IME / ITA ⇒ (Simuado-Ime/Ita) Equação

[AngelitaB]

Se a, b, p e q raízes de uma equação biquadrada, tal que: a+b=p+q=0 e -[tex3]\frac{2}{h}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{2}{h}[/tex3]

(ab+pq)=[tex3]\frac{4abpq}{h+3}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{4abpq}{h+3}[/tex3]

=1.Sabendo que uma das raízes assume como valor: [tex3]\frac{\sqrt{\sqrt{k}-1}}{h} + \frac{\sqrt{\sqrt{k}+1}i}{h}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{\sqrt{\sqrt{k}-1}}{h} + \frac{\sqrt{\sqrt{k}+1}i}{h}[/tex3]

; h, k [tex3]\in [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\in [/tex3]

N*.Indicar um valor de: [tex3]\sqrt{k³-h²}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sqrt{k³-h²}[/tex3]

.
a)5
b)7
c)9
d)11
e)13

Resposta

---

d

[rcompany]

[tex3]

\text{Se }r\text{ é raiz imaginaria, então }\overline{r}\text{ é raiz também}\\
\text{Equação biquadrada: se }r\text{ é raiz então}-r\text{ é raiz também.}\\
\text{Temos }a=\dfrac{\sqrt{\sqrt{k}-1}}{h}+i\dfrac{\sqrt{\sqrt{k}+1}}{h}\\
k,h\in\mathbb{R}^*\implies \dfrac{\sqrt{\sqrt{k}-1}}{h},\dfrac{\sqrt{\sqrt{k}+1}}{h}\in\mathbb{R}\quad\small\text{(em particular }\sqrt{k}-1\geqslant0\implies \sqrt{\sqrt{k}-1}\in\mathbb{R})\\
\text{e então }a\text{ é imaginaria e portanto }\overline{a},-a,-\overline{a}\text{ também são raízes.}\\[12pt]
\text{ou seja }(1)\left\{\begin{array}{}a=a,\text{com }\mathrm{Re}(a)\neq0\\b=-a\\p=\overline{a}\\q=-\overline{a}\end{array}\right.
\text{ou }(2)\left\{\begin{array}{}a=-\overline{a},\text{com }\mathrm{Re}(a)=0,\text{ ie. }k=1\text{ e }a\!=\!i\dfrac{\sqrt{\!\sqrt{2}}}{h}\\b=-a=\overline{a}=-i\dfrac{\sqrt{\!\sqrt{2}}}{h}\\p=p\text{, com }\mathrm{Re}(p)=0\\q=-p=\overline{p}\end{array}\right.\\[60pt]
\text{ou }(3)\left\{\begin{array}{}a=-\overline{a},\text{com }\mathrm{Re}(a)=0\text{ ie. }a=i\dfrac{\sqrt{\!\sqrt{2}}}{h}\\b=-a=\overline{a}=-i\dfrac{\sqrt{\!\sqrt{2}}}{h}\\p=p\text{, }p\in\mathbb{R}\\q=-p\end{array}\right.
\\[48pt]

\begin{array}{}
\left.\begin{array}{r}(1)\\\dfrac{2}{h}(ab+pq)=1\\\dfrac{4abpq}{h+3}=1\end{array}\right\}
&\implies \left\{\begin{array}{l}\dfrac{2}{h}(-a^2-\overline{a}^2)=1\\4(a\overline{a})^2=h+3\end{array}\right.\implies \left\{\begin{array}{l}-\dfrac{4}{h}\mathrm{Re}(a^2)=1 \\4\left(\dfrac{2\sqrt{k}}{h^2}\right)^2=h+3\end{array}\right.\\
&\implies \left\{\begin{array}{l}-\dfrac{4}{h}\left(-\dfrac{2}{h^2}\right)=1 \\k=\dfrac{h^5+3h^4}{16}\end{array}\right.
\implies \left\{\begin{array}{l}h=2 \\k=5\end{array}\right.\\
&\implies \sqrt{k^3-h^2}=\sqrt{125-4}=\sqrt{121}=11
\end{array}\\[96pt]

