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Find two consecutive multiples of 3 whose product is 648.

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Question

Find two consecutive multiples of 3 whose product is 648.

Sum
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Solution

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

According to the given condition, 

3x × 3(x + 1) = 648 

⇒ 9(x2 + x) = 648 

⇒ x2 + x = 72

⇒ x2 + x – 72 = 0 

⇒ 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   ...(Neglecting the negative value) 

When x = 8, 

3x = 3 × 8

= 24 

3(x + 1) = 3 × (8 + 1)

= 3 × 9

= 27 

Hence, the required multiples are 24 and 27. 

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Chapter 4: Quadratic Equations - EXERCISE 4D [Page 224]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 4 Quadratic Equations
EXERCISE 4D | Q 9. | Page 224
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