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.

  1. Rewording the problem and including a constraint $$\frac{\sqrt{x+4}(2x-x^2)-\frac12(\sqrt{x+4})}{x+4}, x>-4$$
  2. Creating an equality, multiplying both sides with the denominator, then subtracting $(2x-x^2)$ to the other side to square both sides $$(\sqrt{x+4}-\frac12\sqrt{x+4})^2=(-(2x-x^2))^2$$
  3. 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"

  • $\begingroup$ What is the equation you are trying to solve? $\endgroup$ – Robert Z Feb 20 '19 at 7:55
  • $\begingroup$ @Robert Z, I included those parentheses because all terms are being multiplied together. I wanted to avoid writing $\times$ to indicate multiplication, so I will be including those parentheses again $\endgroup$ – Lex_i Feb 20 '19 at 7:56
  • $\begingroup$ @RobertZ I'm trying to solve the equation that is quoted $\endgroup$ – Lex_i Feb 20 '19 at 7:58
  • $\begingroup$ In the statement you wrote a function not an equation. No equal sign!! $\endgroup$ – Robert Z Feb 20 '19 at 7:59
  • $\begingroup$ BTW the rewording in 1) is not equivalent to the statement... $\endgroup$ – Robert Z Feb 20 '19 at 8:00

$$\sqrt{x+4}\cdot\sqrt{x+4}=x+4,$$ 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:


  • $\begingroup$ Based on your answer, does this mean $x=2$? Because I went for another attempt and got $-\frac{(x+4)\frac12(2x-x^2)}{x+4}$ which led me to $x(\frac12x-1)=0, x=2$ $\endgroup$ – Lex_i Feb 20 '19 at 8:29
  • $\begingroup$ @Lex_i See please better. Your expression it's exactly that I wrote in the first versa. $\endgroup$ – Michael Rozenberg Feb 20 '19 at 8:33
  • $\begingroup$ Okay good, thanks $\endgroup$ – Lex_i Feb 20 '19 at 8:34
  • $\begingroup$ @Lex_i You are welcome! $\endgroup$ – Michael Rozenberg Feb 20 '19 at 8:35

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