How to solve $\ x^2-19\lfloor x\rfloor+88=0 $

Question

I have no clue on how to solve this. If you guys have, please show me your solution as well.

$$\ x^2-19\lfloor x\rfloor+88=0 $$

Answer 1

First try to figure out the range of $x$. We know that $x-1<\lfloor x\rfloor\le x$. Thus $-19x\le-19\lfloor x\rfloor< -19x+19$. Thus $x^2-19x+88<x^2-19\lfloor x\rfloor+88\le x^2-19x+107$ which gives $x^2-19x+88<0\le x^2-19x+107$. Once you'll solve this, you'll have a finite number of values of $\lfloor x\rfloor$ and solving the equation will become a piece of cake.

Answer 2

$$\ x^2-19\lfloor x\rfloor+88=0 \tag{*}$$

Start by solving $x^2-19x+88=0$. This factors as $(x-8)(x-11)=0$ giving solutions $x\in\{8,11\}$. Since these are integers, they are solutions of (*) as well.

There are no more solutions to (*) for $8<x<9$ since 8 was an ordinary root and $x^2$ is increasing.

For $9\le x<10$, we have $\lfloor x \rfloor = 9$, so that (*) reduces to

$$x^2-19\cdot9+88=0 \implies x^2=171-88=83 \implies x=\sqrt{83}$$

For $10\le x<11$, we have $\lfloor x \rfloor = 10$, so that (*) reduces to

$$x^2-19\cdot10+88=0 \implies x^2=190-88=102 \implies x=\sqrt{102}$$

There are no more solutions to (*) for $11<x<12$ since 11 was an ordinary root and $x^2$ is increasing, no are there any for $x\ge12$.

So the solutions are:

$$x\in\{8,\sqrt{83},\sqrt{102},11\}$$

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For all $x$, we have

$$\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$$

So for $x>0$ and a fixed $\lfloor x \rfloor$, since $x^2$ is increasing in $x$, the expression in (*) is minimised when $x=\lfloor x \rfloor$, i.e. when $x$ is an integer. But then for the factorisation $(x-8)(x-11)$ we have

$$0\le x<8 \implies (x-8)(x-11)>0$$

as both factors are negative. So the LHS in (*) is at least this large, so is always positive for $0\le x<8$, so there are no solutions with $0\le x<8$.

It is trivial to see that there are no solutions for $x<0$, as all terms on the LHS in (*) are then positive.

Answer 3

Edit The first version of this solution was false, and hence I used the same idea with some kind of forcing it to get the result.

As you can see there exists an integer $n$, such that $x=\sqrt n$ hence we have: $$n-19\lfloor \sqrt n \rfloor +88 =0 $$ so let's write $n=k^2+l $ with $0\leq l< 2k+1$ and hence : $$k^2+l-19k+88=0 $$ but this is equivalent to $0\leq (k-8)(11-k)< 2k+1$ and this implies that: $$8\leq k\leq 11 $$ which gives $k=8,9,10,11$ and hence for every solution we compute $l$ and deduce the value of $n$ and finally found that: $$x\in \left\{8,\sqrt{83},\sqrt{102},11\right\}$$

Answer 4

Since $x \geq \lfloor x \rfloor$ for all $x$, you have $x^2 - 19\lfloor x \rfloor + 88 \geq x^2 - 19x + 88$ for all $x$. Hence the graph of $x^2 - 19\lfloor x \rfloor + 88$ lies above that of $x^2 - 19x + 88$. So the function $x^2 - 19\lfloor x \rfloor + 88$ can only be zero where the function $x^2 - 19x + 88 = (x - 11)(x - 8)$ is less than or equal to zero, namely the interval $[8,11]$.

One can then solve $x^2 - 19\lfloor x \rfloor + 88 = 0$ separately on the four ranges in $[8,11]$ where $\lfloor x \rfloor$ is constant: the intervals $[8,9),[9,10],[10,11)$ and the singleton $x = 11$. You get the equations $x^2 - 64 = 0$, $x^2 - 83 = 0$, $x^2 - 102 = 0$, and $x^2 - 121 = 0$ respectively, leading to the four solutions $8, \sqrt{83}, \sqrt{102},$ and $11$.

Answer 5

Let $x = n + \epsilon$ where $n\in \mathbb Z$ and $0 \le \epsilon \lt 1$.

Then $x^2 = (n + \epsilon)^2 = 19n - 88$

Since $0 \le \epsilon \lt 1$, then $n^2 \le 19n - 88 \lt (n+1)^2$

The line $y = 19x - 88$ intersects the parabola $y = x^2$ at $x = 8$ and at $x = 11$. Since the parabola $y = x^2$ is concave up, this implies that

$n \in \{8, 9, 10, 11 \}$ and $x = \sqrt{19n - 88}$

$n = 8 \implies x = 8$
$n = 9 \implies x = \sqrt{83}$
$n = 10 \implies x = \sqrt{102}$
$n = 11 \implies x = 11$