# How can I simplify a radical in the numerator?

I have the following problem to simplify and I am lost on how to proceed:

$$\frac{\sqrt{1-x}+3}{2}$$

I'm aware that I can rewrite $$\sqrt{1-x}$$ as $$(1-x)^\frac{1}{2}$$ but then I don't know where to go from there.

The solution provided is $$\sqrt{\frac{1}{x}+2}$$.

How can I arrive at this solution? More granular baby steps very much appreciated.

• It's not true in general. Take $x=1$ for example Nov 19 '19 at 16:15
• and even worse as $x \to 0+$ Nov 19 '19 at 16:17
• To me, nothing could be simpler than the original expression. Nov 19 '19 at 16:17
• I tried to answer the wrong question in my book. My bad. The solution is the incorrect one - sorry. But, out of curiosity, how would one simplify a radical in the numerator? Nov 19 '19 at 16:20
• @DougFir: When you say "radical in the denominator" (though your title says numerator) do you mean like $\dfrac 1{\sqrt2}$ or $\dfrac 1{\sqrt2 -1}$ or something else? Nov 19 '19 at 16:21

$$\begin{array}{c|cc} x & \frac{\sqrt{1-x}+3}{2} & \sqrt{\frac{1}{x} + 2} \\ \hline 2 & \text{undefined} & \sqrt{\frac{5}{2}} \\ 1 & \frac{3}{2} & \sqrt{3} \\ 0 & 2 & \text{undefined} \end{array}$$
You can write $$\frac{\sqrt{1-x}}{2}+\frac{3}{2}$$