I was solving this equation:

$$x^2+y^2 = 4$$ So I subtracted $x$ squared from both sides $$y^2 = 4 - x^2$$ Now I squared them to isolate $y$ $$\sqrt{y^2} = \sqrt{4 - x^2}$$

Now this is where I got confused, because I originally solved it like this: $$\pm y = xi \pm 2$$ Because when I squared $4$, it is plus or minus $2$, and $-x$ squared gives me an irrational number. Now my teacher corrected me and told me the right answer actually is: $$y = \pm(\sqrt{4 - x^2})$$

I understand what she did, but I just don't think it is simplified enough. So my questions are:

  1. Why is my answer incorrect?

  2. When do you use the plus of minus sign? Why wouldn't you have: $$\pm y = \pm(\sqrt{4 - x^2})$$

  3. When do you use absolute value when using square root?

  • 2
    $\begingroup$ $$\sqrt{a+b}\ne\sqrt{a}+\sqrt{b}$$ $\endgroup$ – Simply Beautiful Art Aug 24 '16 at 21:26
  • $\begingroup$ Amm okay I was probably confusing it with: sqrt{ab} = sqrt{a} + sqrt{b} $\endgroup$ – Pablo Aug 24 '16 at 21:30
  • $\begingroup$ You are not alone en.wikipedia.org/wiki/Freshman%27s_dream (also, I think you mean $\sqrt{ab} = \sqrt{a}\sqrt{b}$) $\endgroup$ – Shai Aug 24 '16 at 21:32

For 1., Simple Art's comment $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$

To respond to 2., technically $\pm y = \pm \sqrt{4-x^{2}}$ is fine, it is just a waste of ink since $$y = \sqrt{4-x^{2}} \Leftrightarrow -y = -\sqrt{4-x^{2}}$$ and $$y = -\sqrt{4-x^{2}} \Leftrightarrow -y = \sqrt{4-x^{2}}$$ Since we are solving for $y$, we ideally want the simplification to begin $y = \ldots$ so the $\pm$ on the left is superfluous

And for 3., it is true that $\sqrt{x^{2}} = |x|$ since $\sqrt{x}$ is always defined to be the positive square root. Note that $x^{2} = y^{2}$ means something subtly different to $x = y$. For the former equation, $x = -2, y = 2$ is a solution, but this is not a solution to the latter. However $x^{2} = y^{2}$ is equivalent to $|x| = |y|$ which is equivalent to $x = \pm y$ or $y = \pm x$

  1. A correct deduction is $\sqrt{y^2} = \sqrt{4-x^2} \implies |y| = \sqrt{4-x^2} \implies y = \pm \sqrt{4-x^2}$

    Note that, for example, $\sqrt{t^2} = 25 \implies |t| = \sqrt{25} \implies t = \pm 5$.

    Warning: very informal explanation.

    You can think it as a composition of functions: $x \stackrel{t^2}{\to} x^2 \stackrel{\sqrt{t}}{\to} \sqrt{x^2} = |x|$. First arrow converts from $\mathbb R$ to positive numbers and powers to 2, and then the square root removes the $t^2$ effect but the sign is kept positive (absolute value).

  2. In general, you should use absolute values always you've got $x^2 = \mathrm{???}$ and you want to isolate $x$.


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