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

Latest revision as of 05:26, 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
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Difference between revisions of "1991 AHSME Problems/Problem 27"

Latest revision as of 05:26, 6 September 2021

Problem

Solution

See also

@@ Line 18: / Line 18: @@
 x^2-1 &= 100-20x+x^2 \\
 x &= 101 \\
-x &= \frac{101}{20}.
+x &= 5.05.
 \end{align*}</cmath>
 We rationalize the denominator in the requested expression, then simplify the result: