$y=\{x\}$ represents greatest integer less than $x$ then solve $x^2-7\{x\}+5=0$.

Question

$y=\{x\}$ represents greatest integer less than $x$, e.g.

$\{1.1\}=1$
$\{2\}=1$
$\{5\}=4$
$\{3.7\}=3$

Then what are the solutions of

$$x^2-7\{x\}+5=0?$$

Note that this is not same as greatest integer function. So its more difficult to solve.

Note that ${x} = \lceil x-1\rceil =\lceil x\rceil -1=-\lfloor -x\rfloor-1$. — Hagen von Eitzen, Jul 21 '13 at 14:13
As it turns out, this problem is hardly different from the related http://math.stackexchange.com/questions/448763/solving-x2-7x5-0-to-find-values-of-x — Hagen von Eitzen, Jul 21 '13 at 14:25
Does anyone know if there is a notation for "greatest integer strictly less than $x$"? It's not a commonly used concept. — Calvin Lin, Jul 21 '13 at 15:10

score 3 · Answer 1 · answered Jul 21 '13 at 14:24

Your problem is equivalent to solving $$ x^2-7\lceil x\rceil +12=0. $$ If $x\le 1$, then $$x^2-7\lceil x\rceil+12\ge -7\lceil x\rceil +12\ge5$$ hence we obtain $x>1$. If $x>7$, then using $\lceil x\rceil < x+1 $ we get $$x^2-7\lceil x\rceil +12>x^2-7x+5=(x-7)x+5>0.$$ We conclude $\lceil x\rceil \in\{2,3,4,5,6,7\}$ and accordingly $x^2=7\lceil x\rceil -12\in\{2,9,16,23,30,37\}$. As $x>0$, we conlcude $$x\in\{\sqrt 2, 3, 4, \sqrt{23},\sqrt{30}\sqrt{37}\}.$$ Checking against the original equation shows that all of these are in fact valid solutions

$y=\{x\}$ represents greatest integer less than $x$ then solve $x^2-7\{x\}+5=0$.

1 Answers1