Question 126736
If one number is x and the other y, their sum, x + y is 17.  The sum of their squares, {{{x^2+y^2=145}}}.


Since {{{x+y=17}}}, {{{y = 17-x}}}, so


{{{x^2+y^2=145}}}
{{{x^2+(17-x)^2=145}}}
{{{x^2+(289-34x+x^2)=145}}}
{{{2x^2-34x+289-145=0}}}
{{{2x^2-34x+144=0}}}
{{{x^2-17x+72=0}}}

{{{-8*-9=72}}} and {{{-8+(-9)= -17}}}


{{{(x-8)(x-9) = 0}}}


So, x is either 8 or 9, and then y is either 9 or 8.  Either way, the larger number is 9.


Check:  {{{8+9=17}}}  and {{{8^2+9^2=64+81=145}}}