DIAMMIBENVERMk1

Answered

2022-07-02

The system of equations ${x}^{2}+{y}^{2}-x-2y=0$ and $x+2y=c$

$(1.)\phantom{\rule{1em}{0ex}}{x}^{2}+{y}^{2}-x-2y=0\phantom{\rule{0ex}{0ex}}(2.)\phantom{\rule{1em}{0ex}}x+2y=c$

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

${c}^{2}-2x-c=0$

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

$y=(c-x)/2$

${x}^{2}+{y}^{2}-x-2y=0$

${x}^{2}+((c-x)/2{)}^{2}-x-2\ast ((c-x)/2)=0$

and i got lost...

$(1.)\phantom{\rule{1em}{0ex}}{x}^{2}+{y}^{2}-x-2y=0\phantom{\rule{0ex}{0ex}}(2.)\phantom{\rule{1em}{0ex}}x+2y=c$

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

${c}^{2}-2x-c=0$

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

$y=(c-x)/2$

${x}^{2}+{y}^{2}-x-2y=0$

${x}^{2}+((c-x)/2{)}^{2}-x-2\ast ((c-x)/2)=0$

and i got lost...

Answer & Explanation

lydalaszq

Expert

2022-07-03Added 11 answers

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 $x=\frac{1}{5}(c\pm 2\sqrt{c(5-c)})$. The other value is given by $y=\frac{1}{2}(c-x)$, 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 ${x}^{2}+{y}^{2}-x-2y=0$ by $y=\frac{1}{2}(c-x)$ (from the second equation). That is, the equation

$\begin{array}{rcl}{x}^{2}+\frac{1}{4}(c-x{)}^{2}-x-(c-x)& =& \frac{1}{4}(4{x}^{2}+(c-x{)}^{2}-4c)\\ & =& \frac{1}{4}(5{x}^{2}-2cx+{c}^{2}-4c)\end{array}$

Then the solutions are (ignoring the $\frac{1}{4}$, of course):

$\begin{array}{rcl}x& =& \frac{1}{10}(2c\pm \sqrt{4{c}^{2}-20({c}^{2}-4c)})\\ & =& \frac{1}{10}(2c\pm \sqrt{16c(5-c)})\\ & =& \frac{1}{5}(c\pm 2\sqrt{c(5-c)})\end{array}$

$5\phantom{\rule{thinmathspace}{0ex}}{x}^{2}-2\phantom{\rule{thinmathspace}{0ex}}c\phantom{\rule{thinmathspace}{0ex}}x+{c}^{2}-4\phantom{\rule{thinmathspace}{0ex}}c=0$

If you solve this, you will get $x=\frac{1}{5}(c\pm 2\sqrt{c(5-c)})$. The other value is given by $y=\frac{1}{2}(c-x)$, 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 ${x}^{2}+{y}^{2}-x-2y=0$ by $y=\frac{1}{2}(c-x)$ (from the second equation). That is, the equation

$\begin{array}{rcl}{x}^{2}+\frac{1}{4}(c-x{)}^{2}-x-(c-x)& =& \frac{1}{4}(4{x}^{2}+(c-x{)}^{2}-4c)\\ & =& \frac{1}{4}(5{x}^{2}-2cx+{c}^{2}-4c)\end{array}$

Then the solutions are (ignoring the $\frac{1}{4}$, of course):

$\begin{array}{rcl}x& =& \frac{1}{10}(2c\pm \sqrt{4{c}^{2}-20({c}^{2}-4c)})\\ & =& \frac{1}{10}(2c\pm \sqrt{16c(5-c)})\\ & =& \frac{1}{5}(c\pm 2\sqrt{c(5-c)})\end{array}$

Bruno Pittman

Expert

2022-07-04Added 4 answers

Great expert answer!

Most Popular Questions