# How to solve this quadratic with floor function?

by Havana Time   Last Updated November 12, 2018 15:20 PM

Solve for $$y \in \mathbb{R}$$ When $$y^2-2\lfloor y\rfloor+9=0$$

In yesterday i try to do as y^2-2y+9=0 Then i carry away floor and try to solve quadratic instead So i got y=1+\sqrt (10),1-sqrt (10) and put floor on y will got \lfloor y\rfloor = 4 and -3 but when i check in the equation i got y is complex so in this question not have the real solution

I can do like this ? Where is false Is have please tell me pls thank you

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$$y^2-2\lfloor y\rfloor+9 \ge y^2-2y+9 >0$$
The first inequality is justified because $$y \ge \lfloor y\rfloor$$ and the second inequality can be justified through quadratic formula.
So the stated problem has no solutions when $$y\in \mathbb{R}$$.