Comparing numbers of the form $c+\sqrt{b}$ (eg, $3+3\sqrt{3}$ and $4+2\sqrt{5}$) without a calculator

Question

It is easy to compare to numbers of the form $a\sqrt{b}$, simply by comparing their squares, for example $3\sqrt{3}$ and $2\sqrt{5}$.

But what if we have $a=3+3\sqrt{3}$ and $b=4+2\sqrt{5}$ for example?

How to compare them without using a calculator? Is there a method that works for all numbers of the form $c+\sqrt{b}$?

I found a similar question to this on the site, but that one is a bit different and I am looking for an answer suitable for elementary students to understand. (but I would love to know any kind of answer even if it does not fit the elementary level).

I was thinking about subtracting $b-a$ and checking wether it is negative or positive,

I got $b-a=1+2\sqrt{5}-3\sqrt{3}$ but $2\sqrt{5}-3\sqrt{3}<0$

(I don't want to continue from here by approximating $\sqrt{5}=2.something$, because we will be somehow using the calculator in our mind; so I am stuck here)

Answer 1

General algorithm:

Let us consider $d>b$. Assume $$a+\sqrt b>c+\sqrt d$$ Then $$(a-c)>(\sqrt d-\sqrt b)$$ Note that we need to proceed further, only if $a>c$, because, if $c>a$ then our assumption is certainly false. If $a>c$: $$(a-c)^2>b+d-2\sqrt {bd}$$ $$2\sqrt {bd}>b+d-(a-c)^2$$ If $(a-c)^2>b+d$, then this is certainly true, hence our assumption is true. Otherwise, we square once more, and compare. This time it's easy to compare, since all radicals are removed.

Example:

Since $27>20$, we assume:

$$4+\sqrt {20}>3+\sqrt {27}$$

We have, $1>\sqrt {27}-\sqrt {20}$

Since $4>3$, we proceed further by squaring:

$$1>47-2\sqrt {540}$$

$$2\sqrt {540}>46$$

Again, we need to proceed further, as $46>0$. So, $$\sqrt {540}>23$$ Squaring the last time, $$540>529$$ Since this is true, our original assumption was true as well.

Answer 2

I think you can use Maclaurin series
For example, $3+3\sqrt{1+2}\approx3+3\cdot(1+2/2)=9$ and $4+2\sqrt{1+4}\approx4+2\cdot (1+4/2)=10$
Clearly, $10>9$ so $4+2\sqrt{5}>3+3\sqrt{3}$

Answer 3

You can use binomial theorem result

$(1\pm x)^n\approx (1\pm nx)$ provided that $x<<1$

Therefore you have $3+3\sqrt3=3+3(4-1)^{\frac{1}{2}}\approx3+6(1-0.25)^{\frac{1}{2}}= 3+6(1-0.125)=9-0.75=8.25$

Similarly $4+2\sqrt5=4+2(4+1)^{\frac{1}{2}}\approx4+4(1+0.25)^{\frac{1}{2}}= 4+4(1+0.125)=8+0.5=8.5$

This would work better when you have large numbers involved inside radical