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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?

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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?

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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

$$ \lfloor (\sqrt{5} + \sqrt{2})^2 \rfloor = 13 $$

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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 $7+2\sqrt10\approx7+6+\frac{1}{3}=13+\frac13$

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  • $\begingroup$ Nice using of differential here, Maisam. +1 $\endgroup$ – mrs Aug 5 '13 at 16:30
  • $\begingroup$ @ Babak S.:thanks dear Babak. $\endgroup$ – M.H Aug 5 '13 at 18:25
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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 $13$

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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 $4$ .

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  • $\begingroup$ If $P$ implies a true statement, it doesn't necessarily follows that $P$ is true. Instead, you should say that since from $40<49$ you can conclude $\sqrt {10} -3<4-\sqrt {10}$ (the steps are given by your proof read backwards), $\sqrt {10} -3<4-\sqrt {10}$ is indeed true. $\endgroup$ – Alraxite Aug 5 '13 at 20:44
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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, $\lfloor (\sqrt{5} + \sqrt{2})^2 \rfloor = 13$

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