Calculate the integer part I have to calculate the  integer part of this:
$$[(\sqrt{2}+\sqrt{5})^2]$$
 I tried to write it like this:
$$[2+5+2\sqrt{10}]=[7+2\sqrt{10}]=7+[2\sqrt{10}]$$
Any ideas?
 A: HINT: Clearly $3<\sqrt{10}<4$, so the integer part of $2\sqrt{10}$ is either $6$ or $7$. How can you tell which it is? And which is it?
A: A direct way of solving this problem is to realize that 
$$ 0 < \sqrt{ 5} - \sqrt{2} < 1 $$
This can be seen by squaring both sides, and using that $ 6 < 2 \sqrt{10} < 7 $ which again is true by squaring.
Since $(\sqrt{5} + \sqrt{2})^2 + (\sqrt{5}- \sqrt{2})^2 = 5 + 2\sqrt{10} + 2 + 5 - 2 \sqrt{10} + 2 = 14 $, hence this allows us to conclude that
$$ \lfloor (\sqrt{5} + \sqrt{2})^2 \rfloor = 13 $$
A: Hint:$$f(x+\Delta x)\approx  f '(x)\Delta x+f(x)\to\sqrt {x+\Delta x}\approx \frac{\Delta x}{2\sqrt x}+\sqrt x$$
$$\text{if } \Delta x=1\to\sqrt10=\sqrt{1+9}\approx\frac{1}{2\sqrt9}+\sqrt9=\frac{1}{6}+3$$
then we have $7+2\sqrt10\approx7+6+\frac{1}{3}=13+\frac13$
A: Find square root of 10 , as $\sqrt{9} = 3 $ , so $\sqrt{10}$ is little greater than $3$  so that should also end in 3 . 
Or calculate square root of $10$ , it is not something impossible . it is $3.16$ 
(Note : use approximations only when finding square root is a tedious time consuming process . Other wise you can find exact answer .Here square root of 10 is doable. Exact answers are preferred compared to approximate answers)
$ 7 + 2 \times 3.16 $
= $ 13.32 $
so integer part of answer is $13$
A: You are on the right track. But as Brian M. Scott said, $\sqrt {10}$ is between $3$ and $4$ , and we want to know to which it is closer. So now we make a guess; let's say I guess it's closer to $3$ . That means that the difference between $3$ and $\sqrt {10}$ is less than the difference between $\sqrt {10}$ and $4$ , or in math language $\sqrt {10} -3<4-\sqrt {10}$ . If this inequality is true, then the guess is right. If it turns out not to be true, then the guess is wrong. So we have
$\sqrt {10} -3<4-\sqrt {10}$
$2\sqrt {10} <7$  Squaring both sides,
$40<49$
Since this is a true statement, our initial inequality is correct so $\sqrt{10}$ is closer to $3$ than to $4$ . 
A: Calvin Lin provided an excellent proof. I am just adding in some details.
Obviously, $\sqrt{ 5} - \sqrt{2} > 0$
Also, $\sqrt{2} > 1$
$2 \sqrt{2} > 2$
$1 + 2 \sqrt{2} + 2 > 5$
$1^2 + 2 \sqrt{2} + (\sqrt{2})^2 > (\sqrt{5})^2$
$(1 + \sqrt{2})^2 > (\sqrt{5})^2$
Taking the positive square root, we have  $1 + \sqrt{2} > \sqrt{5}$
Then, $1 > \sqrt{5} - \sqrt{2}$
Combining the results together, we have $0 < \sqrt{ 5} - \sqrt{2} < 1$
Note that squaring a +ve quantity whose value lies between 0 and 1 is again positive but smaller.
Thus, $0 < (\sqrt{ 5} - \sqrt{2})^2 < \sqrt{ 5} - \sqrt{2} < 1$
Since $(\sqrt{5} + \sqrt{2})^2 + (\sqrt{5}- \sqrt{2})^2 = … = 14$
$(\sqrt{5} + \sqrt{2})^2 = 14 - (\sqrt{5}- \sqrt{2})^2$
Therefore, $\lfloor (\sqrt{5} + \sqrt{2})^2 \rfloor = 13$
