Question 1104242
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Let {{{u = sqrt(x+y)}}}<br>
Then the first equation is
{{{u^2+u = 20}}}
{{{u^2+u-20 = 0}}}
{{{(u+5)(u-4) = 0}}}<br>
Then {{{u = sqrt(x+y) = -5}}} or {{{u = sqrt(x+y) = 4}}}.<br>
Since the square root has to be positive, {{{sqrt(x+y)=4}}}, which means {{{x+y=16}}}  (1).<br>
Similarly, let {{{u = sqrt(x-y)}}}; then the second equation is<br>
{{{u^2+u = 12}}}
{{{u^2+u-12 = 0}}}
{{{(u+4)(u-3) = 0}}}<br>
Then {{{u = sqrt(x-y) = -4}}} or {{{u = sqrt(x-y) = 3}}}.<br>
Then, as before, {{{sqrt(x-y) = 3}}}, which means {{{x-y = 9}}}  (2).<br>
Solving (1) and (2) together gives us x = 25/2, y = 7/2.<br>
Then the product we are looking for is {{{(25/2)*(7/2) = 175/4}}}