Confused about square root

Question

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.

score 1 · Answer 1 · answered Feb 03 '18 at 04:18

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.

Yes my answer should be expressed in y and thats the answer given. Thanks! — johnny, Feb 03 '18 at 04:52

score -1 · Answer 2 · answered Feb 03 '18 at 04:16

-1

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

answered Feb 03 '18 at 04:16

Allawonder

13,327

What does happen if $x+C_2<0$? – Michael Rozenberg Feb 03 '18 at 04:45
Thaaaaaaaaanks! – johnny Feb 03 '18 at 04:53
@MichaelRozenberg Nothing spectacular; the square of a negative is a positive. – Allawonder Feb 03 '18 at 10:35
And if $x+C_2>0$? – Michael Rozenberg Feb 03 '18 at 10:36
@MichaelRozenberg I don't see the point to your questioning. The square of every number is always positive. – Allawonder Feb 03 '18 at 10:37
For example $-1=1$ is wrong, but $(-1)^2=1^2$ is true. I think your step is wrong. – Michael Rozenberg Feb 03 '18 at 10:39
@MichaelRozenberg I don't see how your last comment relates to my answer at all. Of course squaring any quantity gives you a positive quantity, so what? – Allawonder Feb 03 '18 at 10:41
You can not use squaring here without to write before that $x+C_2\leq0.$ – Michael Rozenberg Feb 03 '18 at 10:43
@MichaelRozenberg Of course you can. It's obvious from the original equation. But even without multiplying both sides by $-1$ it's still right, for any negative number squared, even if under a radical, is always positive. – Allawonder Feb 03 '18 at 10:45
No! Your reasoning is wrong! Try the squaring of the both sides to $-1=1$. – Michael Rozenberg Feb 03 '18 at 10:47
1

No, my reasoning is as solid as can be, for for example $(-\sqrt2)^2=2$. – Allawonder Feb 03 '18 at 10:49

Confused about square root

2 Answers2