How do you solve questions like $2^{1/2}$ and can you explain how this works?
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$\begingroup$ "$2^{1\over 2}$" isn't a question ... $\endgroup$– Noah SchweberJan 14, 2020 at 23:09
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2$\begingroup$ What Noah said, but the bigger nitpick here is that expressions can't be solved, only simplified. $2^{1/2}$ can't be simplified any further, but it can be rewritten as $\sqrt 2$. Is there a specific problem you're working on that you need help with? It's unclear what you're asking. $\endgroup$– user307169Jan 14, 2020 at 23:10
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$\begingroup$ Thank you, I am in grade 6 and learning algebra $\endgroup$– VinodJan 14, 2020 at 23:15
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2 Answers
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Let say you have the general problem of $x^{a\over b}$ you can always rewrite this as ${(\sqrt[b] x)^a}$. Just to be clear the b is the bth root not multiplication.
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$\begingroup$ Thank you so much for answering my question. $\endgroup$– VinodJan 14, 2020 at 23:45
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It is known that
x^a × x^b = x^(a+b)
From that, try to input a = b = 0.5 , we get
(x^0.5)×(x^0.5)=x^(0.5+0.5)=x^1=x
(x^0.5)^2=x
By taking the square root of both sides, we obtain
x^0.5= \sqrt{x}
Note that we only take the positive value for the answer.
