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

2 Answers
Jan 27, 2016

Answer:

#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.

Jan 27, 2016

Answer:

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#