Question

The sum of the SQUARES of two consecutive positive integers is 145. How do you find the numbers?

Answer 1

#n²+(n+1)²=145, =n²+n²+2n+1=145, 2n²+2n=144, n²+n=72, n²+n-72=0. n=(-b+-(b²-4*a*c))/2*a,( -1+(1-4*1*-72)^0.5)/2, =(-1+(289)^0.5)/2,=(-1+17)/2=8#. n=8, n+1=9.

Explanation:

given.

Answer 2

I found `8 and 9`

Explanation:

Let us call the numbers:

`n`
and
`n+1`

we get (from our condition) that:

`(n)^2+(n+1)^2=145`

rearrange and solve for `n`:
`n^2+n^2+2n+1-145=0`
`2n^2+2n-144=0`
use the Quadratic Formula:
`n_(1,2)=(-2+-sqrt(4+1152))/4=(-2+-34)/4`
so ve get two values:
`n_1=-9`
`n_2=8`
we chose the positive one so that our numbers will be:
`n=8`
and
`n+1=9`