Question

Solve the following problems :
(i) The product of four consecutive natural
numbers is 840. Find the numbers.​

Answer 1

Answer:

Explanation:

Let four consecutive natural numbers be ( x -2 ) , (x-1) , x and (x+1). Accordingly:-

(x-2).(x-1) .x.(x+1) = 840

(x^2–2x)×(x^2–1)= 840

x^4–2x^3-x^2+2x-840 = 0

On putting x=6

R=1296–432–36+12–840=1308–1308=0

(x-6)is a factor

x^4–2x^3-x^2+2x-840=0

x^3(x-6)+4x^2(x-6)+23x(x-6)+140(x-6)=0

(x-6)(x^3+4x^2+23x+140) = 0

x-6=0 => x =6

1st number =x-2 =6–2 = 4

2nd number =x-1=6–1 = 5

3rd number =x = 6

4th number =x+1 =6+1 = 7

4 numbers are 4 , 5 , 6 and 7 .

Hope it helps :)