Question 1091663
let one number be {{{p}}} and another number {{{q}}}

if one number is {{{4}}} more than another, we have:

{{{p=q+4}}} ....eq.1

if the sum of the squares of the two numbers is {{{72}}}, we have:

{{{p^2+q^2=72}}}....eq.2....substitute {{{p}}} from eq.1

{{{(q+4)^2+q^2=72}}}.....solve for {{{q}}}

{{{q^2+8q+16+q^2=72}}}

{{{2q^2+8q+16-72=0}}}

{{{2q^2+8q-56=0}}}...simplify, divide by {{{2}}}

{{{q^2+4q-28=0}}}


{{{q = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{q = (-4 +- sqrt( 4^2-4*1*(-28) ))/(2*1) }}} 

{{{q = (-4 +- sqrt( 16+112 ))/2}}} 

{{{q = (-4 +- sqrt( 16*4*2 ))/2}}} 

{{{q = (-4 +- (4*2)sqrt(2))/2}}}

{{{q= (-cross(4)2 +- (4*cross(2))sqrt(2 ))/cross(2)}}}

{{{highlight(q = (-2 +- 4*sqrt(2)))}}}

find {{{p}}}

{{{p=q+4}}} ....eq.1

{{{p=(-2 +- 4sqrt(2))+4}}} 

{{{p=(4-2 +- 4sqrt(2))}}} 

{{{highlight(p=(2 +- 4sqrt(2 )))}}}