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Difference between revisions of "1991 AHSME Problems/Problem 27"

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== Problem ==
 
== Problem ==
  
−
If <math>x+\sqrt{x^2-1}+\frac{1}{x-\sqrt{x^2-1}}=20</math> then <math>x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}}=</math>
+
If <cmath>x+\sqrt{x^2-1}+\frac{1}{x-\sqrt{x^2-1}}=20,</cmath> then <cmath>x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}}=</cmath>
  
−
(A) <math>5.05</math> (B) <math>20</math> (C) <math>51.005</math> (D) <math>61.25</math> (E) <math>400</math>
+
<math>\textbf{(A) } 5.05 \qquad
 +
\textbf{(B) } 20 \qquad
 +
\textbf{(C) } 51.005 \qquad
 +
\textbf{(D) } 61.25 \qquad
 +
\textbf{(E) } 400</math>
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{7}</math>
+
We rationalize the denominator in the given equation, then solve for <math>x:</math>
 +
<cmath>\begin{align*}
 +
x+\sqrt{x^2-1}+\frac{x+\sqrt{x^2-1}}{\left(x-\sqrt{x^2-1}\right)\left(x+\sqrt{x^2-1}\right)} &= 20 \\
 +
x+\sqrt{x^2-1}+x+\sqrt{x^2-1} &= 20 \\
 +
x+\sqrt{x^2-1} &= 10 \\
 +
\sqrt{x^2-1} &= 10-x \\
 +
x^2-1 &= 100-20x+x^2 \\
 +
20x &= 101 \\
 +
x &= 5.05.
 +
\end{align*}</cmath>
 +
We rationalize the denominator in the requested expression, then simplify the result:
 +
<cmath>\begin{align*}
 +
x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}} &= x^2+\sqrt{x^4-1}+\frac{x^2-\sqrt{x^4-1}}{\left(x^2+\sqrt{x^4-1}\right)\left(x^2-\sqrt{x^4-1}\right)} \\
 +
&= x^2+\sqrt{x^4-1}+x^2-\sqrt{x^4-1} \\
 +
&= 2x^2 \\
 +
&= \boxed{\textbf{(C) } 51.005}.
 +
\end{align*}</cmath>
 +
 
 +
~Hapaxoromenon (Solution)
 +
 
 +
~MRENTHUSIASM (Reformatting)
  
 
== See also ==
 
== See also ==

Latest revision as of 04:26, 6 September 2021

Problem

If \[x+\sqrt{x^2-1}+\frac{1}{x-\sqrt{x^2-1}}=20,\] then \[x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}}=\]

$\textbf{(A) } 5.05 \qquad \textbf{(B) } 20 \qquad \textbf{(C) } 51.005 \qquad \textbf{(D) } 61.25 \qquad \textbf{(E) } 400$

Solution

We rationalize the denominator in the given equation, then solve for $x:$ \begin{align*} x+\sqrt{x^2-1}+\frac{x+\sqrt{x^2-1}}{\left(x-\sqrt{x^2-1}\right)\left(x+\sqrt{x^2-1}\right)} &= 20 \\ x+\sqrt{x^2-1}+x+\sqrt{x^2-1} &= 20 \\ x+\sqrt{x^2-1} &= 10 \\ \sqrt{x^2-1} &= 10-x \\ x^2-1 &= 100-20x+x^2 \\ 20x &= 101 \\ x &= 5.05. \end{align*} We rationalize the denominator in the requested expression, then simplify the result: \begin{align*} x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}} &= x^2+\sqrt{x^4-1}+\frac{x^2-\sqrt{x^4-1}}{\left(x^2+\sqrt{x^4-1}\right)\left(x^2-\sqrt{x^4-1}\right)} \\ &= x^2+\sqrt{x^4-1}+x^2-\sqrt{x^4-1} \\ &= 2x^2 \\ &= \boxed{\textbf{(C) } 51.005}. \end{align*}

~Hapaxoromenon (Solution)

~MRENTHUSIASM (Reformatting)

See also

1991 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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
All AHSME Problems and Solutions

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