Wikipedia:Reference desk/Archives/Mathematics/2012 July 28

= July 28 =

Very important question
If $$x=7+\sqrt{40}$$, then, find $$\sqrt{x}+1/\sqrt{x}$$. 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, $$\sqrt{7+2\sqrt{10}}$$ 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 $$\sqrt{7+\sqrt{40}} = \sqrt{a} + \sqrt{b}$$. The first step towards doing this is to square both sides giving you $$7+\sqrt{40} = a + b + \sqrt{4ab}$$. Can you find a and b such that a + b = 7 and 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 $$\sqrt x = \sqrt a + \sqrt b$$
 * then
 * $$\sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{a} + \sqrt{b} + \frac{1}{\sqrt{a} + \sqrt{b}}$$
 * $$ = \sqrt{a} + \sqrt{b} + \frac{ \sqrt{a} - \sqrt{b} }{a - b}$$
 * $$ = \frac{(a - b + 1)\sqrt{a} + (a - b - 1)\sqrt{b}}{a - b}$$ . 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)

Tests
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)