A quadratic equation can be solved by many methods. In this exercise, students will be solving quadratic equations by factorisation method. For further clarifications regarding this method of solving the Selina Solutions for Class 10 Maths can be used by students. Also, the solutions to this exercise are available in the Concise Selina Solutions for Class 10 Maths Chapter 5 Quadratic Equations Exercise 5(C) PDF in the links provided below.
Concise Selina Solutions for Class 10 Maths Chapter 5 Quadratic Equations Exercise 5(C) Download PDF
Access other exercises of Selina Solutions Concise Maths Class 10 Chapter 5 Quadratic Equations
Access Selina Solutions Concise Maths Class 10 Chapter 5 Quadratic Equations Exercise 5(C)
Solve equations, number 1 to 20, given below, using factorization method:
1. x2 – 10x – 24 = 0
Solution:
Given equation, x2 – 10x – 24 = 0
x2 – 12x + 2x – 24 = 0
x(x – 12) + 2(x – 12) = 0
(x + 2)(x – 12) = 0
So, x + 2 = 0 or x – 12 = 0
Hence,
x = -2 or x = 12
2. x2 – 16 = 0
Solution:
Given equation, x2 – 16 = 0
x2 + 4x – 4x + 16 = 0
x(x + 4) -4(x + 4) = 0
(x – 4) (x + 4) = 0
So, (x – 4) = 0 or (x + 4) = 0
Hence,
x = 4 or x = -4
3. 2x2 – ½ x = 0
Solution:
Given equation, 2x2 – ½ x = 0
4x2 – x = 0
x(4x – 1) = 0
So, either x = 0 or 4x – 1 = 0
Hence,
x = 0 or x = ¼
4. x(x – 5) = 24
Solution:
Given equation, x(x – 5) = 24
x2 – 5x = 24
x2 – 5x – 24 = 0
x2 – 8x + 3x – 24 = 0
x(x – 8) + 3(x – 8) = 0
(x + 3)(x – 8) = 0
So, x + 3 = 0 or x – 8 = 0
Hence,
x = -3 or x = 8
5. 9/2 x = 5 + x2
Solution:
Given equation, 9/2 x = 5 + x2
On multiplying by 2 both sides, we have
9x = 2(5 + x2)
9x = 10 + 2x2
2x2 – 9x + 10 = 0
2x2 – 4x – 5x + 10 = 0
2x(x – 2) – 5(x – 2) = 0
(2x – 5)(x -2) = 0
So, 2x – 5 = 0 or x – 2 = 0
Hence,
x = 5/2 or x = 2
6. 6/x = 1 + x
Solution:
Given equation, 6/x = 1 + x
On multiplying by x both sides, we have
6 = x(1 + x)
6 = x + x2
x2 + x – 6 = 0
x2 + 3x – 2x – 6 = 0
x(x + 3) – 2(x + 3) = 0
(x – 2) (x + 3) = 0
So, x – 2 = 0 or x + 3 = 0
Hence,
x = 2 or x = -3
7. x = (3x + 1)/ 4x
Solution:
Given equation, x = (3x + 1)/ 4x
On multiplying by 4x both sides, we have
4x(x) = 3x + 1
4x2 = 3x + 1
4x2 – 3x – 1 = 0
4x2 – 4x + x – 1 = 0
4x(x – 1) + 1(x – 1) = 0
(4x + 1) (x – 1) = 0
So, 4x + 1 = 0 or x – 1 = 0
Hence,
x = -1/4 or x = 1
8. x + 1/x = 2.5
Solution:
Given equation, x + 1/x = 2.5
x + 1/x = 5/2
Taking LCM on L.H.S, we have
(x2 + 1)/ x = 5/2
2(x2 + 1) = 5x
2x2 + 2 = 5x
2x2 – 5x + 2 = 0
2x2 – 4x – x + 2 = 0
2x(x – 2) -1(x – 2) = 0