\begin{array}{}
\left.\begin{array}{r}(2)\\\dfrac{2}{h}(ab+pq)=1\\\dfrac{4abpq}{h+3}=1\end{array}\right\}&\implies \left\{\begin{array}{l}a=i\dfrac{\sqrt{\!\sqrt{2}}}{h},\,b=-i\dfrac{\sqrt{\!\sqrt{2}}}{h},\,p=i\mathrm{Im}(p),\,q=-i\mathrm{Im}(p)\\\dfrac{2}{h}(ab+pq)=1\\\dfrac{4abpq}{h+3}=1\end{array}\right.\\
&\implies \left\{\begin{array}{l}a=i\dfrac{\sqrt{\!\sqrt{2}}}{h},\,b=-i\dfrac{\sqrt{\!\sqrt{2}}}{h},\,p=i\mathrm{Im}(p),\,q=-i\mathrm{Im}(p)\\a\overline{a}+p\overline{p}=\dfrac{h}{2}\\4a\overline{a}p\overline{p}=h+3\end{array}\right.\\
&\implies \left\{\begin{array}{l}p\overline{p}=\dfrac{h}{2}-\dfrac{\sqrt{2}}{h^2}=\dfrac{h^3-2\sqrt{2}}{2h^2}\\p\overline{p}=\dfrac{h^3+3h^2}{4\sqrt{2}}\end{array}\right.\\
&\implies 2h^5+6h^4-4\sqrt{2}h^3+16=0\quad\text{que não tem raiz em }\mathbb{N}\begin{array}{l}\small\text{ (estudar as variações da função e notar }\\\small\text{ que não se anula para h positivo)}\end{array}
\end{array}\\[120pt]

\begin{array}{}
\left.\begin{array}{r}(3)\\\dfrac{2}{h}(ab+pq)=1\\\dfrac{4abpq}{h+3}=1\end{array}\right\}&\implies \left\{\begin{array}{l}a=i\dfrac{\sqrt{\!\sqrt{2}}}{h},\,b=-i\dfrac{\sqrt{\!\sqrt{2}}}{h},\,p\in\mathbb{R},\,q=-p\\p^2=\dfrac{2\sqrt{2}-h^3}{2h^2}\\-p^2=\dfrac{h+3}{4a\overline{a}}\quad\text{impossível já que }h > 0\text{ e }a\overline{a}>0\end{array}\right.\\
\end{array}\\[36pt]

[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]

\text{Se }r\text{ é raiz imaginaria, então }\overline{r}\text{ é raiz também}\\
\text{Equação biquadrada: se }r\text{ é raiz então}-r\text{ é raiz também.}\\
\text{Temos }a=\dfrac{\sqrt{\sqrt{k}-1}}{h}+i\dfrac{\sqrt{\sqrt{k}+1}}{h}\\
k,h\in\mathbb{R}^*\implies \dfrac{\sqrt{\sqrt{k}-1}}{h},\dfrac{\sqrt{\sqrt{k}+1}}{h}\in\mathbb{R}\quad\small\text{(em particular }\sqrt{k}-1\geqslant0\implies \sqrt{\sqrt{k}-1}\in\mathbb{R})\\
\text{e então }a\text{ é imaginaria e portanto }\overline{a},-a,-\overline{a}\text{ também são raízes.}\\[12pt]
\text{ou seja }(1)\left\{\begin{array}{}a=a,\text{com }\mathrm{Re}(a)\neq0\\b=-a\\p=\overline{a}\\q=-\overline{a}\end{array}\right.
\text{ou }(2)\left\{\begin{array}{}a=-\overline{a},\text{com }\mathrm{Re}(a)=0,\text{ ie. }k=1\text{ e }a\!=\!i\dfrac{\sqrt{\!\sqrt{2}}}{h}\\b=-a=\overline{a}=-i\dfrac{\sqrt{\!\sqrt{2}}}{h}\\p=p\text{, com }\mathrm{Re}(p)=0\\q=-p=\overline{p}\end{array}\right.\\[60pt]
\text{ou }(3)\left\{\begin{array}{}a=-\overline{a},\text{com }\mathrm{Re}(a)=0\text{ ie. }a=i\dfrac{\sqrt{\!\sqrt{2}}}{h}\\b=-a=\overline{a}=-i\dfrac{\sqrt{\!\sqrt{2}}}{h}\\p=p\text{, }p\in\mathbb{R}\\q=-p\end{array}\right.
\\[48pt]

