# Solve for x $\left\lfloor x \right\rfloor^3+2x^2=x^3+2\left\lfloor x\right\rfloor^2$

Solve for x $$\left\lfloor x \right\rfloor^3+2x^2=x^3+2\left\lfloor x\right\rfloor^2$$ where $\left\lfloor t \right\rfloor$ denotes the largest integer not exceeding t

$X \in \mathbb{Z}$ is a solution. Is there other root? Thanks.

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Where do differential equations come in? – Chris Eagle Nov 23 '12 at 17:41
Also, most integers are not solutions. $2$, for example. – Chris Eagle Nov 23 '12 at 17:43
Sorry for that mistake. I just correct it – tangkhaihanh Nov 23 '12 at 17:51
How to have that root? – tangkhaihanh Nov 23 '12 at 17:53
I have solved it already. Thanks very much – tangkhaihanh Nov 23 '12 at 18:03

Write the equation as $x^3 - 2x^2 = \lfloor x \rfloor^3 - 2\lfloor x \rfloor^2$. Clearly this has solutions in $\mathbb{Z}$.

On most intervals of the form $[n, n+1]$ with $n \in \mathbb{Z}$, the function $x^3 - 2x^2$ is monotone. You will only get interesting values on the intervals where it isn't.

To find those, take the derivativve $3x^2 - 4x$ and set it to zero - you get $0$ and $\frac{4}{3}$ as critical points. So the only possible zero outside of $\mathbb{Z}$ will occur where $\lfloor x \rfloor = 1$.

Solve the equation now: $x^3 - 2x^2 = -1$ has three real solutions, one of which lies in the interval $(1, 2)$ and therefore satisfies $\lfloor x \rfloor = 1$.

This solution is $\frac{1}{2}(1 + \sqrt{5})$.

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As @Cocopuffs explained you can analyse the function $f(x)=x^3-x^2$. Note that if $f$ is strictly monotone on $[n,n+1]$ then $f$ won't attains $f(n)$ on $(n,n+1)$. then you analyse the intervals where $f$ isnt monotone.

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