Difference between revisions of "1961 AHSME Problems/Problem 13"
Rockmanex3 (talk | contribs) (Solution to Problem 13) |
Rockmanex3 (talk | contribs) m (→Solution) |
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<cmath>\sqrt{t^2 \cdot (t^2 + 1)}</cmath> | <cmath>\sqrt{t^2 \cdot (t^2 + 1)}</cmath> | ||
<cmath>\sqrt{t^2} \cdot \sqrt{t^2 + 1}</cmath> | <cmath>\sqrt{t^2} \cdot \sqrt{t^2 + 1}</cmath> | ||
| − | Remember that <math>\sqrt{t^2} = |t|</math> because square rooting a real number will always result in a [[nonnegative number]]. | + | Remember that <math>\sqrt{t^2} = |t|</math> because square rooting a nonnegative real number will always result in a [[nonnegative number]]. |
<cmath>|t|\sqrt{t^2 + 1}</cmath> | <cmath>|t|\sqrt{t^2 + 1}</cmath> | ||
The answer is <math>\boxed{\textbf{(E)}}</math>. | The answer is <math>\boxed{\textbf{(E)}}</math>. | ||
Latest revision as of 08:53, 31 May 2018
Problem
The symbol 
means 
is a positive number or zero, and 
if is a negative number.
For all real values of 
the expression 
is equal to?
Solution
Factor out the 
inside the square root.
![\[\sqrt{t^2 \cdot (t^2 + 1)}\]](http://latex.artofproblemsolving.com/a/3/a/a3a2805faef09f91e4976bfeea21f5e7d85c01da.png)
![\[\sqrt{t^2} \cdot \sqrt{t^2 + 1}\]](http://latex.artofproblemsolving.com/b/3/9/b39b63892d6f4c522c2304f21c54c56315b779b7.png)
Remember that 
because square rooting a nonnegative real number will always result in a nonnegative number.
![\[|t|\sqrt{t^2 + 1}\]](http://latex.artofproblemsolving.com/1/d/b/1dbad9c24e74e519aad83e9b1c3f174f046253d7.png)
The answer is 
.
See Also
| 1961 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
