Book: RS Aggarwal - Mathematics

Subject: Maths - Class 10th

Q. No. 72 of Exercise 10A

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72
Solve each of the following quadratic equations:4(x + 1) + 4(1 - x) = 10

Given: 4(x + 1) + 4(1 - x) = 10

- - - - - - - (1)

Let 4x = y - - - - - - - - - - (2)

substituting for y in (1)

4y2 - 10y + 4 = 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 = - 10 c = 4

= 4.4 = 16

And either of their sum or difference = b

= - 10

Thus the two terms are - 8 and - 2

Sum = - 8 - 2 = - 10

Product = - 8. - 2 = 16

4y2 - 10y + 4 = 0

4y2 - 8y - 2y + 4 = 0

4y(y - 2) - 2(y - 2) = 0

(y - 2) (4y - 2) = 0

(y - 2) = 0 or (4y - 2) = 0

y = 2 or y = 1/2

substituting the value of y in (2)

4x = 2 or 4x = 2 - 1

22x = 21 or 22x = 2 - 1

2x = 1 or 2x = - 1

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
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15
16
17
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21
22
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26
27
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31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
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51
52
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55
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72
73