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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 ... $\endgroup$ Jan 14, 2020 at 23:09
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    $\begingroup$ 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. $\endgroup$
    – user307169
    Jan 14, 2020 at 23:10
  • $\begingroup$ Thank you, I am in grade 6 and learning algebra $\endgroup$
    – Vinod
    Jan 14, 2020 at 23:15

2 Answers 2

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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. $\endgroup$
    – Vinod
    Jan 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.

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  • $\begingroup$ This is a good answer and i kind of understand. $\endgroup$
    – Vinod
    Jan 14, 2020 at 23:32
  • $\begingroup$ I am really glad to hear that. Sorry if i didnt manage to make it looks good. $\endgroup$
    – sentheta
    Jan 15, 2020 at 10:58

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