# How to find the sum of all possible values of $\lfloor {k} \rfloor$ of the quadratic equation $x^2-2kx+k^2-1=0$?

If the roots of the quadratic equation $x^2-2kx+k^2-1=0$ lie in the interval $(–4, 5)$, how to find the sum of all possible values of $\lfloor {k} \rfloor$?

Attempt:

$$x^2-2kx+k^2-1=0$$ $$\Rightarrow (x-k)^2=1$$ $$\Rightarrow k=x \mp 1$$

From this we could say that $k \in (-3,6)$ when $k=x+1$ and $k \in (-5,4)$ when $k=x-1$, but then how to do the rest?

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You want the sum of all possible $\lfloor k\rfloor$ such that both roots of $x^2-2kx+k^2-1=0$ lie in the interval $(-4,5)$. You’ve correctly determined that if $r$ is a root of the quadratic, then $r=k\pm 1$. For what values of $k$ are $k-1$ and $k+1$ both in the interval $(-4,5)$? You need to find the $k$ for which $$-4<k-1<5$$ and $$-4<k+1<5\;.$$ Once you’ve solved those inequalities simultaneously, you’ll have an interval of possible values of $k$, and you should have no trouble determining the possible values of $\lfloor k\rfloor$.
Assuming, of course, that OP understands the meaning of the $\lfloor x\rfloor$ notation. – Gerry Myerson Nov 7 '11 at 0:43
so satisfying the both constraints $k \in (-5,6)$,a and as $\lfloor -4.f\rfloor = -5$ and $\lfloor 5.f \rfloor = 5$,we could conclude that the sum is zero,right? – Quixotic Nov 7 '11 at 2:09
$k=5$ is in $(-5,6)$. Does it satisfy both constraints? – Gerry Myerson Nov 7 '11 at 3:09
@MaX: You have not identified correctly the $k$ that satisfy both inequalities simultaneously. Maybe you could observe that $k+1$ is automatically bigger than $k-1$. – André Nicolas Nov 7 '11 at 7:05