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How can a square root function containing a variable and a constant be reexpressed as a square root function containing only a variable?

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    Why could it? ${}{}$ – 2017-01-11
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    This cannot be done. For an approximate solution, however, note that $x+1\approx x+1+\frac1{4x}=(\sqrt x+\frac1{2\sqrt x})^2$, which means $\sqrt{x+1}\approx\sqrt x+\frac1{2\sqrt x}$. – 2017-01-11

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It cannot, since $\sqrt{x+b}$ is defined for $x=-b$, but $a\sqrt x+c$ is undefined at $x=-b$.

Now, you can approximate as follows:

$$\sqrt{x+b}\approx\sqrt x+\frac b{2\sqrt x}$$

which is the two term binomial expansion for large $x$. For small $x$, to avoid division by $0$, one should invert it:

$$\sqrt{x+b}\approx\sqrt b+\frac x{2\sqrt b}$$

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    For your approximation: it may be worth mentioning it is only valid for $x\to \infty$ (when $x\to0$, it'd be $\sqrt{b}+\frac{x}{2\sqrt{b}}$) – 2017-01-11
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    @ClementC. Good pointers. – 2017-01-11
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    What about if $b$ is negative ? – 2017-01-11
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    @openspace Then the first expansion is recommended. – 2017-01-11
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    A closer $ $approximation could be derived from:\begin{align}\sqrt{x+1}&\approx\sqrt{x+1-\frac1{1+8x+16x^2}}\\&=\sqrt x+\frac{2\sqrt x}{1+4x}\end{align} – 2017-01-12
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    @AkivaWeinberger Ah goodness, you always strive to some strange extreme. 'Tis better perhaps, but probably not necessary. – 2017-01-12
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    @SimpleArt Not necessary, indeed… but one can also note that closer approximations can be obtained by truncating $\sqrt{x+1}=\sqrt x+\cfrac1{2\sqrt x+\cfrac1{2\sqrt x+\dotsb}}$ at further places. – 2017-01-12