Difference between revisions of "1956 AHSME Problems/Problem 42"
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...") |
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<cmath>x + 4 = x - 3 -2\sqrt{x-3} + 1</cmath> | <cmath>x + 4 = x - 3 -2\sqrt{x-3} + 1</cmath> | ||
<cmath>6 = -2\sqrt{x-3}</cmath> | <cmath>6 = -2\sqrt{x-3}</cmath> | ||
| − | We don't need to solve any further. The square root term is negative, therefore there will be no real roots. The answer is <math>\boxed{ | + | We don't need to solve any further. The square root term is negative, therefore there will be no real roots. The answer is <math>\boxed{D}.</math> |
-coolmath34 | -coolmath34 | ||
Revision as of 03:58, 8 March 2021
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 | |
| 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
