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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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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

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

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)[reply]

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)[reply]

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)[reply]