# 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, 2019 at 21:43
• For the same reason that $\sqrt{3^2+4^2}=5$ is not the same as $\sqrt{3^2}+\sqrt{4^2}=7$ Apr 17, 2019 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! Apr 17, 2019 at 21:53
• @Ethan makes a good point. Another point to consider is that, in general, $$\left(x^2\right)^\frac12=|x|.$$ Apr 17, 2019 at 23:10

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.

Taking a power of something isn't a linear operation

• +1, absolutely correct, but I am afraid will be totally lost on the OP... Apr 17, 2019 at 21:45

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. Apr 17, 2019 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). Apr 17, 2019 at 22:10
• Then they should be taught that $(a+b)^2=(a+b)(a+b)$ and how to expand that out. Apr 17, 2019 at 22:25
• @RhysHughes "Multiply that out" using the distributive law, not "factor that out". Apr 17, 2019 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 Apr 17, 2019 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.