find out the fractional part for $\sqrt{25} + \sqrt{24}$

Question

I would like to know if there is any more elegant solution than the extraction of the root or the approximation.

I tried something like this:

$\sqrt{25}$ + $\sqrt{24}$ = $( \sqrt3 + \sqrt2 )^2$

But how can I get the fractional part from here?

Answer 1

This response is a continuation of a comment that I made to the OP's query. Although this response can in no way be construed as an answer, I feel that the response is worthwhile. I also think that providing this response as a series of comments (instead) would have made this response much less legible.

To the OP:
If (after 24-48 hours) you find that no one has focused on the question that you are trying to ask then (regardless of whether the interpretation in my comment was accurate), I suggest that you do the following:

(1)
As another comment suggested, review How to ask a good question.

(2)
Create a 2nd mathSE query. In this 2nd query, provide a link to this query, and explain that you are re-posting your question as a new query, because the old query was unclear. Also, in this (original) query, add an Addendum that explains that you now realize that this query has caused confusion, and that you have consequently re-posted your question. In this Addendum, provide a link to the new mathSE query.

---

Let $\{x\}$ denote the fractional part of $x$.

(3)
Your new query should begin with a very precise goal.

One possible goal might be:
given any positive integers $n$ and $m$,
how does one simplify the computation of $\{\sqrt{n} + \sqrt{m}\}$?

If this is what you are attempting to accomplish, then you are going to have to make a serious effort to describe (perhaps) with examples exactly what you mean by the word simplify.

If this is not your actual goal, then (as I say) you must clearly indicate the mathematical problem (i.e. goal) that you are trying to solve.

Please try to make it impossible for anyone in your audience to be confused as to what you are asking.

---

The remainder of my response examines the confusion that I experienced when reading your query.

(A)
Where did the LHS of :
$[E_1]: ~\{\sqrt{25} + \sqrt{24}\} ~=~ \{~(\sqrt{3} + \sqrt{2})^2~\}$
originate?

Was the computation of $\{\sqrt{25} + \sqrt{24}\}$ an assigned problem from a book or a class, or a problem that you made up on your own?

(B)
How did you calculate the RHS of $[E_1].$
Although $[E_1]$ is accurate, that doesn't explain your thinking in associating the positive integers $3$ and $2$ with the positive integers $24$ and $25$.

Did you start with the positive integers $25$ and $24$, and employ some mathematical algorithm to compute the positive integers $3$ and $2$? If so, what algorithm?

Alternatively, did you simply presume that $\{\sqrt{25}\} = 0,$ notice that $\{\sqrt{24}\} = \{2\sqrt{6}\}$, and choose the positive integers $2$, and $3$ because you happened to notice that $2\sqrt{6} = 2\left(\sqrt{2}\sqrt{3}\right)$?

(C)
Your query finishes with the question:
"But how can I get the fractional part from here?"

$[Q_1]:~$ What type of expression are you ultimately looking for?

In your new query, this question, can be associated with your stating a clear mathematical goal. Further, if I were you, I would attack the confusion around $[Q_1]$ by providing two to three examples that illustrate exactly what you mean by the phrase:
get the fractional part.

Are you looking for a way of expressing tight lower and upper bounds on the fractional part?

Are you looking for a mathematical formula that expresses the fractional part exactly?
I am guessing that if this is your goal, then, given $x$,
you are looking for a mathematical expression for $y$,
where $0 \leq y < 1,$ and $y = \{x\}.$

If this is your goal, you should state this explicitly, and the examples that you provide should illustrate your (at least attempting to) compute $y$, given $x$.

I was very confused here.

Answer 2

$$\{5+\sqrt{24}\}=\sqrt{24}-4.$$ Is it the question?

Answer 3

$$x=\sqrt{25} + \sqrt{24},\\ [x]=[\sqrt{25} + \sqrt{24}]=5+4=9\\ \{x\}=x-[x]=5+\sqrt{24}-9$$

Answer 4

As described in other answers the number required is $\sqrt{24}-4$.

If we set $x+4 = \sqrt{24}$, squaring and rearranging gives $x^2+8x-8 = 0$.

This can be solved using numerical methods to your required level of accuracy.

The Newton-Raphson method gives the recursion $x_{n+1} = x_n - \dfrac{x_n^2-8x_n+8}{2x_n+8}$ or $x_{n+1} = \dfrac{x_n^2+8}{2x_n+8}$. Starting with $x_0=1$ we get $x_1 = \dfrac{9}{10}$, $x_2 = \dfrac{881}{980}$ and so on.

Note $\left (\dfrac{881}{980}+4 \right )^2 = 24.00000104 \dots$