# 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.

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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=5.2m-2` ``

`(2m+2)^2=3(10m-2) , `

`4m^2+8m+4=30m-6`

`4m^2-22m+10=0`

`4m^2-20m-2m+10=0`

`4m(m-5)-2(m-5)=0`

`(m-5)(4m-2)=0`

`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 `

`10,12 `

`ans.`

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