1956 AHSME Problems/Problem 42
Revision as of 21:43, 12 February 2021 by Coolmath34 (talk | contribs) (Created page with "== Problem 42== The equation <math>\sqrt {x + 4} - \sqrt {x - 3} + 1 = 0</math> has: <math>\textbf{(A)}\ \text{no root} \qquad \textbf{(B)}\ \text{one real root} \\ \text...")
Problem 42
The equation 
has:
Solution
Simplify.
![\[\sqrt {x + 4} = \sqrt{x-3} -1\]](http://latex.artofproblemsolving.com/7/3/4/7347c678f9d6f849f3b5154b4e718a6b97912962.png)
![\[x + 4 = x - 3 -2\sqrt{x-3} + 1\]](http://latex.artofproblemsolving.com/4/0/b/40b95a346721b6b18e0521b940ecb561b0a70405.png)
![\[6 = -2\sqrt{x-3}\]](http://latex.artofproblemsolving.com/a/d/5/ad5a887225adc4fbd4f2d379a853bfe3377b53b0.png)
We don't need to solve any further. The square root term is negative, therefore there will be no real roots. The answer is 
-coolmath34
See Also
| 1956 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 41 |
Followed by Problem 43 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
