-1
$\begingroup$

$$ \sqrt[]{(\sqrt{7-2\sqrt[]{10}} + \sqrt[]{2})\cdot 2\sqrt[]{5}} $$

I know that the answer is $\sqrt[]{10}$, but how do I calculate it mathematically if I don't have access to a calculator?

$\endgroup$

closed as off-topic by Travis, Watson, user99914, user228113, Daniel W. Farlow Mar 30 '16 at 14:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, Watson, Community, Community, Daniel W. Farlow
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What are you trying to do exactly? This can be algebraically manipulated but it's debatable whether or not any other form you'll get is any better than the current one. $\endgroup$ – tilper Mar 30 '16 at 12:41
  • 2
    $\begingroup$ Next time show efforts $\endgroup$ – Archis Welankar Mar 30 '16 at 12:42
7
$\begingroup$

first note $\sqrt{7-2\sqrt{10}}=\sqrt{5+2-2\sqrt5\sqrt2}=\sqrt{(\sqrt5-\sqrt2)^2}=\sqrt5-\sqrt2$

hence

$$\sqrt[]{\big(\sqrt{7-2\sqrt[]{10}} + \sqrt[]{2}\big)\times2\sqrt[]{5}}=\sqrt5\sqrt2$$

$\endgroup$
3
$\begingroup$

Hint:

$$\begin{align} \sqrt{7-2\sqrt{10}} & =\sqrt{2+5-2\sqrt{10}} \\ &=\sqrt{(\sqrt{5}-\sqrt{2})^2} \\ \therefore \sqrt{\sqrt{5}\times 2\sqrt{5}} &=\sqrt{10} \end{align}$$

:)

$\endgroup$
  • $\begingroup$ The $\LaTeX$ command for a "centered" dot (multiplication) is \cdot. $\endgroup$ – hardmath Mar 30 '16 at 12:55
  • $\begingroup$ That has been answered above $\endgroup$ – Theodoros Mpalis Mar 30 '16 at 13:05
  • $\begingroup$ See the time difference i was typing while the answer was given $\endgroup$ – Archis Welankar Mar 30 '16 at 13:06
  • $\begingroup$ @hardmath we only use \cdot if we have a multi-variable equation. For example, we won’t write $2a\times b$ because $\times$ looks like another letter, namely $x$. Instead, we write $2ab$ or $2a\cdot b$ but the former is most common, unless $a$ and/or $b$ are represented as two seperate quotients. If we had something like $3\times 4$ then this would be ok, but the expression $3\cdot 4$ could come across as a decimal in certain countries which would arouse confusion $\endgroup$ – Mr Pie Dec 15 '17 at 6:51

Not the answer you're looking for? Browse other questions tagged or ask your own question.