Question 214333
let the numbers be {{{j}}} and {{{k}}}
given:
(1) {{{k - j = 3}}}
(2) {{{j^2 + k^2 = 65}}}
From (1),
{{{k = j + 3}}}
{{{j^2 + (j + 3)^2 = 65}}}
{{{j^2 + j^2 + 6j + 9 = 65}}}
{{{2j^2 + 6j - 56 = 0}}}
{{{j^2 + 3j - 28 = 0}}}
solve using quadratic equation
{{{j = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = 1}}}
{{{b = 3}}}
{{{c = -28}}}
{{{j = (-3 +- sqrt( 3^2-4*1*(-28) ))/(2*1) }}}
{{{j = (-3 +- sqrt( 9 + 112 ))/2 }}}
{{{j = (-3 +- sqrt( 121 ))/2 }}}
{{{j = (-3 + 11)/2}}}
{{{j = 4}}}
and
{{{j = (-3 - 11)/2}}}
{{{j = -7}}} (can't use negative result)
{{{k = j + 3}}}
{{{k = 7}}}
The numbers are 4 and 7
check:
(2) {{{j^2 + k^2 = 65}}}
{{{4^2 + 7^2 = 65}}}
{{{16 + 49 = 65}}}
{{{65 = 65}}} OK