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## myininaya 2 years ago Tutorial: $\sqrt{x^2}$ vs. $(\sqrt{x})^2$

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1. myininaya

Ways to write: $\sqrt{x^2}$ $\sqrt{x^2}=((x)^2)^\frac{1}{2}$ $\sqrt{x^2}=|x|$ $\sqrt{x^2}=x, x>0$ $\sqrt{x^2}=-x, x<0$ $\sqrt{x^2}=0, x=0$ $\sqrt{x^2}=\pm x$ The most confusing way to write it though is the first because you will think that you can just do $((x)^2)^\frac{1}{2}=x^{1}=x$ But by doing this you miss what it can be if x<0 Ways to write: $(\sqrt{x})^2$ $(\sqrt{x})^2=(x^\frac{1}{2})^2$ $(\sqrt{x})^2=\sqrt{x} \cdot \sqrt{x}=x$ Now lets go back to $\sqrt{x^2}$ : …. $\sqrt{x^2}=x, x>0$ Example x=3 $\sqrt{3^2}=3, x>0$ …. $\sqrt{x^2}=-x, x<0$ Example x=-3 $\sqrt{(-3)^2}=-(-3), x<0$ …. So we should understand now that: $\sqrt{x^2}=\pm x$ We should also understand that : $\sqrt{x^2} \neq (\sqrt{x})^2$

2. myininaya

and again the talk about $\sqrt{x} \cdot \sqrt{x} =x$ we are assuming there that x>0

3. ganeshie8

crystal clear :P

4. apoorvk

I had trouble regarding this a couple of years back, and @myininaya has explained it in her impeccable style! Imma FAN of her's ALL OVER AGAIN!!! :O \m/

5. FoolForMath

One interesting question is "why do we choose $$\sqrt{x^2} = |x|$$?" In other terms why do we always want the positive value? One reason is only then we can define define composition in a meaningful way: $\sqrt{\sqrt{x}}=\sqrt[4]{x}$ Now consider that we had chosen $$\sqrt x<0$$, then $$\sqrt{\sqrt x}$$ would be imaginary but $$\sqrt[4]x$$ is still a real number (positive of negative). You may ask why can't we choose both positive and negative value? The answer is sqrt() is a function so it must return a unique value. These are some of the many reason for which we have defined root of even order to be always positive. It is worth mentioning that for $$x\ge 0$$, $$(\sqrt{x})^2=\sqrt{x^2}=|x|$$ PS: Special thanks to M.SE users Asaf Karagila and Arturo Magidin to help me understand this concept.

6. mathslover

great tutorial

7. ParthKohli

Arturo Magidin :'( RIP

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