Q. 95.0( 1 Vote )

# Find two consecut

Let the required consecutive multiples of 3 be 3x and 3(x + 1)

According to given condition,

3x.3(x + 1) = 648

9(x2 + x) = 648

x2 + x = 72

x2 + x – 72 = 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 = 1 c = – 72

= 1. – 72 = – 72

And either of their sum or difference = b

= 1

Thus the two terms are 9 and – 8

Difference = 9 – 8 = 1

Product = 9. – 8 = – 72

x2 + 9x – 8x – 72 = 0

x (x + 9) – 8(x + 9) = 0

(x + 9) (x – 8) = 0

(x + 9) = 0 or (x – 8) = 0

x = – 9 or x = 8

x = 8 (rejecting the negative values)

3x = 3.8 = 24

3(x + 1) = 3(8 + 9) = 3.9 = 27

Hence, the required numbers are 24 and 27

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