How do I calculate $\sqrt{ x / \sqrt{x}}$?

Question

$\sqrt{ x / \sqrt{x}}$. Can anyone please help I need to solve it for my homework? Thanks a lot! I've tried multiplying a √(​x)/​√(​x) to the equation but then I'm stuck... EDIT: So is it true that I can simplify it to $x^\frac14$ ?

Answer 1

Hint:

Write $$\sqrt x= x^\frac12$$

and use the indices rules $$\frac{x^m}{x^n}=x^{m-n}\text{ and }(x^m)^n=x^{mn}$$

Answer 2

We have: $$A=\sqrt{\frac{x}{\sqrt{x}}}$$ We know also that: $$\frac{x}{\sqrt{x}}=\sqrt{x}$$ So, substituing, we have: $$A=\sqrt{\frac{x}{\sqrt{x}}}=\sqrt{\sqrt{x}}=\sqrt[4]{x}$$

Answer 3

If we take $x^{\frac 1k} = \sqrt[k]{x}$ as a definition and if we verifi that $x^{a+b} = x^ax^b$ and $(x^a)^b = x^{ab}$ hold not just for integers but for rational numbers then:

$\sqrt{\frac x{\sqrt x}} = (\frac {x^1}{x^{\frac 12}})^{\frac 12} = (x^{1-\frac 12})^{\frac 12} = x^{\frac 12(1-\frac 12)} = x^{\frac 14}$.

Altenativly you can take it in steps.

$\sqrt x$ exists. So $x \ge 0$ and $x = \sqrt x^2$.

So $\frac{x}{\sqrt x}=\frac {\sqrt{x}^2}{\sqrt x} = \sqrt x$.

So $\sqrt{\frac{x}{\sqrt x}} = \sqrt{\sqrt x}$.

Now by definition $\sqrt{\sqrt x}^2 = \sqrt x$ and so $(\sqrt{\sqrt x})^4 = (\sqrt{\sqrt x})^2(\sqrt{\sqrt x})^2 = \sqrt x\sqrt x = x$. So $\sqrt{\sqrt x} =\sqrt[4]x$.

or ----

$\sqrt{\frac {x}{\sqrt x} }= $

$\sqrt{\frac {x}{\sqrt x}}\cdot 1=$

$\sqrt{\frac {x}{\sqrt x}}\cdot \sqrt 1=$

$\sqrt{\frac {x}{\sqrt x}}\cdot \sqrt \frac {\sqrt x}{\sqrt x}=$

$\sqrt{\frac {x\sqrt x}{\sqrt x\sqrt x}}=$

$\sqrt{\frac {x\sqrt x}{x}}=$

$\sqrt{\sqrt x} = \sqrt[4] x$

...... Or .....

$m = \sqrt{\frac x{\sqrt x}}$

$m^2 = \frac x{\sqrt x}$

$m^4 = \frac {x^2}{x} = x$.

So $m$ is one of the fourth roots of $x$. As square roots are non-negative, $m$ is the principal non-negative fourth root of $x$.