# Why isn't $(2x+x^2)^{1/2}$ the same as $(2x)^{1/2}+x$? [duplicate]

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

• Why should they be the same? To convince yourself that they are different, try particular values of $x$, like $x=1$. – lulu Apr 17 at 21:43
• For the same reason that $\sqrt{3^2+4^2}=5$ is not the same as $\sqrt{3^2}+\sqrt{4^2}=7$ – Henry Apr 17 at 21:43
• 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! – PierreCarre Apr 17 at 21:53
• @Ethan makes a good point. Another point to consider is that, in general, $$\left(x^2\right)^\frac12=|x|.$$ – Cameron Buie Apr 17 at 23:10

Taking a power of something isn't a linear operation

• +1, absolutely correct, but I am afraid will be totally lost on the OP... – gt6989b Apr 17 at 21:45

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.

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

• law of universal linearity is a great phrase ... I'm stealing that. – Don Thousand Apr 17 at 23:20

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...

• 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). – Clayton Apr 17 at 22:10
• Then they should be taught that $(a+b)^2=(a+b)(a+b)$ and how to expand that out. – Rhys Hughes Apr 17 at 22:25
• @RhysHughes "Multiply that out" using the distributive law, not "factor that out". – Ethan Bolker Apr 17 at 22:26
• I corrected it to "expand" before your comment. Factor is the inverse and I realised this just after I wrote the comment – Rhys Hughes Apr 17 at 22:27

Maybe a picture convinces you. 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.

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 $$|a+b|$$.

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.