0
$\begingroup$

So i have this equation $$-\sqrt{C_1-y^2}=x+C_2$$

I want to get rid of radical by squaring both sides, the negative sign in front of the first term will become positive right?

I know this is a silly question but i need to make sure, my background at algebra is terrible.

$\endgroup$
1
$\begingroup$

Yes it is, and your question is not very terrible since after your calculation, you will reach weird things.

$$C_1-y^2=(x+C_2)^2$$

You can of course change the subject and get a nice equation.

$$y^2=-(x+C_2)^2+C_1$$

Even though the equation seems nice, we need to remember that criteria from the equations are $C_1-y^2\geq 0$ and $x+C_2\leq 0$ since square root always take positive root, and things inside positive should be positive (I assume you are working with real variables.

So that's it. You can erase the negative sign (by squaring both sides), but the negative sign does play a role when you try to draw the curve out.

$\endgroup$
  • 1
    $\begingroup$ Yes my answer should be expressed in y and thats the answer given. Thanks! $\endgroup$ – johnny Feb 3 '18 at 4:52
-1
$\begingroup$

Yes, so you have $$C_1-y^2=(x+C_2)^2.$$

$\endgroup$
  • $\begingroup$ What does happen if $x+C_2<0$? $\endgroup$ – Michael Rozenberg Feb 3 '18 at 4:45
  • $\begingroup$ Thaaaaaaaaanks! $\endgroup$ – johnny Feb 3 '18 at 4:53
  • $\begingroup$ @MichaelRozenberg Nothing spectacular; the square of a negative is a positive. $\endgroup$ – Allawonder Feb 3 '18 at 10:35
  • $\begingroup$ And if $x+C_2>0$? $\endgroup$ – Michael Rozenberg Feb 3 '18 at 10:36
  • 1
    $\begingroup$ No, my reasoning is as solid as can be, for for example $(-\sqrt2)^2=2$. $\endgroup$ – Allawonder Feb 3 '18 at 10:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.