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# does 2=-2?

I was thinking recently and got stuck on a strange problem:

$$(-2)^{2}=4=(-2)^{0.5\times4}=((-2)^{0.5})^{4}=4=\sqrt{2}^{4}$$

Take the fourth route if both sides:

$$(-2)^{0.5}=\sqrt{2}$$

$$\sqrt{-2}=\sqrt{2}$$

$$-2=2$$

I don't know what the problem is: have I made a mistake? Or is their a way of explaining this that I am not aware of? Please leave a comment if you have any ideas!

Note by Katie Marsden
1 year, 12 months ago

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$$4=x^{4}$$ has $$4$$ solutions for $$x$$. They are like roots for polynomials. These roots are $$\pm\sqrt{2}$$ and $$\pm\sqrt{-2}$$.

btw I edited the Latex of your note. · 1 year, 12 months ago

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Thank! I still won't pretend I undestand how that rule accounts for the other possible answers to that (i.e. Wouldn't that mean there were 4 answers to root-2, one of which being root 2?). But thanks that's really interesting-and thanks for the editing I'm really bad at that kind of thing! · 1 year, 12 months ago

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I'll give a simpler example: $$x^{2} = 4$$ would have 2 solutions for $$x$$, namely $$-2$$ and $$2$$. Even though $$(-2)^{2}=4$$ and $$(2)^{2}=4$$, $$-2\neq 2$$ · 1 year, 12 months ago

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