# Thread: flloor and ceiling question

1. ## flloor and ceiling question

Hi, I think the answer of this question x= $\sqrt{1/2}$. But I am not sure for this answer. What do you think about for this problem ?

2. Originally Posted by isiksoy7
Hi, I think the answer of this question x= $\sqrt{1/2}$. But I am not sure for this answer. What do you think about for this problem ?
That is correct.
There is a second answer.

3. Originally Posted by Plato
That is correct.
There is a second answer.
What is the second answer ?

4. Originally Posted by isiksoy7
What is the second answer ?
Graph the function $x^2 - \left\lfloor x \right\rfloor$ from 0 to 3.

5. ... a very easy alternative to find the solutions of the equation...

$x^{2} - \lfloor {x} \rfloor = \frac{1}{2}$ (1)

... is to set $\lfloor {x} \rfloor = k$ with k integer and verify if the solution of the 'standard' second order equation...

$x^{2} - k =\frac{1}{2}$ (2)

... is also solution of (1)...

Kind regards

$\chi$ $\sigma$

6. Originally Posted by chisigma
... a very easy alternative to find the solutions of the equation...

$x^{2} - \lfloor {x} \rfloor = \frac{1}{2}$ (1)

... is to set $\lfloor {x} \rfloor = k$ with k integer and verify if the solution of the 'standard' second order equation...

$x^{2} - k =\frac{1}{2}$ (2)

... is also solution of (1)...

Kind regards

$\chi$ $\sigma$
if I request , you can explain more please I am young at this topic and hard for me.

7. All right!... it is self evident that $\lfloor {x} \rfloor$ is an integer that we can indicate with k. Because $x^{2} \ge 0$ and $\frac{1}{2} >0$ and is...

$x^{2} - \lfloor {x} \rfloor = x^{2} - k = \frac{1}{2}$ (1)

... we have to suppose that is $k \ge 0$. Now we proceed 'step by step' to verify with increasing value of k starting from 0...

$k=0 \rightarrow x^{2} = \frac{1}{2} \rightarrow x= \sqrt{\frac{1}{2}}$ (2)

and because $\lfloor {x} \rfloor =0 =k$ , $x= \sqrt{\frac{1}{2}}$ is solution of (1)...

$k=1 \rightarrow x^{2} = \frac{3}{2} \rightarrow x= \sqrt{\frac{3}{2}}$ (3)

and because $\lfloor {x} \rfloor = 1 =k$ , $x= \sqrt{\frac{3}{2}}$ is also solution of (1)...

For k>1 there are no more solution , so that the (1) has two solutions... like a 'standard' second order equation ...

Kind regards

$\chi$ $\sigma$