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I just don't get why this isn't true.

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    $\begingroup$ Why should they be the same? To convince yourself that they are different, try particular values of $x$, like $x=1$. $\endgroup$ – lulu Apr 17 at 21:43
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    $\begingroup$ For the same reason that $\sqrt{3^2+4^2}=5$ is not the same as $\sqrt{3^2}+\sqrt{4^2}=7$ $\endgroup$ – Henry Apr 17 at 21:43
  • $\begingroup$ Why don't you try to do the opposite? Solve the equation $\sqrt{2x+x^2} = \sqrt{2x} + x$ and determine for which values of $x$ does the equation hold... You will see that $x=0$ is the only solution! $\endgroup$ – PierreCarre Apr 17 at 21:53
  • $\begingroup$ @Ethan makes a good point. Another point to consider is that, in general, $$\left(x^2\right)^\frac12=|x|.$$ $\endgroup$ – Cameron Buie Apr 17 at 23:10
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Taking a power of something isn't a linear operation

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    $\begingroup$ +1, absolutely correct, but I am afraid will be totally lost on the OP... $\endgroup$ – gt6989b Apr 17 at 21:45
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let $y=(2x+x^2)^{0.5}$. And $z=(2x)^{0.5} +x$. Then $y^2=2x+x^2$ and $z^2 =((2x)^{0.5}+x)^{2}$. Then it should be clear that $z$ can not be equal to $y$ since the well defined operations do not match. Therefore, you can not define it that way.

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Why should it be true. other than wishful thinking? When you first learn algebra you remember the distributive law for "multiplication by a fixed constant $a$", namely $$ a(x + y) = ax +ay . $$ That is true because you can prove it, first for integers and then for any real numbers.

But, sadly, $$ (x + y)^2 \ne x^2 + y ^2 \\ $$ $$ \sqrt{x + y} \ne \sqrt{x} + \sqrt{y} $$ $$ a^{x + y} \ne a^{x} + a^{y} $$ $$ \log(x + y) \ne \log{x} + \log{y} $$ $$ \sin(x + y) \ne \sin{x} + \sin{y} $$

See

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  • $\begingroup$ law of universal linearity is a great phrase ... I'm stealing that. $\endgroup$ – Don Thousand Apr 17 at 23:20
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You have $$ \sqrt{2x + x^2} = \sqrt{x(2+x)} = \sqrt{x} \sqrt{2+x} $$ but in general, $$ \sqrt{a+b} \ne \sqrt{a} + \sqrt{b}. $$ Suppose this "equality" indeed held, then squaring both sides yields $$ a+b = a + b + 2\sqrt{ab} \iff \sqrt{ab} = 0 $$ So you can see either $a=0$ or $b=0$ are needed for your "equality" to hold...

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  • $\begingroup$ I used to give this kind of argument, but what if the student mistakenly thinks $(a+b)^2=a^2+b^2$? Then your argument just shows $a+b=a+b$ and the student will think they are correct... (Of course, some assumptions are being made on the parts of $a$ and $b$, but when thinking like a student, I don't find it unwarranted). $\endgroup$ – Clayton Apr 17 at 22:10
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    $\begingroup$ Then they should be taught that $(a+b)^2=(a+b)(a+b)$ and how to expand that out. $\endgroup$ – Rhys Hughes Apr 17 at 22:25
  • $\begingroup$ @RhysHughes "Multiply that out" using the distributive law, not "factor that out". $\endgroup$ – Ethan Bolker Apr 17 at 22:26
  • $\begingroup$ I corrected it to "expand" before your comment. Factor is the inverse and I realised this just after I wrote the comment $\endgroup$ – Rhys Hughes Apr 17 at 22:27
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Maybe a picture convinces you.

enter image description here

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Square both, you see:

$$(\sqrt{2x+x^2})^2\equiv(\sqrt{2x}+x)^2$$

$$2x+x^2\equiv(\sqrt{2x}+x)(\sqrt{2x}+x)\equiv2x+x^2+2x\sqrt{2x}$$ $$\implies 0\equiv2x\sqrt 2x$$ which is clearly false.

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Hint: Test the equation with $x=1$. You will see that your equations give different results. The square root is the inverse of the square operator. It is clear that $(a+b)^2\neq a^2 + b^2$ hence the inversion $\sqrt{a^2+b^2}$ is in general not equal to $|a+b|$.

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This is (a version of) the "freshman's binomial". While it is true in characteristic $p$ that $(x+y)^p=x^p+y^p$, this is false in general.

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