2021 GMC 10B Problems/Problem 24

Problem

Find the range $x$ lies in such that $\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}}}=1+x^8$ and $x$ is a positive number.

$\textbf{(A)} ~x=0 \qquad\textbf{(B)} ~0<x<\frac{1}{2} \qquad\textbf{(C)} ~x=\frac{1}{2} \qquad\textbf{(D)} ~\frac{1}{2}<x<1 \qquad\textbf{(E)} ~x=1$

Solution

Substituting $1+x^8$ into $\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}}$ , we have $\begin{align*} \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}}} &= 1+x^8 \\ \sqrt{x^8 + x + 1} &= x^8 + 1 \\ x^8 + x + 1 &= x^{16} + 2x^8 + 1 \\ 0 &= x^{16} + x^8 - x. \end{align*}$

By the Zero-Product Property, we have $x^{15} + x^7 -1 = 0$ as the positive solution to our equation. Approximating the solutions to this equation, we have $x\approx 0.93$ , which is in the range of $\boxed{\textbf{(D)}~\dfrac12 < x < 1}\hskip 0.5pt$ .

~pineconee

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2021 GMC 10B Problems/Problem 24

Problem

Solution