DIAMMIBENVERMk1

2022-07-02

The system of equations ${x}^{2}+{y}^{2}-x-2y=0$ and $x+2y=c$
$\left(1.\right)\phantom{\rule{1em}{0ex}}{x}^{2}+{y}^{2}-x-2y=0\phantom{\rule{0ex}{0ex}}\left(2.\right)\phantom{\rule{1em}{0ex}}x+2y=c$
Solving for $y$ in $\left(2.\right)$ gives
$y=\left(c-x\right)/2$
Is there a way to simplify equation $\left(1.\right)$?
Because at the end I arrive at
${c}^{2}-2x-c=0$
and can't proceed. Need to get typical form of square equation $A{x}^{2}+Bx+C=0$.
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) $\phantom{\rule{1em}{0ex}}{x}^{2}+{y}^{2}-x-2y=0,\phantom{\rule{1em}{0ex}}x+2y=c$, 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 ${B}^{2}-4AC$ we can get parameter.
So
$y=\left(c-x\right)/2$
${x}^{2}+{y}^{2}-x-2y=0$
${x}^{2}+\left(\left(c-x\right)/2{\right)}^{2}-x-2\ast \left(\left(c-x\right)/2\right)=0$
and i got lost...

lydalaszq

Expert

If you substitute correctly, you will get the equation
$5\phantom{\rule{thinmathspace}{0ex}}{x}^{2}-2\phantom{\rule{thinmathspace}{0ex}}c\phantom{\rule{thinmathspace}{0ex}}x+{c}^{2}-4\phantom{\rule{thinmathspace}{0ex}}c=0$
If you solve this, you will get $x=\frac{1}{5}\left(c±2\sqrt{c\left(5-c\right)}\right)$. The other value is given by $y=\frac{1}{2}\left(c-x\right)$, but you don't need this to answer your question.
If the two sets of solutions have the same values, then we must have $c\left(5-c\right)=0$, hence $c=0$ or $c=5$.
Addendum: To get the above equation, replace $y$ in ${x}^{2}+{y}^{2}-x-2y=0$ by $y=\frac{1}{2}\left(c-x\right)$ (from the second equation). That is, the equation
$\begin{array}{rcl}{x}^{2}+\frac{1}{4}\left(c-x{\right)}^{2}-x-\left(c-x\right)& =& \frac{1}{4}\left(4{x}^{2}+\left(c-x{\right)}^{2}-4c\right)\\ & =& \frac{1}{4}\left(5{x}^{2}-2cx+{c}^{2}-4c\right)\end{array}$
Then the solutions are (ignoring the $\frac{1}{4}$, of course):
$\begin{array}{rcl}x& =& \frac{1}{10}\left(2c±\sqrt{4{c}^{2}-20\left({c}^{2}-4c\right)}\right)\\ & =& \frac{1}{10}\left(2c±\sqrt{16c\left(5-c\right)}\right)\\ & =& \frac{1}{5}\left(c±2\sqrt{c\left(5-c\right)}\right)\end{array}$

Bruno Pittman

Expert