadratic formula.
**Key Concepts:
- Roots of the Quadratic Equation: These are the values of x that satisfy the equation ax2+bx+c=0.
- Discriminant: Given by Ξ=b2 β4ac it determines the nature of the roots.
Question 21. (16/x) - 1 = 15/(x + 1), x β 0, -1.
**Solution:
We have equation,
(16/x) - 1 = 15/(x + 1)
(16 - x)/x = 15/(x + 1)
15x = (x + 1) (16 - x)
15x = 16x - x2 + 16 - x
15x - 16x + x2 -16 + x = 0
x2 - 16 = 0
(x - 4) (x + 4) = 0
Therefore, roots of the equation are 4 or -4.
**Question 22. (x + 3)/(x + 2) = (3x - 7)/(2x - 3), x β -2, 3/2.
**Solution:
We have equation,
(x + 3) (2x - 3) = (3x - 7) (x + 2)
2x2 - 3x + 6x - 9 = 3x2 + 6x -7x -14
2x2 + 3x -9 = 3x2 -x -14
x2 -4x -5 = 0
We can factorize this equation as:
x2 - 5x + x -5 = 0
x (x - 5) + 1 (x - 5) = 0
(x + 1) (x - 5) = 0
Therefore, roots of the equation are 5 or -1.
Question 23. (2x/(x - 4)) + ((2x - 5)/(x - 3)) = 25/3, x β 3, 4
**Solution:
We have equation,
(2x/(x - 4)) + ((2x - 5)/(x - 3)) = 25/3
((2x)(x - 3) + (2x - 5) (x - 4))/((x - 4) (x - 3)) = 25/3
25x2 - 175x + 300 = 12x2 - 57x + 60
13x2 - 118x + 240 = 0
We can factorize this equation as:
13x2 - 78x - 40x + 240 = 0
13x (x - 6) - 40 (x - 6) = 0
(13x - 40) (x - 6) = 0
Therefore, roots of the equation are 6 or 40/13.
Question 24. ((x + 3)/(x - 2)) - ((1 - x)/(x)) = 17/4, x β 0, 2
**Solution:
We have equation,
((x + 3)/(x - 2)) + ((1 - x)/(x)) = 17/4
8x2 + 8 = 17x2 - 34x
9x2 - 34x - 8 = 0
We can factorize this equation as:
9x2 - 36x + 2x - 8 = 0
9x(x - 4) + 2(x - 4) = 0
(9x + 2) (x - 4) = 0
Therefore, roots of the equation are 4 or -2/9.
Question 25. ((x - 3)/(x + 3)) - ((x + 3)/(x - 3)) = 48/7, x β 3, -3
**Solution:
We have equation,
((x - 3)/(x + 3)) - ((x + 3)/(x - 3)) = 48/7
-84x = 48x2 - 432
48x2 + 84x - 432 = 0
4x2 + 7x -36 = 0
We can factorize this equation as:
4x2 + 16x - 9x - 36 = 0
4x (x + 4) - 9 (x + 4) = 0
(4x - 9) (x + 4) = 0
Therefore, roots of the equation are-4 or 9/4.
Question 26. (1/(x - 2)) + (2/(x - 1)) = 6/x, x β 0
**Solution:
We have equation,
(1/(x - 2)) + (2/(x - 1)) = 6/x
3x2 - 5x = 6x2 - 18x + 12
3x2 - 13x + 12 = 0
We can factorize this equation as:
3x2 - 9x -4x + 12 = 0
3x (x - 3) - 4 (x - 3) = 0
(3x - 4) (x - 3) = 0
Therefore, roots of the equation are 3 or 4/3.
Question 27. ((x + 1)/(x - 1)) - ((x - 1)/(x + 1)) = 5/6, x β 1, -1
**Solution:
We have equation,
((x + 1)/(x - 1)) - ((x - 1)/(x + 1)) = 5/6
5x2 - 5 = 24x
5x2 - 24x - 5 = 0
We can factorize this equation as:
5x2 - 25x + x - 5 = 0
5x (x - 5) + 1 (x - 5) = 0
(5x + 1) (x - 5) = 0
Therefore, roots of the equation are 5 or -1/5.
