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

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)

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

Latest revision as of 04:26, 6 September 2021

Problem

Solution

See also

@@ Line 1: / Line 1: @@
 == 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 ==
-We have <math>\frac{1}{x-\sqrt{x^2-1}} = \frac{x+\sqrt{x^2-1}}{x^2-(x^2-1)} = x+\sqrt{x^2-1}.</math> Hence the given equation becomes <math>2(x+\sqrt{x^2-1}) = 20 \implies x+\sqrt{x^2-1} = 10.</math> Thus <math>\frac{1}{x-\sqrt{x^2-1}} = 10</math>, so <math>x-\sqrt{x^2-1} = \frac{1}{10}</math>. Adding this to <math>x+\sqrt{x^2-1} = 10</math> gives <math>2x = 10 + \frac{1}{10} \implies x = \frac{101}{20}.</math> Now the quantity we need to find is <math>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}}{x^4 - (x^4 - 1)} = 2x^2 = 2(\frac{101}{20})^2 = 51.005</math>, which is <math>\boxed{C}.</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 \\
+x &= 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 ==
 {{AHSME box|year=1991|num-b=26|num-a=28}}