Difference between revisions of "1956 AHSME Problems/Problem 23"
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The discriminant is equal to 0, so this simplifies to <math>x = \frac{2\sqrt{2}}{2a}=\frac{\sqrt{2}}{a}.</math> Because we are given that <math>a</math> is real, <math>x</math> is always real, and the answer is <math>\boxed{\textbf{(C)}}.</math> | The discriminant is equal to 0, so this simplifies to <math>x = \frac{2\sqrt{2}}{2a}=\frac{\sqrt{2}}{a}.</math> Because we are given that <math>a</math> is real, <math>x</math> is always real, and the answer is <math>\boxed{\textbf{(C)}}.</math> | ||
| − | ~ cxsmi (significant edits) | + | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] (significant edits) |
==See Also== | ==See Also== | ||
Latest revision as of 13:47, 4 April 2024
Problem 23
About the equation 
, with 
and 
real constants,
we are told that the discriminant is zero. The roots are necessarily:
Solution
Plugging into the quadratic formula, we get
![\[x = \frac{2\sqrt{2} \pm \sqrt{8-4ac}}{2a}.\]](http://latex.artofproblemsolving.com/c/8/d/c8d83dfb1fbba5bc64524dbfbf0aa38fe0a5325d.png)
The discriminant is equal to 0, so this simplifies to 
Because we are given that is real, 
is always real, and the answer is 
~ cxsmi (significant edits)
See Also
| 1956 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
