Difference between revisions of "1953 AHSME Problems/Problem 23"
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| − | + | ==Problem== | |
| − | |||
| − | We multiply both sides by <math>\sqrt{x+10}</math> to get | + | The equation <math>\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5</math> has: |
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| + | <math>\textbf{(A)}\ \text{an extraneous root between } - 5\text{ and } - 1 \\ | ||
| + | \textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6\\ \textbf{(C)}\ \text{a true root between }20\text{ and }25\qquad | ||
| + | \textbf{(D)}\ \text{two true roots}\\ \textbf{(E)}\ \text{two extraneous roots} </math> | ||
| + | |||
| + | ==Solution== | ||
| + | We can multiply both sides by <math>\sqrt{x+10}</math> to get <math>x+4=5\sqrt{x+10}</math>. We can now square both sides to get <math>x^2+8x+16=25x+250</math>, which yields <math>x^2-17x-234=0</math>. We can factor the quadratic as <math>(x+9)(x-26)=0</math>, giving us roots of <math>-9</math> and <math>26</math>. Plugging in these values, we find that <math>-9</math> is an extraneous root and <math>26</math> is a true root. This gives us the final answer of <math>\boxed{\textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6}</math>. | ||
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| + | ==See Also== | ||
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| + | {{AHSME 50p box|year=1953|num-b=22|num-a=24}} | ||
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| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:30, 27 November 2019
Problem
The equation 
has:
Solution
We can multiply both sides by 
to get 
. We can now square both sides to get 
, which yields 
. We can factor the quadratic as 
, giving us roots of 
and 
. Plugging in these values, we find that is an extraneous root and is a true root. This gives us the final answer of 
.
See Also
| 1953 AHSC (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 • 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. 
