2021 AMC 10B Problem 2

1. Start with a small example to see what $\sqrt{t^2}$ does. With $t = -5$: $\sqrt{(-5)^2} = \sqrt{25} = 5$, not $-5$. A square root is never negative, so $\sqrt{t^2}$ is $t$ with its minus sign removed. In other words, $\sqrt{t^2} = |t|$.
2. So the expression is
$$|3 - 2\sqrt{3}| + |3 + 2\sqrt{3}|.$$
3. Now decide the sign of $3 - 2\sqrt{3}$. Compare the two parts by squaring them: $3^2 = 9$ and $(2\sqrt{3})^2 = 12$. Since $12 > 9$, the number $2\sqrt{3}$ is bigger than $3$ (about $3.46$). So $3 - 2\sqrt{3}$ is negative.
4. Removing its minus sign gives $|3 - 2\sqrt{3}| = 2\sqrt{3} - 3$.
5. The second number $3 + 2\sqrt{3}$ is clearly positive, so it stays as it is.
6. Add:
$$(2\sqrt{3} - 3) + (3 + 2\sqrt{3}) = 4\sqrt{3}.$$

The answer is $\boxed{\textbf{(D)}\ 4\sqrt{3}}$.

The $\sqrt{\ }$ sign always means the positive root, so taking the square root of a square cannot give a negative number. Whenever a problem squares something and then takes the square root, stop and ask whether the inside is negative. Here the whole problem comes down to comparing $2\sqrt{3}$ with $3$.

For any real $t$, $\sqrt{t^2} = |t|$. So the expression equals
$$|3 - 2\sqrt{3}| + |3 + 2\sqrt{3}|.$$

Since $2\sqrt{3} = \sqrt{12} > \sqrt{9} = 3$, the number $3 - 2\sqrt{3}$ is negative, so its absolute value is $2\sqrt{3} - 3$. The second number is clearly positive, so its absolute value is $3 + 2\sqrt{3}$.

Adding: $(2\sqrt{3} - 3) + (3 + 2\sqrt{3}) = 4\sqrt{3}$.

The answer is $\boxed{\textbf{(D)}\ 4\sqrt{3}}$.