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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.

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    $\begingroup$ It's not true in general. Take $x=1$ for example $\endgroup$ – J. W. Tanner Nov 19 at 16:15
  • $\begingroup$ and even worse as $x \to 0+$ $\endgroup$ – Henry Nov 19 at 16:17
  • $\begingroup$ To me, nothing could be simpler than the original expression. $\endgroup$ – Allawonder Nov 19 at 16:17
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    $\begingroup$ 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? $\endgroup$ – Doug Fir Nov 19 at 16:20
  • $\begingroup$ @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? $\endgroup$ – J. W. Tanner Nov 19 at 16:21
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These two expressions are not equal.

\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}

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

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