Find how many positive real values of $x$ satisfy the equation $2[x]^2=5x+2$, where $[x]$ denotes greatest integer less than or equal to $x$.

$x=f+I$ where $f$ is the fractional part and $I$ is the integer part.

Solving above equation I got




This gave two values of $I$ $I=3.5$ and $I=-1$


Solving this I got $I=3$ and $f=0.5$, $I=0$ and $f=-0.85$

But the answer is only one value of $x$ and that is $x=3.2$

This is a gmat exam question.


1 Answer 1


Solving for integer values of I in $$0 \le 2I^2 - 5I - 2 < 5$$

Let $2I^2 - 5I - 2 = c$ where $c$ is a real number between $0$ and $5$. Then note $I$ is an integer iff $$\frac{5 \pm \sqrt{25 + 8(2+c)}}{4} = I$$ is an integer.

The only $c$ that satisfies this is $1/8$.

If $c = 1/8$ then $I = 3$.

$$5f = 2(9)- 15 - 2$$ $$f = 0.2$$

Therefore $x =3 + 0.2 = 3.2 $

EDIT: Due to OP's request here is a note on the thought process I went through to evaluate $c$.

First note if $I$ is an integer then $5 \pm \sqrt{25 + 8(2+c)}$ must divide $4$. Since $5$ is odd then $\sqrt{25 + 8(2+c)}$ must be odd aswell for there to exist an integer solution.

We also know $0 \le c < 5$. So plugging in $0$ and $5$ for $c$ gives $\sqrt{25 + 8(2+c)}$ the values$\sqrt{41}$ and $\sqrt{81} =9$ so $\sqrt{25 + 8(2+c)}$ must be between $6$ and $9$

The only possible integers between $6$ and $9$ are $7$ and $8$. Since $7$ is odd it is the answer. Soling for $c$ would then result in $1/8$

  • $\begingroup$ why you did 2I^2-5I-2=c? @ZiadFakhoury? $\endgroup$ Aug 21, 2017 at 20:20
  • $\begingroup$ Well we know it is equal to some number between $0$ to $5$, yes? So i assigned that number $c$. So then we have $2I^2 - 5I -2 = c \implies 2I^2 -5I - 2- c = 0$. You then use the quadratic formula to solve for $I$ given the constraints A) $I$ must be an integer B) $c$ is a real number between $0$ to $5$ $\endgroup$ Aug 21, 2017 at 20:32
  • $\begingroup$ How did you evaluate c so easily? Putting values takes time @ZiadFakhoury $\endgroup$ Aug 21, 2017 at 20:47
  • $\begingroup$ @SakuziMarkel I've edited my answer as to your request. Hope this helps :) $\endgroup$ Aug 21, 2017 at 20:57

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