1991 AHSME Problems/Problem 27

Problem

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

(A) $5.05$ (B) $20$ (C) $51.005$ (D) $61.25$ (E) $400$

Solution

We have $\frac{1}{x-\sqrt{x^2-1}} = \frac{x+\sqrt{x^2-1}}{x^2-(x^2-1)} = x+\sqrt{x^2-1}.$ Hence the given equation becomes $2(x+\sqrt{x^2-1}) = 20 \implies x+\sqrt{x^2-1} = 10.$ Thus $\frac{1}{x-\sqrt{x^2-1}} = 10$, so $x-\sqrt{x^2-1} = \frac{1}{10}$. Adding this to $x+\sqrt{x^2-1} = 10$ gives $2x = 10 + \frac{1}{10} \implies x = \frac{101}{20}.$ Now the quantity we need to find is $x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}} = x^2 + \sqrt{x^4-1} + \frac{x^2 - \sqrt{x^4-1}}{x^4 - (x^4 - 1)} = 2x^2 = 2(\frac{101}{20})^2 = 51.005$, which is $\boxed{C}.$

See also

1991 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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