Question 28. ((x - 1)/(2x + 1)) + ((2x + 1)/(x - 1)) = 5/2, x β 1, -1/2
**Solution:
We have equation,
((x - 1)/(2x + 1)) + ((2x + 1)/(x - 1)) = 5/2
2 (5x2 + 2x + 2) = 5 (2x2 - x - 1)
9x + 9 = 0
9 (x + 1) = 0
Therefore, roots of the equation are -1.
Question 29. (4/x) - 3 = 5/(2x + 3), x β 0, -3/2
**Solution:
We have equation,
(4/x) - 3) = 5/(2x + 3)
5x = (2x + 3) (4 - 3x)
5x = 8x - 6x2 + 12 - 9x
6x2 - 6x -12 = 0
x2 - x -2 = 0
We can factorize this equation as:
x2 + 2x - x - 2 = 0
x (x + 2) -1 (x + 2) = 0
(x - 1) (x + 2) = 0
Therefore, roots of the equation are 1 or -2.
Question 30. ((x - 4)/(x - 5)) + ((x - 6)/(x - 7)) = 10/3, x β 5, 7
**Solution:
We have equation,
((x - 4)/(x - 5)) + ((x - 6)/(x - 7)) = 10/3
4x2 - 54x + 176 = 0
2x2 - 27x + 88 = 0
We can factorize this equation as:
2x2 - 16x -11x + 88 = 0
2x (x - 8) - 11 (x - 8) = 0
(2x - 11) (x - 8) = 0
Therefore, roots of the equation are 8 or 11/2.
Question 31. ((x - 2)/(x - 3)) + ((x - 4)/(x - 5)) = 10/3, x β 3, 5.
**Solution:
We have equation,
((x - 2)/(x - 3)) + ((x - 4)/(x - 5)) = 10/3
We can rewrite it as :
((x - 3 + 1)/(x - 3)) + ((x - 5 + 1)/(x - 5)) = 10/3
1 + 1 + (1/(x - 3)) + (1/( x - 5)) = 10 3
4 (x2 - 8x + 15) = 6x - 24
4x2 - 38x + 84 = 0
2x2 -19x + 42 = 0
We can factorize this equation as:
2x2 - 12x - 7x + 42 = 0
2x (x - 6) - 7 (x - 6) = 0
(2x - 7) (x - 6) = 0
Therefore, roots of the equation are 6 or 7/2.
Question 32. (( 5 + x)/(5 - x)) - ((5 - x)/(5 + x)) = 15/4, x β 5, -5.
**Solution:
We have equation,
(( 5 + x)/(5 - x)) - ((5 - x)/(5 + x)) = 15/4
80x = 375 - 15x2
15x2 +80x -375 = 0
3x2 +16x - 75 = 0
We can factorize this equation as:
3x2 + 25x - 9x - 75 = 0
x (3x + 25) - 3 (3x + 25) = 0
(3x + 25) (x - 3) = 0
Therefore, roots of the equation are 3 or -25/3.
Question 33. (3/(x + 1)) - (1/2) = 2/(3x - 1), x β -1, 1/3
**Solution:
We have equation,
(3/(x + 1)) - (1/2) = 2/(3x - 1)
2 (2x + 2) = (5 - x) (3x - 1)
4x + 4 = 15x - 5 - 3x2 + x
3x2 - 12x + 9 = 0
x2 - 4x + 3 = 0
We can factorize this equation as:
x2 - 3x - x + 3 = 0
x (x - 3) - 1 (x - 3) = 0
(x - 3) (x - 1) = 0
Therefore, roots of the equation are 1 or 3.
