Represent $x_1=\frac{1-\sqrt{37}}{3}$ and $x_2=\frac{1+\sqrt{37}}{3}$ on a number line In which case the numbers $x_1=\dfrac{1-\sqrt{37}}{3}$ and $x_2=\dfrac{1+\sqrt{37}}{3}$ are correctly represented on the number line?

I did the calculations with $\sqrt{36}=6$ and then we have $x_1\approx\dfrac{1-6}{3}=-\dfrac{5}{3}=-1\frac23$. So (Б) and (Г) are wrong answers. Now for $x_2\approx\dfrac{1+6}{3}=\dfrac{7}{3}=2\frac13$. The correct answer is В). I am not very proud of this solution so what else can we do? Thank you in advance!
 A: Symmetries  in the 4 options hint to an alternative way to solve the problem.  Given $x_1=\dfrac{1-a}{3}$ and $x_2=\dfrac{1+a}{3}$ , the center is $\frac {1}{2}(x_1+x_2)=\frac{1}{3}$ independent of $a$,(B) is the only choice.
A: The visual nature of the answer options means there very likely is no more rigorous way to solve this. Figure out which three can't be right, and keep the one remaining option as your answer. You did that very well, and you've completed the problem.
A: You don't need to approximate $\sqrt{37}\approx 6$ or to wonder .... "gee $\sqrt{42}$ is between $6$ and $7$ which should I use?"
Not that $36 < 37 < 49$ so $6 < \sqrt {37} < 7$.
So $-6 > -\sqrt{37} > -7$
$-5 = 1-6 > 1-\sqrt{37} > 1-7 = -6$
So $-\frac 53 >\frac {1-\sqrt{37}}3 > \frac {-6}3 = -2$.
So A) and B) is the best picture for that.
Likewise if the question was for $y_1 = \frac {1 - \sqrt {42}}3$ you'd do similar and not have pick choice over the other.
$36 < 42 < 49$ so.... well... the exact same thing happens.
$-\frac 53 >\frac { 1-\sqrt{42}}3 > -2$ and A) or B) are the best picture.
(BTW $-2 < y_1 < x_1 < -\frac 53$ where $y_1 =\frac {1-\sqrt{42}}3$ and $x_1 = \frac {1-\sqrt {37}}3$ in case you were curious or were ever asked to draw something like that.)
