Solve the equation $$x=1-5(1-5x^2)^2$$
###My work
Let $f(x)=1-5x^2$. Then we have tha equation $f(f(x))=x$. But in this case we don't use the equation $f(x)=x$ because $f(x)$ is not monotonic function
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Here's a SVG graph of the relevant functions. – PM 2Ring Feb 10 '21 at 11:12
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A related problem is https://math.stackexchange.com/questions/2866985/solution-to-putnam-2007-a-1 – Maverick Feb 16 '21 at 03:39
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You are on the right track; a modification of your approach will get you the rest of the way.
Let $y = f(x) = 1 - 5x^2$. Then $f(f(x)) = x$ implies $f(y) = x$, hence the solution to the simultaneous system $$\begin{align} y &= 1 - 5x^2, \\ x &= 1 - 5y^2, \end{align}$$ will yield the desired $x$-values. To this end, we take the difference to obtain $$y - x = 5(y^2 - x^2) = 5(y-x)(x+y),$$ hence $$(y-x)(5(x+y) - 1) = 0.$$ Then $x = y$ amounts to solving $x = 1 - 5x^2$, which are the "obvious" solutions. The other set corresponds to $x + y = \frac{1}{5}$, which are the "nontrivial" solutions, and require solving the quadratic $$5x^2 - x - \frac{4}{5} = 0.$$
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Another option is to divide $x-(1-5(1-5x^2)^2)$ by $x-(1-5x^2)$, which gives $25x^2-5x-4$, then solve both quadratics. But your way is nicer. :) – PM 2Ring Feb 10 '21 at 11:33