\begin{array}{}
\left.\begin{array}{r}(1)\\\dfrac{2}{h}(ab+pq)=1\\\dfrac{4abpq}{h+3}=1\end{array}\right\}
&\implies \left\{\begin{array}{l}\dfrac{2}{h}(-a^2-\overline{a}^2)=1\\4(a\overline{a})^2=h+3\end{array}\right.\implies \left\{\begin{array}{l}-\dfrac{4}{h}\mathrm{Re}(a^2)=1 \\4\left(\dfrac{2\sqrt{k}}{h^2}\right)^2=h+3\end{array}\right.\\
&\implies \left\{\begin{array}{l}-\dfrac{4}{h}\left(-\dfrac{2}{h^2}\right)=1 \\k=\dfrac{h^5+3h^4}{16}\end{array}\right.
\implies \left\{\begin{array}{l}h=2 \\k=5\end{array}\right.\\
&\implies \sqrt{k^3-h^2}=\sqrt{125-4}=\sqrt{121}=11
\end{array}\\[96pt]

\begin{array}{}
\left.\begin{array}{r}(2)\\\dfrac{2}{h}(ab+pq)=1\\\dfrac{4abpq}{h+3}=1\end{array}\right\}&\implies \left\{\begin{array}{l}a=i\dfrac{\sqrt{\!\sqrt{2}}}{h},\,b=-i\dfrac{\sqrt{\!\sqrt{2}}}{h},\,p=i\mathrm{Im}(p),\,q=-i\mathrm{Im}(p)\\\dfrac{2}{h}(ab+pq)=1\\\dfrac{4abpq}{h+3}=1\end{array}\right.\\
&\implies \left\{\begin{array}{l}a=i\dfrac{\sqrt{\!\sqrt{2}}}{h},\,b=-i\dfrac{\sqrt{\!\sqrt{2}}}{h},\,p=i\mathrm{Im}(p),\,q=-i\mathrm{Im}(p)\\a\overline{a}+p\overline{p}=\dfrac{h}{2}\\4a\overline{a}p\overline{p}=h+3\end{array}\right.\\
&\implies \left\{\begin{array}{l}p\overline{p}=\dfrac{h}{2}-\dfrac{\sqrt{2}}{h^2}=\dfrac{h^3-2\sqrt{2}}{2h^2}\\p\overline{p}=\dfrac{h^3+3h^2}{4\sqrt{2}}\end{array}\right.\\
&\implies 2h^5+6h^4-4\sqrt{2}h^3+16=0\quad\text{que não tem raiz em }\mathbb{N}\begin{array}{l}\small\text{ (estudar as variações da função e notar }\\\small\text{ que não se anula para h positivo)}\end{array}
\end{array}\\[120pt]

\begin{array}{}
\left.\begin{array}{r}(3)\\\dfrac{2}{h}(ab+pq)=1\\\dfrac{4abpq}{h+3}=1\end{array}\right\}&\implies \left\{\begin{array}{l}a=i\dfrac{\sqrt{\!\sqrt{2}}}{h},\,b=-i\dfrac{\sqrt{\!\sqrt{2}}}{h},\,p\in\mathbb{R},\,q=-p\\p^2=\dfrac{2\sqrt{2}-h^3}{2h^2}\\-p^2=\dfrac{h+3}{4a\overline{a}}\quad\text{impossível já que }h > 0\text{ e }a\overline{a}>0\end{array}\right.\\
\end{array}\\[36pt]

[/tex3]

[tex3]
\boxed{\hspace{1cm}\\[8pt]\hspace{1cm}h=2,\,k=5,\, a=\dfrac{\sqrt{\sqrt{5}-1}}{2}+i\dfrac{\sqrt{\sqrt{5}+1}}{2},\,b=-a,\,p=\overline{a},\,q=-\overline{a}\hspace{1cm}\\\hspace{1cm}\text{e }\sqrt{k^3-h^2}=11\\[12pt]}
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]
\boxed{\hspace{1cm}\\[8pt]\hspace{1cm}h=2,\,k=5,\, a=\dfrac{\sqrt{\sqrt{5}-1}}{2}+i\dfrac{\sqrt{\sqrt{5}+1}}{2},\,b=-a,\,p=\overline{a},\,q=-\overline{a}\hspace{1cm}\\\hspace{1cm}\text{e }\sqrt{k^3-h^2}=11\\[12pt]}
[/tex3]