Question 34. (3/(x + 1)) + (4/(x - 1)) = 29/(4x - 1), x β -1, 1, 1/4
**Solution:
We have equation,
(3/(x + 1)) + (4/(x - 1)) = 29/(4x - 1)
(7x + 1) (4x - 1) = 29 (x2 - 1)
28x2 - 7x + 4x - 1 = 29x2 - 29
x2 + 3x -28 = 0
We can factorize this equation as:
x2 + 7x - 4x - 28 = 0
x (x + 7) - 4 (x + 7) = 0
(x - 4) (x + 7) = 0
Therefore, roots of the equation are 4 or -7.
Question 35. (2/(x + 1)) + (3/(2(x - 2))) = 23/5x, x β 0, -1, 2
**Solution:
We have equation,
(2/(x + 1)) + (3/(2(x - 2))) = 23/5x
35x2 - 25x = 46 (x2 - x - 2)
11x2 - 21x - 92 = 0
We can factorize this equation as:
11x2 - 44x + 23x - 92 = 0
11x (x - 4) + 23 (x - 4) = 0
(11x + 23) ( x - 4) = 0
Therefore, roots of the equation are 4 or -23/11.
Question 36. x2 - (β3 + 1) x + β3 = 0
**Solution:
We have equation,
x2 - (β3 + 1) x + β3 = 0
x2 - β3x - x + β3 = 0
We can factorize this equation as:
x (x - β3) - 1 (x - β3) = 0
(x -1) (x - β3) = 0
Therefore, roots of the equation are 1 or β3.
Question 37. 3β5 x2 + 25x - 10β5 = 0
**Solution:
We have equation,
3β5 x2 + 25x - 10β5 = 0
β5 (3x2 + (25/β5) x - (10β5/β5)) = 0
β5 (3x2 + 5β5x - 10) = 0
We can factorize this equation as:
3x2 - β5x + 6β5x - 10 = 0
x (3x - β5) + 2β5 (3x - β5) = 0
(x + 2β5) (3x - β5) = 0
Therefore, roots of the equation are -2β5 or β5/3.
Question 38. β3x2 - 2β2 x - 2β3 = 0
**Solution:
We have equation,
β3x2 - 2β2 x - 2β3 = 0
Here a = β3, b = -2β2 and c = -2β3
Since, Discriminant D = b2 - 4ac and x = (-b Β± βD)/2a
Therefore,
D = 8 + 24 = 32, and
x = (-(-2β2) Β± β32)/2β3
x = ( 2β2 Β± 4β2)/2β3
Therefore, roots of the equations are (2β2 + 4β2)/2β3 or (2β2 - 4β2)/2β3.
Question 39. 4β3x2 + 5x - 2β3 = 0
**Solution:
We have equation,
4β3x2 + 5x - 2β3 = 0
Here a = 4β3, b = 5 and c = -2β3
Since, Discriminant D = b2 - 4ac and x = (-b Β± βD)/2a
Therefore,
D = 25 + 96 = 121, and
x = (-(5) Β± β121)/8β3
x = (-5 Β± 11)/8β3
Therefore, roots of the equations are -2 /β3 or β3/ 4.
Question 40. β2x2 - 3x - 2β2 = 0
**Solution:
We have equation,
β2x2 - 3x - 2β2 = 0
Here a = β2, b = -3 and c = -2β2
Since, Discriminant D = b2 - 4ac and x = (-b Β± βD)/2a
Therefore,
D = 9 + 16 = 25, and
x = (-(-3) Β± β25)/2β2
x = (3 Β± 5)/2β2
Therefore, roots of the equations are 2β2 or -1/ β2.
Conclusion
The Quadratic equations are essential in algebra and are crucial for the solving various mathematical problems. Exercise 8.3 | Set 2 from RD Sharmaβs Class 10 book provides the valuable practice in solving these equations through different methods. Mastery of these techniques will enhance problem-solving skills and provide a solid foundation for the further mathematical study.