1. Find two consecutive even numbers such that one-third of the square of the larger one is two less than five times the smaller one.
Let the first even number be n+2. Then the next will be n+2.
It is given that one-third of the square of the larger one is two less than five times the smaller one.
Definitely (n+2) is the larger one.
`(n+2)^2/3 = 5n-2`
`(n+2)^2 = 3(5n-2)`
`n^2+4n+4 = 15n-6`
`n^2-11n+10 = 0`
`n^2-10n-n+10 = 0`
`n(n-10)-1(n-10) = 0`
`(n-10)(n-1) = 0`
n = 10 or n = 1
Since n is an even number n should be 10. So the next number will be 10+2 = 12.
So the two even numbers are 10 and 12.
let cosecutive even nos. are 2m,2m+2 , m is an integer. by given condition , let 2m+2 is greter than 2m.
`(2m+2)^2=3(10m-2) , `
`m-5=0 or 4m-2=0`
`m=5 or m=1/2`
`if m=5 then Nos. are 2.5. , 2.5+2 i.e 10 ,12`
`m=1/2 then numbers are 1,3 `