Question 924373
Let's use lower-case for the variables and try a solution.


{{{(p-q)^2=25}}}, {{{pq=14}}}, and you want to find {{{(p+q)^2}}}.


{{{p^2-2pq+q^2=25}}}, first equation partially simplified;
{{{p^2+q^2-2pq=25}}}
Substitute using the second equation:
{{{p^2+q^2-2*14=25}}}
{{{p^2+q^2=25+28}}}
{{{p^2+q^2=53}}}
Another substitution again using from the second equation:
{{{p^2+(14/p)^2=53}}}
{{{p^4+14^2=53p^2}}}
{{{p^4-53p^2+14^2=0}}}
Solution first for p^2.----------------, this method will be long and very detailed...



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<b>Maybe another way.</b>


{{{p-q=0+- 5}}}  from the first equation.
{{{-q=-p+- 5}}}
{{{highlight_green(q=p+- 5)}}}.
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Using this q in the next equation of pq=14,
{{{p(p-5)=14}}}
{{{p^2-5p-14=0}}}
{{{(p+2)(p-7)=0}}}
p=-2  OR  p=7
OR
{{{p(p+5)=14}}}
{{{p^2+5p-14=0}}}
{{{(p-2)(p+7)=0}}}
p=2  OR  p=-7


Again, looking at the pq=14 equation, p and q must be of the same sign so that their product is POSITIVE 14.  This means that their solutions must be:
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p=-2 and q=-7
OR
p=2  and q=7.
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The exercise question asked for was, find {{{(p+q)^2}}}.
Using either combination of p and q found, this expression will be {{{highlight((2+7)^2=9^2=highlight(81))}}}.