I have the following problem
$$f(x)=\frac{(x+4)^{\frac12}(2x-x^2)(-\frac12(x+4)-^\frac{1}{2})}{x+4}, x>-4$$
Here's how far I got. I actually went a little farther, but after a couple extra steps I wasn't going anywhere.
- Rewording the problem and including a constraint $$\frac{\sqrt{x+4}(2x-x^2)-\frac12(\sqrt{x+4})}{x+4}, x>-4$$
- Creating an equality, multiplying both sides with the denominator, then subtracting $(2x-x^2)$ to the other side to square both sides $$(\sqrt{x+4}-\frac12\sqrt{x+4})^2=(-(2x-x^2))^2$$
- Ending up with this, unable to continue $$\frac52x+10=(5x^3+4x^2-\frac32x+6)^2$$
I'm guessing I probably made some mistake from the start, and I'm guessing I need to work on simplifying roots and exponents better, like, instead of writing $x^2$ I write $xx$, I know things like that makes factoring, for example, easier. I'm still working on that.
Anyway, I'd appreciate some help here.
UPDATE: My mistake, I overlooked the directions. Apparently, I am to "write the expression as a single quotient in which only positive exponents and/or radicals appear"
$$\sqrt{x+4}\cdot\sqrt{x+4}=x+4,$$ which for your previous problem gives $$f(x)=\frac{\sqrt{x+4}(2x-x^2)\left(-\frac{1}{2}\sqrt{x+4}\right)}{x+4}=\frac{x^2-2x}{2}$$ and for your new problem gives:
$$f(x)=\frac{\sqrt{x+4}(2x-x^2)\left(-\frac{1}{2}\cdot\frac{1}{\sqrt{x+4}}\right)}{x+4}=\frac{x^2-2x}{2(x+4)}.$$