-2
$\begingroup$

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?

$\endgroup$
  • 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
1
$\begingroup$

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 $y = \pm x$

$\endgroup$
0
$\begingroup$
  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 $x$.

$\endgroup$

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.