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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?

1 Answers 1

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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 $$-4 and $$-4$k$, and you should have no trouble determining the possible values of $\lfloor k\rfloor$.

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    Assuming, of course, that OP understands the meaning of the $\lfloor x\rfloor$ notation.2011-11-07
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    @Gerry Myerson:I do,I do .. I do!:)2011-11-07
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    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?2011-11-07
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    $k=5$ is in $(-5,6)$. Does it satisfy both constraints?2011-11-07
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    @MaX: You have not identified correctly the $k$ that satisfy *both* inequalities simultaneously. Maybe you could observe that $k+1$ is automatically bigger than $k-1$.2011-11-07
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    André Nicolas:I draw the number-line,perhaps this time it's correct,$k \in(-3,4)$ but the answer is still zero,isn't?2011-11-07
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    @MaX: Right: $k-1$ has to bigger than $-4$, so $k$ has to be bigger than $-3$, but $k+1$ has to be smaller than $5$, so $k$ has to be smaller than $4$. Thus, $k$ must be in $(-3,4)$. Now what are the possible values of $\lfloor k\rfloor$? It can certainly be $-2$, for instance, or $3$; can it be $-3$? $4$?2011-11-07
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    @Brian M. Scott:As this is open interval,so I think $k$ can't be $-3$ or $4$ but $\lfloor k\rfloor$ could be $-3$ (say $\lfloor -2.f\rfloor$ but not $4$, precisely $-3\lt \lfloor k\rfloor \le 3$,right?2011-11-07
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    @MaX: Looks good. So you just need $(-3)+(-2)+\dots+3$. (Nice of them to make the arithmetic so easy!)2011-11-07
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    Yeah,and the answer is my favorite number.2011-11-07