Find the number of real roots $x$ such that $\frac{x}{x+4} = \frac{5[x]-7}{7[x]-5}$

Question

Here $[.]$ represents the greatest integer function

Let $x=[x] + \{x\}= t+u$

Here t is GIF and u is fractional part function

$$\frac{t+u}{t+u+4} = \frac{5t-7}{7t-5}$$ $$u=-\frac{2t^2 -18t +28}{9t+2}$$

So now $$0\le u<1$$

Solving the first inequality gives $$t\in (-\infty,-\frac 29)\cup [2,7]$$

The second one gives $$t>-\frac 29$$

So clearly $[x] \in [2,7]$

So $$x\in [2,8)$$

This is giving infinite real roots but we need a finite number. Also only 2 and 7 seem to satisfy the problem. What is wrong with this solution?

Answer 1

First, your expression for $u$ is wrong. According to WolframAlpha, the correct expression is:

$$ u = - \frac{t^2 - 9t + 14}{t + 1} \qquad (t \notin \{-1, \tfrac57\}) $$

Solving for $u \in [0, 1)$ then gives $t \in [2, 3) \cup (5, 7]$.

Moreover, note that $t = [x]$ is by definition an integer, so $t \in \{2, 6, 7\}$, giving the three solutions

$$ x_1 = 2, \qquad x_2 = \frac{46}{7}, \qquad x_3 = 7. $$

Answer 2

Your mistake is from $$\frac{t+u}{t+u+4} = \frac{5t-7}{7t-5}$$ to $$u=-\frac{2t^2 -18t +28}{9t+2}$$

The correct method would be do cross multiplication and simplify again.

$ (t+u)(7t-5)=(t+u+4)(5t-7) \implies 7 t^2 + 7 t u - 5 t - 5 u = 5 t^2 + 5 t u + 13 t - 7 u - 28 $ This simplifies to $$ u = - \frac{t^2 - 9t + 14}{t + 1} \qquad $$