Wikipedia:Reference desk/Archives/Mathematics/2012 July 28#Very important question

If $x=7+\backslash sqrt\{40\}$, then, find $\backslash sqrt\{x\}+1/\backslash sqrt\{x\}$. I found this question in a competition book. I solved almost every question of that book but I could not solve this one. As I don't take any tuition class, will you, please, solve the given question. Sunny Singh (DAV) (talk) 15:40, 28 July 2012 (UTC)

:What have you tried so far? As a hint, $\backslash sqrt\{7+2\backslash sqrt\{10\}\}$ is of a form that denests quite nicely. --Kinu ^t/_c 16:09, 28 July 2012 (UTC)

:To denest, look for a and b such that $\backslash sqrt\{7+\backslash sqrt\{40\}\}\; =\; \backslash sqrt\{a\}\; +\; \backslash sqrt\{b\}$. The first step towards doing this is to square both sides giving you $7+\backslash sqrt\{40\}\; =\; a\; +\; b\; +\; \backslash sqrt\{4ab\}$. Can you find a and b such that {{nowrap|1=a + b = 7}} and {{nowrap|1=4ab = 40}}? — Fly by Night (talk) 22:15, 28 July 2012 (UTC)

:The obvious solution is just to use a calculator to find that x = 13.32455532, and plug that in to find your answer, but I assume they want you to solve this in a more convoluted manner. BTW, please try to use more descriptive titles. In this case, "Square root problem" might work. StuRat (talk) 22:48, 28 July 2012 (UTC)

::I think the question wants you to simplify the radicals rather than evaluate them as decimals, using the suggestions above if $\backslash sqrt\; x\; =\; \backslash sqrt\; a\; +\; \backslash sqrt\; b$

::then

::$\backslash sqrt\{x\}\; +\; \backslash frac\{1\}\{\backslash sqrt\{x\}\}\; =\; \backslash sqrt\{a\}\; +\; \backslash sqrt\{b\}\; +\; \backslash frac\{1\}\{\backslash sqrt\{a\}\; +\; \backslash sqrt\{b\}\}$

::$=\; \backslash sqrt\{a\}\; +\; \backslash sqrt\{b\}\; +\; \backslash frac\{\; \backslash sqrt\{a\}\; -\; \backslash sqrt\{b\}\; \}\{a\; -\; b\}$

::$=\; \backslash frac\{(a\; -\; b\; +\; 1)\backslash sqrt\{a\}\; +\; (a\; -\; b\; -\; 1)\backslash sqrt\{b\}\}\{a\; -\; b\}$ . 83.100.173.200 (talk) 09:44, 29 July 2012 (UTC)

::: I'm skeptical of your last step there. —Tamfang (talk) 10:05, 29 July 2012 (UTC)

::::Thank you, I've hopefully corrected it! 83.100.173.200 (talk) 10:23, 29 July 2012 (UTC)

::::: Better! —Tamfang (talk) 05:06, 1 August 2012 (UTC)

There is a name for tests that require an absolute level of ability and you are not judged against others, such as a driving test, and tests where your result if indicative of your ability relative to others taking the test, such as A-levels, but I cannot remember what the names are. Can someone help? Thanks. 92.14.213.234 (talk) 17:04, 28 July 2012 (UTC)

:Tests where you are measured relative to others are "graded on a curve". However, the grade before the curve is also your absolute competence level.

:One problem with just measuring absolute competence level is that it's highly dependent on the test. Thus, a poorly worded test or one which tests skills not needed in real life ("What was the name of Benjamin Harrison's childhood goldfish ?") does not produce useful results.

:A problem with grading on a curve results from too small of a sample size. Thus, a student measured against one class may rank better than if the same student is ranked against another class. The solution, of course, is to compare with a larger pool. StuRat (talk) 22:35, 28 July 2012 (UTC)

: I assume that the distinction is between norm-referenced assessment and criterion-referenced assessment.→86.139.64.77 (talk) 17:01, 29 July 2012 (UTC)