Difference between revisions of "1957 AHSME Problems/Problem 22"
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== Solution == | == Solution == | ||
| − | <math>\ | + | |
| + | By repeatedly rearranging the equation and squaring both sides, we can solve for <math>x</math>: | ||
| + | \begin{align*} | ||
| + | \sqrt{x-1}-\sqrt{x+1}+1 &= 0 \\ | ||
| + | \sqrt{x-1}+1 &= \sqrt{x+1} \\ | ||
| + | x-1+2\sqrt{x-1}+1 &= x+1 \\ | ||
| + | 2\sqrt{x-1} &= 1 \\ | ||
| + | \sqrt{x-1} &= \frac{1}{2} \\ | ||
| + | x-1 &= \frac{1}{4} \\ | ||
| + | x &= \frac{5}{4} | ||
| + | \end{align*} | ||
| + | After checking for [[extraneous solutions]], we see that <math>x=\tfrac{5}{4}</math> does indeed solve the equation. Thus, <math>4x=5</math>, and so our answer is <math>\boxed{\textbf{(A) }5}</math>. | ||
== See Also == | == See Also == | ||
Latest revision as of 10:08, 25 July 2024
Problem
If 
, then 
equals:
Solution
By repeatedly rearranging the equation and squaring both sides, we can solve for 
:
\begin{align*}
\sqrt{x-1}-\sqrt{x+1}+1 &= 0 \\
\sqrt{x-1}+1 &= \sqrt{x+1} \\
x-1+2\sqrt{x-1}+1 &= x+1 \\
2\sqrt{x-1} &= 1 \\
\sqrt{x-1} &= \frac{1}{2} \\
x-1 &= \frac{1}{4} \\
x &= \frac{5}{4}
\end{align*}
After checking for extraneous solutions, we see that 
does indeed solve the equation. Thus, 
, and so our answer is 
.
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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. 
