Question 1202304
<pre>
The solution above is incomplete for there are two solutions.

Let the smaller number be x and the larger number be x+k where k > 0

{{{system(x^2+(x+k)^2=218,(k+x-x)x=42)}}}

{{{system(x^2+x^2+2kx+k^2=218,kx=42)}}}

{{{system(2x^2+2kx+k^2=218,kx=42)}}}

{{{x=42/k}}}

{{{2(42/k)^2+2k(42/k)+k^2=218}}}

{{{2(1764^""/k^2)+84+k^2=218}}}

{{{3528^""/k^2+k^2=134}}}

{{{3528+k^4=134k^2}}}

{{{k^4-134k^2+3528=0}}}

Believe it or not, that factors!:

{{{(k^2-36)(k^2-98)=0}}}

{{{matrix(5,3,  k^2-36=0,"",k^2-98=0,
              k^2=36,"",k^2=98,
                k=6,"",k=sqrt(98),
                  "","",k=sqrt(49*2),
                  "","",k=7sqrt(2))}}}  <--[Remember that k > 0]


Using k = 6, substituting in

{{{x=42/k}}}
{{{x=42/6=7}}}

smaller = 7, larger = 7+6 = 13   <--one answer

Using {{{k=7sqrt(2)}}}

{{{x=42/k}}}
{{{x=42/(7sqrt(2))}}}
{{{x=6/sqrt(2)}}}
{{{x=(6/sqrt(2))(sqrt(2)/sqrt(2))}}}
{{{x=6sqrt(2)/2}}}
{{{x=3sqrt(2)}}}

{{{smaller=3sqrt(2)}}}    {{{larger=3sqrt(2)+7sqrt(2)=10sqrt(2)}}}  <--second answer

Edwin</pre>