step by step solution:
the given equation is
x=3+\frac{1}{2+\frac{1}{3+\frac{1}{2+\frac{1}{3+\dots }}}}let’s consider:
\textcolor{primary}{3+\frac{1}{2+\frac{1}{3+\dots }}=x}after substituting the above value of x into the initial equation we get
x=\textcolor{primary}{3+\frac{1}{2+\frac{1}{{x} }}} \implies x=3+\frac{1}{\frac{2x+1}{x}}x=3+\frac{1}{\frac{2x+1}{x}}\implies x=3+\frac{x}{2x+1} x=3+\frac{x}{2x+1}\implies x=\frac{3(2x+1)+x}{2x+1}x=\frac{3(2x+1)+x}{2x+1}\implies x(2x+1)=6x+3+xx(2x+1)=6x+3+x\implies 2x^{2}+x=7x+3\implies 2x^2-6x-3=02x^2-6x-3=0
Solve the equation for x we get:
x=\frac{3-\sqrt{15}}{2}~~,~~\frac{3+\sqrt{15}}{2}Similar posts:
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