Question 1126756
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<pre>
The original equation is


    {{{x^2}}} - {{{y^2}}} = 2xy.


Divide both sides by {{{y^2}}}   (which is not zero !).  You will get


    {{{(x/y)^2}}} - 1 = {{{2*(x/y)}}}.


Introduce new variable  t = {{{x/y}}}.   Then the last equation takes the form


    {{{t^2}}} - 2t - 1 = 0.


It is a quadratic equation on t. Solve it using the quadratic formula


    {{{t[1,2]}}} = {{{(2 +- sqrt(2^2 -4*(-1)))/2}}} = {{{(2 +- sqrt(8))/2}}} = {{{1 +- sqrt(2)}}}.


<U>Answer</U>.  The ratio  {{{x/y}}},  where x > 0 and y > 0 satisfy the given equation, is  {{{1 + sqrt(2)}}}.
</pre>

Solved.


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Naturally, &nbsp;this solution has &nbsp;<U>N O T H I N G &nbsp;&nbsp; C O M M O N</U>&nbsp; with the answer and solution by &nbsp;@josgarithmetic, 


whose post is &nbsp;&nbsp;<U>T O T A L L Y  &nbsp;&nbsp;I R R E L E V A N T</U>&nbsp; to the given problem.



As usual.


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@josgarithmetic is VERY well known &nbsp;(or better to say, &nbsp;very badly known, &nbsp;unfortunately) &nbsp;in this forum,
for his inability to solve problems correctly and to make calculations accurately.



So be very careful when you obtain "solutions" from him.


In half of all cases &nbsp;(if not at most cases)&nbsp; they are wrong.


Do not trust him and always wait for reaction of other tutors.



In the normal school environment he would be fired in the second day.



I think about thousands of visitors and students who do understand almost nothing in Math and who were misguided 
by his false and wrong quasi- and pseudo-"solutions".



People, &nbsp;be careful about &nbsp;@josgarithmetic &nbsp;! &nbsp;&nbsp;! &nbsp;&nbsp;&nbsp;!