RS Aggarwal - Mathematics

Book: RS Aggarwal - Mathematics

Chapter: 10. Quadratic Equations

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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