Question

S. Two natural numbers differ by 3 and their product is 504. Find the
numbers.
9. Find two consecutive multiples of 3 whose product is 648.​

Answer 1

Answer:

Let two numbers be x and y

x - y = 3 ------(1)

and xy = 504 ------(2)

then from first x = y + 3 -------(3)

Putting value of eq(3) in eq (2)

(y + 3) y = 504

y2 + 3y - 504 = 0

y2 + ( 24 - 21 ) y -504 = 0

y2 + 24y -21y -504 = 0

y ( y + 24 ) - 21 ( y + 24 ) = 0

( y - 21 ) ( y + 24 )

y = 21 or y = -24

then putting this value of y in eq(1)

we get x = 24 or x = - 21

Step-by-step explanation:

1 question is above

2 question is below

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.