0

How do you solve questions like $2^{1/2}$ and can you explain how this works?

Vinod
  • 105
  • 3
  • "$2^{1\over 2}$" isn't a question ... – Noah Schweber Jan 14 '20 at 23:09
  • 2
    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. –  Jan 14 '20 at 23:10
  • Thank you, I am in grade 6 and learning algebra – Vinod Jan 14 '20 at 23:15

2 Answers2

2

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.

0

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.

sentheta
  • 581
  • 3
  • 13