The system of equations x 2 + y 2 − x − 2 y =...

DIAMMIBENVERMk1

DIAMMIBENVERMk1

Answered

2022-07-02

The system of equations x 2 + y 2 x 2 y = 0 and x + 2 y = c
( 1. ) x 2 + y 2 x 2 y = 0 ( 2. ) x + 2 y = c
Solving for y in ( 2. ) gives
y = ( c x ) / 2
Is there a way to simplify equation ( 1. )?
Because at the end I arrive at
c 2 2 x c = 0
and can't proceed. Need to get typical form of square equation A x 2 + B x + C = 0.
The solution for c is 0 or 5.
For what real values of the parameter do the common solutions of the following pairs of simultaneous equations became identical?
(g) x 2 + y 2 x 2 y = 0 , x + 2 y = c, Ans. c = 0 or 5.
The process is to solve for y, then to substitute that into second equation. From that we get A , B and C. Delta being B 2 4 A C we can get parameter.
So
y = ( c x ) / 2
x 2 + y 2 x 2 y = 0
x 2 + ( ( c x ) / 2 ) 2 x 2 ( ( c x ) / 2 ) = 0
and i got lost...

Answer & Explanation

lydalaszq

lydalaszq

Expert

2022-07-03Added 11 answers

If you substitute correctly, you will get the equation
5 x 2 2 c x + c 2 4 c = 0
If you solve this, you will get x = 1 5 ( c ± 2 c ( 5 c ) ). The other value is given by y = 1 2 ( c x ), but you don't need this to answer your question.
If the two sets of solutions have the same values, then we must have c ( 5 c ) = 0, hence c = 0 or c = 5.
Addendum: To get the above equation, replace y in x 2 + y 2 x 2 y = 0 by y = 1 2 ( c x ) (from the second equation). That is, the equation
x 2 + 1 4 ( c x ) 2 x ( c x ) = 1 4 ( 4 x 2 + ( c x ) 2 4 c ) = 1 4 ( 5 x 2 2 c x + c 2 4 c )
Then the solutions are (ignoring the 1 4 , of course):
x = 1 10 ( 2 c ± 4 c 2 20 ( c 2 4 c ) ) = 1 10 ( 2 c ± 16 c ( 5 c ) ) = 1 5 ( c ± 2 c ( 5 c ) )
Bruno Pittman

Bruno Pittman

Expert

2022-07-04Added 4 answers

Great expert answer!

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