solve the following problem - (any 1)
a) The product of four consecutive natural numbers
is 840. find the numbers?
Answers
Answer:
Let the consecutive numbers be n, (n+1) , (n + 2 ) (n+ 3).
Given that their product = 840
then n(n+1)(n + 2 )(n+ 3) = 840.
this can also written as n(n+3)⋅(n+1)(n+2) = 840
Add and subtarct 1 to the first number of the product.
((n^2+3n+1)−1)((n2+3n+1)+1) = 840
(n^2+3n+1)^2−1 = 840
(n^2+3n+1)^2 = 841
(n^2+3n+1)^2 = 29 ^2
(n^2+3n+1) = 29 implies n^2+3n+1–29=0 = n^2 + 3n - 28
(n + 7) (n - 4) = 0
if n = 4 we have 4, 5, 6, 7 n = -7 is not possible -7 does not belongs to the natural number.
or
If n is the smallest of the four consecutive positive integers, then
840=n(n+3)⋅(n+1)(n+2)
=((n2+3n+1)−1)((n2+3n+1)+1)
=(n2+3n+1)2−1 .
Hence, n2+3n+1=29 , so that n=4 ( we may neglect n=−7 , which is negative ) . We may verify that 4⋅5⋅6⋅7=840 .
Step-by-step explanation:
Mark my answer as brainliest
Answer:
Let four consecutive natural numbers are( x -2 ) , (x-1) , x and (x+1). Acordingly:-
(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
Euther 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 . Answe
Step-by-step explanation: