## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 41 of Exercise 10A

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41
##### Solve each of the following quadratic equations:4x2 - 4a2x + (a4 - b4) = 0

4x2 - 4a2x + (a4 - b4) = 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 = - 4a2 ; c = (a4 - b4)

= 4. (a4 - b4)

= 4a4 - 4b4

And either of their sum or difference = b

= - 4a2

Thus the two terms are - 2(a2 + b2) and - 2(a2 - b2)

Difference = - 2(a2 + b2) - 2(a2 - b2)

= - 2a2 - 2b2 - 2a2 + 2b2

= - 4a2

Product = - 2(a2 + b2). - 2(a2 - b2)

= 4(a2 + b2)(a2 - b2)

= 4. (a4 - b4)

( using a2 - b2 = (a + b) (a - b))

4x2 - 4a2x + (a4 - b4) = 0

4x2 - 4a2x + ((a2)2 – (b2)2) = 0

( using a2 - b2 = (a + b) (a - b))

4x2 - 2(a2 + b2) x - 2(a2 - b2) x + (a2 + b2) (a2 - b2) = 0

2x [2x - (a2 + b2)] - (a2 - b2) [2x - (a2 + b2)] = 0

[2x - (a2 + b2)] [2x - (a2 - b2)] = 0

[2x - (a2 + b2)] = 0 or [2x - (a2 - b2)] = 0

Hence the roots of given equation are

1
1
1
1
1
1
1
1
1
1
1
2
3
4
5
6
7
8
9
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