Show which of $6-2\sqrt{3}$ and $3\sqrt{2}-2$ is greater without using calculator How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)
Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).
 A: We have $\sqrt{3}\leq 1.8$ so $6-2\sqrt{3}\geq 2.4$, whereas $\sqrt{2}\leq 1.42$ so $3\sqrt{2}-2\leq 2.26$.
A: $$
6-2√3 \sim 3√2-2\\
8 \sim 3√2 +2√3 \\
64 \sim 30+12√6\\
34 \sim 12√6\\
17 \sim 6√6\\
289 \sim 36 \cdot 6\\
289 > 216
$$
A: Define
$a=6-2\sqrt 3>0$
$b=3\sqrt 2-2>0$
$a-b = 8 - (2\sqrt 3 + 3\sqrt 2)$
$(2\sqrt 3 + 3\sqrt 2)^2 = 30+12\sqrt 6 = 6×(5+2\sqrt 6)
 < 60 < 64$
because $6=2×3 < (5/2)^2$
$a-b > 8-8=0, a>b$
A: $6-2\sqrt 3 \gtrless 3\sqrt 2-2$  
Rearrange: $8 \gtrless 3\sqrt 2 + 2\sqrt 3$  
Square: $64 \gtrless 30+12\sqrt 6$
Rearrange: $34 \gtrless 12\sqrt 6$
Square: $1156 \gtrless 864$
A: I'll use >=< to represent the unknown comparison.
$ 6-2\sqrt{3} >=< 3\sqrt{2}-2$
Lets start by adding two to both sides to reduce the number of numbers. This doesn't change the comparison result.
$ 8-2\sqrt{3} >=< 3\sqrt{2}$
Both sides are clearly positive ( $ 2\sqrt{3} < 6 $ ) so we can square both sides without changing the comparison result.
In a more maginal case where we were unsure if the left hand side was positive we could have compared the two terms in the left hand side by squaring both of them and hence determined whether the left hand side was positive or negative.
$ 64 -32\sqrt{3} + 12 >=< 18$
Now lets collect terms.
$ 60 >=< 32\sqrt{3}$
Divide by four.
$ 15 >=< 8\sqrt{3}$
Square again.
$ 225 >=< 64 \times 3 $
$ 225 > 192 $
Therefore 
$ 6-2\sqrt{3} > 3\sqrt{2}-2$
A: Here's yet another way, for those who aren't comfortable with the $\gtrless$ or $\sim$ notation.
We can use crude rational approximations to $\sqrt 2$ and $\sqrt 3$.
$$\begin{align}
\left(\frac{3}{2}\right)^2 = \frac{9}{4} & \gt 2\\
\frac{3}{2} & \gt \sqrt 2\\
\frac{9}{2} & \gt 3\sqrt 2
\end{align}$$
And
$$\begin{align}
\left(\frac{7}{4}\right)^2 = \frac{49}{16} & \gt 3\\
\frac{7}{4} & \gt \sqrt 3\\
\frac{7}{2} & \gt 2\sqrt 3
\end{align}$$
Adding those two approximations, we get
$$\begin{align}
\frac{9}{2} + \frac{7}{2} = 8 & \gt 3\sqrt 2 + 2\sqrt 3\\
6 + 2 & \gt 3\sqrt 2 + 2\sqrt 3\\
6 - 2\sqrt 3 & \gt 3\sqrt 2 - 2
\end{align}$$
A: $6-2√3 \approx2(1.732) = 6 - 3.464 = 2.536$
$3√2-2 \approx 3(1.414) - 2 = 4.242 - 2 = 2.242$
This implies that $\dfrac 52$ is between the two quantities:
\begin{align}
   \dfrac{49}{4} &> 12 \\
   \dfrac 72 &> 2\sqrt 3 \\
   \dfrac{12}{2} &> 2\sqrt 3 + \dfrac 52 \\
   6 - 2\sqrt 3 &> \dfrac 52
\end{align}
and
\begin{align}
   \dfrac{81}{4} &> 18 \\
   \dfrac 92 &> 3\sqrt 2 \\
   \dfrac 52 &> 3\sqrt 2 - 2
\end{align}
It follows that $6-2√3 > 3\sqrt 2 - 2$.
