Find three consecutive positive even integers such that the sum of the squares of the first two integers is 48 more than 14 times the third integer

Question

Consecutive numbers 

 

Answer 1

Let n be the first of 3 consecutive positive even integers; then the others would be n + 2 & n + 4.

 

The equation resulting from the given scenario is n² + (n+2)² = 48 + 14(n+4).

 

Now, solve this quadratic & make sure that the answer is positive & even.

Answer 2

Let 3 no. are 

 

x,x+2, x+4

 

given that  

 

x²+(x+2)²=48+14(x+4)

 

x²+x²+4x+4=48+14x+56

 

2x²  - 10 x + 4 =104

 

2x² - 10 x - 100=0 

 

x²-5x-50=0

 

(x-10)(x+5)=0

 

x=10  and x=-5

 

we need even integer

 

so x=10

3 consecutive  integers are 10,12,14