## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 11 of Exercise 10E

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11
##### Find two consecutive positive even integers whose product is 288.

Let the two consecutive positive even integers be x and (x + 2)

According to given condition,

x (x + 2) = 288

x2 + 2x – 288 = 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 = 2 c = – 288

= 1. – 288 = – 288

And either of their sum or difference = b

= 2

Thus the two terms are 18 and – 16

Difference = 18 – 16 = 2

Product = 18. – 16 = – 288

x2 + 18x – 16x – 288 = 0

x (x + 18) – 16(x + 18) = 0

(x + 18) (x – 16) = 0

(x + 18) = 0 or (x – 16) = 0

x = – 18 or x = 16

x = 16 (x is a positive odd integer)

x + 2 = 16 + 2 = 18

Hence, the required integers are 16 and 18

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