Three consecutive even natural numbers are such that the product of the first and the third is greater than 7 times the middle by 4. Find the numbers.
Answers
Three consecutive even natural numbers are such that the product of the first and the third is greater than 7 times the middle by 4. Find the numbers.
Step by step explanation :
Let the three consecutive even natural numbers be (x - 2), (x) and (x+2).
From the given condition,
(x - 2) (x + 2) = 7x + 4
Using the identity that :
(a - b)(a + b) = a^2 - b^2
.•. x^2 - 4 = 7x + 4
x^2 - 7x - 4 - 4 = 0
x^2 - 7x - 8 = 0
x^2 - 8x - x - 8 = 0
x( x - 8 ) + 1 ( x - 8 ) = 0
(x + 1) (x - 8) = 0
x = -1 or x = 8
x = -1 is not considerable because natural numbers are never negative.
x = 8.
So, Three consecutive even numbers are :
x - 2 = 8 - 2 = 6
x = 8
x + 2 = 8 + 2 = 10
6, 8, 10 are three consecutive even natural numbers.
Three consecutive even natural numbers are such that the product of the first and the third is greater than 7 times the middle by 4. Find the numbers.
Answer is 1 or 8.
Let the given 3 numbers be ( a + 2 ) , a ,( a - 2 )
From the given condition, we can say that...
(a + 2) (a - 2) = 7a + 4
a^2 - 2^2 = 7a+4
a^2 - 4 - 4 - 7a = 0
a^2 - 7a - 8 = 0
a^2 -8a + a - 8 = 0
a(a-8) -1( a - 8) = 0
Therefore, We get :
a = 1 or a = 8
Hope it helps you.