Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 GMC 10B Problems/Problem 24"

(Created page with "==Problem== Find the range <math>x</math> lies in such that <math>\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}}}=1+x^8</math> and <math>x</math> is a positive numbe...")
 
m (→‎Solution)
Line 14: Line 14:
  
 
By the Zero-Product Property, we have <math>x^{15} + x^7 -1 = 0</math> as the positive solution to our equation. Approximating the solutions to this equation, we have <math>x\approx 0.93</math>, which is in the range of <math>\boxed{\textbf{(D)}~\dfrac12 < x < 1}\hskip 0.5pt</math>.
 
By the Zero-Product Property, we have <math>x^{15} + x^7 -1 = 0</math> as the positive solution to our equation. Approximating the solutions to this equation, we have <math>x\approx 0.93</math>, which is in the range of <math>\boxed{\textbf{(D)}~\dfrac12 < x < 1}\hskip 0.5pt</math>.
 +
 
~pineconee
 
~pineconee

Revision as of 01:04, 6 March 2022

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_GMC_10B_Problems/Problem_24&oldid=172364"