1)Simplify for x>0
sqrt(x^4-4x^2)/x
i have tried
sqrt((x+2)(x-2))
sqrt x^2-4
but they are incorrect, can anyone help me?
2) rationalize the denominator and simplify
x-4/sqrt(x)-4
Hi lemontea.
When factoring, you left out an $\displaystyle x^2$ (Edit: No you didn't; you just canceled it--I'm too tired to read properly). It should be:
$\displaystyle \frac{\sqrt{x^4 - 4x^2}}x$
$\displaystyle =\frac{\sqrt{x^2\left(x^2 - 4\right)}}x$
Now, remember that $\displaystyle \sqrt{ab} = \sqrt{a}\sqrt{b}$, and that $\displaystyle \sqrt{a^2} = |a|$, and you should be able to simplify this.
I am assuming you mean $\displaystyle \frac{x - 4}{\sqrt{x} - 4}$ and not $\displaystyle x - \frac4{\sqrt{x}} - 4$ as it is written. If so, please be more careful about your notation in the future.
Anyway, when rationalizing a binomial denominator, multiply the numerator and denominator of the fraction by the conjugate binomial. For example:
$\displaystyle \frac{a}{\sqrt{2} + \sqrt{3}}$
$\displaystyle =\frac{a\color{red}\,\left(\sqrt{2} - \sqrt{3}\right)}{\left(\sqrt{2} + \sqrt{3}\right)\color{red}\left(\sqrt{2} - \sqrt{3}\right)}$
$\displaystyle =\frac{a\,\sqrt{2} - a\,\sqrt{3}}{\left(\sqrt{2}\right)^2 + \sqrt{2}\,\sqrt{3} - \sqrt{2}\,\sqrt{3} - \left(\sqrt{3}\right)^2}$
$\displaystyle =\frac{a\,\sqrt{2} - a\,\sqrt{3}}{2 - 3}$
$\displaystyle =\frac{a\,\sqrt{2} - a\,\sqrt{3}}{-1}$
$\displaystyle =a\,\sqrt{3} - a\,\sqrt{2}$
I misread your first post. The answers you got seem correct to me. Have you tried separating the radical into $\displaystyle \sqrt{x - 2}\sqrt{x + 2}$? I don't see what else you could do here.
Have you tried multiplying out the numerator? It sounds like you are trying to submit these to an online system, and it is probably very finicky about how your answers are formatted.