SOLUTION: Ix x>0, y>0 and {{{x^2 -y^2 =2xy}}}, find the exact value of {{{x/y}}}. Thanks.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Ix x>0, y>0 and {{{x^2 -y^2 =2xy}}}, find the exact value of {{{x/y}}}. Thanks.      Log On


   



Question 1126756: Ix x>0, y>0 and x%5E2+-y%5E2+=2xy, find the exact value of x%2Fy.

Thanks.

Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39616) About Me  (Show Source):
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
The original equation is


    x%5E2 - y%5E2 = 2xy.


Divide both sides by y%5E2   (which is not zero !).  You will get


    %28x%2Fy%29%5E2 - 1 = 2%2A%28x%2Fy%29.


Introduce new variable  t = x%2Fy.   Then the last equation takes the form


    t%5E2 - 2t - 1 = 0.


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


    t%5B1%2C2%5D = %282+%2B-+sqrt%282%5E2+-4%2A%28-1%29%29%29%2F2 = %282+%2B-+sqrt%288%29%29%2F2 = 1+%2B-+sqrt%282%29.


Answer.  The ratio  x%2Fy,  where x > 0 and y > 0 satisfy the given equation, is  1+%2B+sqrt%282%29.

Solved.

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Naturally,  this solution has  N O T H I N G    C O M M O N  with the answer and solution by  @josgarithmetic,

whose post is   T O T A L L Y   I R R E L E V A N T  to the given problem.


As usual.

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@josgarithmetic is VERY well known  (or better to say,  very badly known,  unfortunately)  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  (if not at most cases)  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,  be careful about  @josgarithmetic  !   !    !