7
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ What about starting by solving $-19 \leq x^2 - 19x + 88 \leq 19$? $\endgroup$ – Clement C. Sep 20 '15 at 20:26
13
$\begingroup$

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

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:

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


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.

$\endgroup$
  • $\begingroup$ great job, only person that post actual answer $\endgroup$ – jack Sep 20 '15 at 20:40
  • $\begingroup$ You should say why there are no solutions for $x < 8$. $\endgroup$ – Zarrax Sep 20 '15 at 21:02
4
$\begingroup$

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 $11$.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ I like the way you solved it let things easier rather than using inequalities, but there's something wrong somwhere because you're missing $\sqrt{83}$ and $\sqrt{102}$ that Marconius found and which are indeed solutions. $\endgroup$ – Scientifica Sep 20 '15 at 20:43
  • $\begingroup$ don't delete it this fast $\endgroup$ – Scientifica Sep 20 '15 at 20:45
  • $\begingroup$ Also, -8 and -11 are not solutions. If $x=-8$, the expression is equal to 304. $\endgroup$ – Bernard Masse Sep 20 '15 at 20:45
  • $\begingroup$ @BernardMasse they are removed it's ok $\endgroup$ – Scientifica Sep 20 '15 at 20:46
  • 2
    $\begingroup$ I think the error is in assuming $\sqrt{n} = \lfloor\sqrt{n}\rfloor$ $\endgroup$ – Winther Sep 20 '15 at 20:46
2
$\begingroup$

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 $x = \sqrt{19n - 88}$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Off Topic: math.stackexchange مرحبا بك في ;) $\endgroup$ – Scientifica Sep 20 '15 at 20:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.