Difference between revisions of "1954 AHSME Problems/Problem 21"
Katzrockso (talk | contribs) (Created page with "== Problem 21== The roots of the equation <math>2\sqrt {x} + 2x^{ - \frac {1}{2}} = 5</math> can be found by solving: <math> \textbf{(A)}\ 16x^2-92x+1 = 0\qquad\textbf{(B)}...") |
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Make the substitution <math>t=\sqrt{x}</math>. Then <math>2t+\frac{2}{t}=5\implies 2t^2-5t+2=0</math>. Then <math>2x-5\sqrt{x}+2=0\implies 2x+2=5\sqrt{x}\implies (2x+2)^2=25x\implies 4x^2+8x+4=25x\implies 4x^2-17x+4=0</math>, <math>\fbox{C}</math> | Make the substitution <math>t=\sqrt{x}</math>. Then <math>2t+\frac{2}{t}=5\implies 2t^2-5t+2=0</math>. Then <math>2x-5\sqrt{x}+2=0\implies 2x+2=5\sqrt{x}\implies (2x+2)^2=25x\implies 4x^2+8x+4=25x\implies 4x^2-17x+4=0</math>, <math>\fbox{C}</math> | ||
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| + | ==See Also== | ||
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| + | {{AHSME 50p box|year=1954|num-b=20|num-a=22}} | ||
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| + | {{MAA Notice}} | ||
Latest revision as of 01:29, 28 February 2020
Problem 21
The roots of the equation 
can be found by solving:
Solution
Make the substitution 
. Then 
. Then 
, 
See Also
| 1954 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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. 
