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myininayaBest ResponseYou've already chosen the best response.17
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 : \[\sqrt{x^2} \neq (\sqrt{x})^2\]
 one year ago

myininayaBest ResponseYou've already chosen the best response.17
and again the talk about \[\sqrt{x} \cdot \sqrt{x} =x \] we are assuming there that x>0
 one year ago

apoorvkBest ResponseYou've already chosen the best response.0
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/
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
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.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Arturo Magidin :'( RIP
 10 months ago
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