## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 40 of Exercise 10A

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40
##### Solve each of the following quadratic equations:4x2 + 4bx - (a2 - b2) = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 4 b = 4b c = - (a2 - b2)

= 4. - (a2 - b2)

= - 4a2 + 4b2

And either of their sum or difference = b

= 4b

Thus the two terms are 2(a + b) and - 2(a - b)

Difference = 2a + 2b - 2a + 2b = 4b

Product = 2(a + b). - 2(a - b) = - 4(a2 - b2)

using

2x[2x + (a + b)]-(a-b) [2x + (a + b)] = 0

[2x + (a + b)] [2x-(a-b)] = 0

[2x + (a + b)] = 0 or [2x-(a-b)] = 0

Hence the roots of equation are

1
1
1
1
1
1
1
1
1
1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
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68
69
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71
72
73