(2x – 1)(x – 2) = 0
So, 2x – 1 = 0 or x – 2 = 0
Hence,
x = ½ or x = 2
9. (2x – 3)2 = 49
Solution:
Given equation, (2x – 3)2 = 49
Expanding the L.H.S, we have
4x2 – 12x + 9 = 49
4x2 – 12x – 40 = 0
Dividing by 4 on both side
x2 – 3x – 10 = 0
x2 – 5x + 2x – 10 = 0
x(x – 5) + 2(x – 5) = 0
(x + 2) (x – 5) = 0
So, x + 2 = 0 or x – 5 = 0
Hence,
x = -2 or 5
10. 2(x2 – 6) = 3(x – 4)
Solution:
Given equation, 2(x2 – 6) = 3(x – 4)
2x2 – 12 = 3x – 12
2x2 = 3x
x(2x – 3) = 0
So, x = 0 or (2x – 3) = 0
Hence,
x = 0 or x = 3/2
11. (x + 1) (2x + 8) = (x + 7) (x + 3)
Solution:
Given equation, (x + 1) (2x + 8) = (x + 7) (x + 3)
2x2 + 2x + 8x + 8 = x2 + 7x + 3x + 21
2x2 + 10x + 8 = x2 + 10x + 21
x2 = 21 – 8
x2 – 13 = 0
(x – √13) (x + √13) = 0
So, x – √13 = 0 or x + √13 = 0
Hence,
x = – √13 or x = √13
12. x2 – (a + b)x + ab = 0
Solution:
Given equation, x2 – (a + b)x + ab = 0
x2 – ax – bx + ab = 0
x(x – a) – b(x – a) = 0
(x – b) (x – a) = 0
So, x – b = 0 or x – a = 0
Hence,
x = b or x = a
13. (x + 3)2 – 4(x + 3) – 5 = 0
Solution:
Given equation, (x + 3)2 – 4(x + 3) – 5 = 0
(x2 + 9 + 6x) – 4x – 12 – 5 = 0
x2 + 2x – 8 = 0
x2 + 4x – 2x – 8 = 0
x(x + 4) – 2(x – 4) = 0
(x – 2)(x + 4) = 0
So, x – 2 = 0 or x + 4 = 0
Hence,
x = 2 or x = -4
14. 4(2x – 3)2 – (2x – 3) – 14 = 0
Solution:
Given equation, 4(2x – 3)2 – (2x – 3) – 14 = 0
Let substitute 2x – 3 = y
Then the equation becomes,
4y2 – y – 14 = 0
4y2 – 8y + 7y – 14 = 0
4y(y – 2) + 7(y – 2) = 0
(4y + 7)(y – 2) = 0
So, 4y + 7 = 0 or y – 2 = 0
Hence,
y = -7/4 or y = 2
But we have taken y = 2x – 3
Thus,
2x – 3 = -7/4 or 2x – 3 = 2
2x = 5/ 4 or 2x = 5
x = 5/8 or x = 5/2
15. 3x – 2/ 2x- 3 = 3x – 8/ x + 4
Solution:
Given equation, 3x – 2/ 2x- 3 = 3x – 8/ x + 4
On cross-multiplying we have,
(3x – 2)(x + 4) = (3x – 8)(2x – 3)
3x2 – 2x + 12x – 8 = 6x2 – 16x – 9x + 24
3x2 + 10x – 8 = 6x2 – 25x + 24
3x2 – 35x + 32 = 0
3x2 – 3x – 32x + 32 = 0
3x(x – 1) – 32(x – 1) = 0
(3x – 32)(x – 1) = 0
So, 3x – 32 = 0 or x – 1 = 0
Hence,
x = 32/3 or x = 1
16. 2x2 – 9x + 10 = 0, when:
(i) x ∈ N (ii) x ∈ Q
Solution:
Given equation, 2x2 – 9x + 10 = 0
2x2 – 4x – 5x + 10 = 0
2x(x – 2) – 5(x – 2) = 0
(2x – 5)(x – 2) = 0
So, 2x – 5 = 0 or x – 2 = 0
Hence,
x = 5/2 or x = 2
(i) When x ∈ N
x = 2 is the solution.
(ii) When x ∈ Q
x = 2, 5/2 are the solutions
17.
Solution:
2(2x2 + 18) = 5(x2 – 9)
4x2 + 36 = 5x2 – 45
x2 – 81 = 0
(x – 9)(x + 9) = 0
So, x – 9 = 0 or x + 9 = 0
Hence,
x = 9 or x = -9
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