Book: RS Aggarwal - Mathematics

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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