(Mackenzie-SP) A equação kx³+x²+1=0 , com k E Z, admite raiz inteira.Então, k pode ser:
a) -1 b)-3 c)3 d)1 e)2
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![\displaystyle
a\in \mathbb Z\\ \\
(x-a)(kx^2+rx-1/a) =kx^3 + (r - ak)x^2 - \frac{a^2r + 1}{a}x + 1\\ \\
r-ak=1 \Longrightarrow \boxed{k=\frac{r-1}{a}}\\ \\
\frac{a^2r + 1}{a}=0\Longrightarrow \boxed{r=-\frac{1}{a^2}}\\ \\
k=-\frac{a^2+1}{a^3}\\ \\
\text{adem\'as: si }a\ \textgreater \ 1 \text{ entonces }a^3\ \textgreater \ a^2+1\ \textgreater \ 0 \iff 0\ \textless \ \frac{a^2+1}{a^3}\ \textless \ 1\\ \\
\text{Si } a\ \textless \ -1\Longrightarrow 0\ \textless \ -\frac{a^2+1}{a^3}\ \textless \ 1\\ \\
\text{Por lo tanto }a\in[-1,1] \text{ y como }a\in \mathbb Z \Longrightarrow a\in\{-1,1\}](https://tex.z-dn.net/?f=%5Cdisplaystyle%0Aa%5Cin+%5Cmathbb+Z%5C%5C+%5C%5C%0A%28x-a%29%28kx%5E2%2Brx-1%2Fa%29+%3Dkx%5E3+%2B+%28r+-+ak%29x%5E2+-+%5Cfrac%7Ba%5E2r+%2B+1%7D%7Ba%7Dx+%2B+1%5C%5C+%5C%5C%0Ar-ak%3D1+%5CLongrightarrow+%5Cboxed%7Bk%3D%5Cfrac%7Br-1%7D%7Ba%7D%7D%5C%5C+%5C%5C%0A%5Cfrac%7Ba%5E2r+%2B+1%7D%7Ba%7D%3D0%5CLongrightarrow++%5Cboxed%7Br%3D-%5Cfrac%7B1%7D%7Ba%5E2%7D%7D%5C%5C+%5C%5C%0Ak%3D-%5Cfrac%7Ba%5E2%2B1%7D%7Ba%5E3%7D%5C%5C+%5C%5C%0A%5Ctext%7Badem%5C%27as%3A+si+%7Da%5C+%5Ctextgreater+%5C+1+%5Ctext%7B+entonces+%7Da%5E3%5C+%5Ctextgreater+%5C+a%5E2%2B1%5C+%5Ctextgreater+%5C+0+%5Ciff+0%5C+%5Ctextless+%5C+%5Cfrac%7Ba%5E2%2B1%7D%7Ba%5E3%7D%5C+%5Ctextless+%5C+1%5C%5C+%5C%5C%0A%5Ctext%7BSi+%7D+a%5C+%5Ctextless+%5C+-1%5CLongrightarrow+0%5C+%5Ctextless+%5C+-%5Cfrac%7Ba%5E2%2B1%7D%7Ba%5E3%7D%5C+%5Ctextless+%5C+1%5C%5C+%5C%5C%0A%5Ctext%7BPor+lo+tanto+%7Da%5Cin%5B-1%2C1%5D+%5Ctext%7B+y+como+%7Da%5Cin+%5Cmathbb+Z+%5CLongrightarrow+a%5Cin%5C%7B-1%2C1%5C%7D%0A "\displaystyle
a\in \mathbb Z\\ \\
(x-a)(kx^2+rx-1/a) =kx^3 + (r - ak)x^2 - \frac{a^2r + 1}{a}x + 1\\ \\
r-ak=1 \Longrightarrow \boxed{k=\frac{r-1}{a}}\\ \\
\frac{a^2r + 1}{a}=0\Longrightarrow \boxed{r=-\frac{1}{a^2}}\\ \\
k=-\frac{a^2+1}{a^3}\\ \\
\text{adem\'as: si }a\ \textgreater \ 1 \text{ entonces }a^3\ \textgreater \ a^2+1\ \textgreater \ 0 \iff 0\ \textless \ \frac{a^2+1}{a^3}\ \textless \ 1\\ \\
\text{Si } a\ \textless \ -1\Longrightarrow 0\ \textless \ -\frac{a^2+1}{a^3}\ \textless \ 1\\ \\
\text{Por lo tanto }a\in[-1,1] \text{ y como }a\in \mathbb Z \Longrightarrow a\in\{-1,1\}")
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