## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 73 of Exercise 10A

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73
##### Solve each of the following quadratic equations:22x - 3.2(x + 2) + 32 = 0

Given: 22x - 3.2(x + 2) + 32 = 0

(2x)2 - 3.2x.22 + 32 = 0 - - - - (1)

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

substituting for y in (1)

y2 - 12y + 32 = 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 = 1 b = - 12 c = 32

= 1.32 = 32

And either of their sum or difference = b

= - 12

Thus the two terms are - 8 and - 4

Sum = - 8 - 4 = - 12

Product = - 8. - 4 = 32

y2 - 8y - 4y + 32 = 0

y(y - 8) - 4(y - 8) = 0

(y - 8) (y - 4) = 0

(y - 8) = 0 or (y - 4) = 0

y = 8 or y = 4

2x = 8 or 2x = 4

substituting the value of y in (2)

2x = 23 or 2x = 22

x = 2 or x = 3

Hence the roots of equation are 2, 3

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