Square root of floor of square root of sum $\sqrt{\lfloor\sqrt{\left|x+y\right|}\rfloor}$

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Let $x,y$ $\in\mathbb{R}$ and $\lfloor x\rfloor = 10$ and $\lfloor y\rfloor = 14$.

Need to compute: $\sqrt{\lfloor\sqrt{\left|x+y\right|}\rfloor}= \; ?$

MY IDEA:

$$10\le x < 11$$

$$14\le y < 15$$

So $$24\le x+y < 26.$$

But then i don't know what to say about:

$$\lfloor\sqrt{|x+y|}\rfloor = \ldots$$

Because i don't get what is the use here for the absolute part: what his role here?

And how can i proceed?

Thanks.

2

There are 2 best solutions below

1
On BEST ANSWER

What you did so far is good. Now justify these steps: \begin{align*} 24 &\le x + y < 26 \\ 24 &\le |x + y| < 26 \\ \sqrt{24} &\le \sqrt{|x + y|} < \sqrt{26} \\ 4 &\le \left\lfloor{\sqrt{|x + y|}} \right\rfloor \le 5 \\ \sqrt{\left\lfloor{\sqrt{|x + y|}} \right\rfloor} &= 2 \text{ or } \sqrt{5} \end{align*} Indeed, as Ross Millikan states, there are two possible answers. If you expected a single answer, you must not have written the question exactly right.

21
On

Since you know $x+y > 0$, you know $|x+y|=x+y$ and you can remove the absolute value bars. They were just there to confuse you. You won't get a single answer.