Difference between revisions of "1953 AHSME Problems/Problem 8"
Katzrockso (talk | contribs) (Created page with "==Problem 8== The value of <math>x</math> at the intersection of <math>y=\frac{8}{x^2+4}</math> and <math>x+y=2</math> is: <math>\textbf{(A)}\ -2+\sqrt{5} \qquad \textbf{(B...") |
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== Solution == | == Solution == | ||
<math>x+y=2\implies y=2-x</math>. Then <math>2-x=\frac{8}{x^2+4}\implies (2-x)(x^2+4)=8</math>. We now notice that <math>x=0\implies (2)(4)=8</math>, so <math>\fbox{C}</math> | <math>x+y=2\implies y=2-x</math>. Then <math>2-x=\frac{8}{x^2+4}\implies (2-x)(x^2+4)=8</math>. We now notice that <math>x=0\implies (2)(4)=8</math>, so <math>\fbox{C}</math> | ||
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| + | ==See Also== | ||
| + | |||
| + | {{AHSME 50p box|year=1953|num-b=7|num-a=9}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:40, 1 April 2017
Problem 8
The value of 
at the intersection of 
and 
is:
Solution
. Then 
. We now notice that 
, so 
See Also
| 1953 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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. 
