RS Aggarwal - Mathematics

Book: RS Aggarwal - Mathematics

Chapter: 10. Quadratic Equations

Subject: Maths - Class 10th

Q. No. 18 of Exercise 10E

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18

Divide 57 into two parts whose product is 680.

Let the two consecutive positive even integers be x and (57 – x)

According to given condition,


x (57 – x) = 680


57x – x2 = 680


x2 – 57x – 680 = 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 = – 57 c = – 680


= 1. – 680 = – 680


And either of their sum or difference = b


= – 57


Thus the two terms are – 40 and – 17


Sum = – 40 – 17 = – 57


Product = – 40. – 17 = – 680


x2 – 57x – 680 = 0


x2 – 40x – 17x – 680 = 0


x (x – 40) – 17(x – 40) = 0


(x – 40) (x – 17) = 0


(x – 40) = 0 or (x – 17) = 0


x = 40 or x = 17


When x = 40


57 – x = 57 – 40 = 17


When x = 17


57 – x = 57 – 17 = 40


Hence the required parts are 17 and 40.


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