a)5
b)-13
c)18
d)12
e)17
Resposta
a
IME / ITA ⇒ (Simulado-Ime/Ita) Equação Tópico resolvido
Moderador: [ Moderadores TTB ]
Set 2021
30
12:01
(Simulado-Ime/Ita) Equação
Se as equações 2 [tex3]x^{4}[/tex3]
+([tex3]a^{2} + b^{2}[/tex3]
+13)[tex3]x^{2}[/tex3]
-c=0 e [tex3]x^{4}[/tex3]
+(2a-3b)[tex3]x^{2}[/tex3]
-ab=0,tem o mesmo conjunto solução. Calcule a+b+c.
a)5
b)-13
c)18
d)12
e)17
a
a)5
b)-13
c)18
d)12
e)17
a
Out 2021
02
19:26
Re: (Simulado-Ime/Ita) Equação
Faltou indicar no enunciado que [tex3]\underline{a,b,c\in\mathbb{R}}[/tex3]
e o resultado do gabarito está errado.
[tex3]
\begin{array}{rl}
\left.\begin{array}{rl}
2x^4+(a^2+b^2+13)x^2-c=2(x^2-r_1^2)(x^2-r_2^2)\\
x^4+(2a-3b)x^2-ab=(x^2-r_1^2)(x^2-r_2^2)
\end{array}\right\}
\implies
\left\{\begin{array}{ll}
a^2+b^2+13=4a-6b&(1)\\
2ab=c&(2)
\end{array}\right.
\end{array}\\[24pt]
\begin{array}{}
(1)&\iff b^2+6b+a^2-4a+13=0\\
&\iff\left\{\begin{array}{ll}b=\dfrac{-6+i\sqrt{-36+4(a^2-4a+13)}}{2}=\dfrac{-6+i\sqrt{4a^2-16a+16}}{2}=-3+i(a-2)\\
\text{ ou }\\b=-3-i(a-2)\end{array}\right.\end{array}\\[48pt]
(1)\text{ e }(2)\implies \left\{\begin{array}{}\left\{\begin{array}{}c=2a(-3+i(a-2))=-6a+i(2a^2-4a)\\b=-3+i(a-2)\end{array}\right.\\\text{ou}\\\left\{\begin{array}{}c=2a(-3-i(a-2))=-6a-i(2a^2-4a)\\b=-3-i(a-2)\end{array}\right.\\\end{array}\right.\\[48pt]
\text{E então}\\
\left\{\begin{array}{}a+b+c=a-3+i(a-2)-6a+i(2a^2-4a)=-5a-3+i(2a^2-3a-2)\\
\text{ou}\\
a+b+c=a-3-i(a-2)-6a-i(2a^2-4a)=-5a-3-i(2a^2-3a-2)
\end{array}\right.\\[24pt]
a,b,c\in\mathbb{R}\implies \left\{\begin{array}{}a\in\mathbb{R}\\a-2=0\\2a(a-2)=0\end{array}\right.\implies a=2\implies\left\{\begin{array}{}a=2\\b=-3\\c=-12\end{array}\right.\implies a+b+c=2-3-12=-13\\
[/tex3]
\text{e as raízes de }x^4+13x^2+6=0\text{ são as raízes de }\left\{\begin{array}{}x^2=\dfrac{-13-\sqrt{145}}{2}\\\text{ou}\\x^2=\dfrac{-13+\sqrt{145}}{2}\end{array}\right.\\
\text{ou seja:}
\left\{\begin{array}{m}x=i\sqrt{\dfrac{13-\sqrt{145}}{2}}\\\text{ou}\\x=-i\sqrt{\dfrac{13-\sqrt{145}}{2}}\\\text{ou}\\x=i\sqrt{\dfrac{13+\sqrt{145}}{2}}\\\text{ou}\\x=-i\sqrt{\dfrac{13+\sqrt{145}}{2}}\\\end{array}\right.\text{, todas raízes imaginárias puras.}
[/tex3]
[tex3]
\begin{array}{rl}
\left.\begin{array}{rl}
2x^4+(a^2+b^2+13)x^2-c=2(x^2-r_1^2)(x^2-r_2^2)\\
x^4+(2a-3b)x^2-ab=(x^2-r_1^2)(x^2-r_2^2)
\end{array}\right\}
\implies
\left\{\begin{array}{ll}
a^2+b^2+13=4a-6b&(1)\\
2ab=c&(2)
\end{array}\right.
\end{array}\\[24pt]
\begin{array}{}
(1)&\iff b^2+6b+a^2-4a+13=0\\
&\iff\left\{\begin{array}{ll}b=\dfrac{-6+i\sqrt{-36+4(a^2-4a+13)}}{2}=\dfrac{-6+i\sqrt{4a^2-16a+16}}{2}=-3+i(a-2)\\
\text{ ou }\\b=-3-i(a-2)\end{array}\right.\end{array}\\[48pt]
(1)\text{ e }(2)\implies \left\{\begin{array}{}\left\{\begin{array}{}c=2a(-3+i(a-2))=-6a+i(2a^2-4a)\\b=-3+i(a-2)\end{array}\right.\\\text{ou}\\\left\{\begin{array}{}c=2a(-3-i(a-2))=-6a-i(2a^2-4a)\\b=-3-i(a-2)\end{array}\right.\\\end{array}\right.\\[48pt]
\text{E então}\\
\left\{\begin{array}{}a+b+c=a-3+i(a-2)-6a+i(2a^2-4a)=-5a-3+i(2a^2-3a-2)\\
\text{ou}\\
a+b+c=a-3-i(a-2)-6a-i(2a^2-4a)=-5a-3-i(2a^2-3a-2)
\end{array}\right.\\[24pt]
a,b,c\in\mathbb{R}\implies \left\{\begin{array}{}a\in\mathbb{R}\\a-2=0\\2a(a-2)=0\end{array}\right.\implies a=2\implies\left\{\begin{array}{}a=2\\b=-3\\c=-12\end{array}\right.\implies a+b+c=2-3-12=-13\\
[/tex3]
[tex3]
\boxed{\hspace{2cm}\\[6pt]\hspace{1cm}a+b+c=-13\hspace{1cm}\\\hspace{2cm}}\\[24pt]
[/tex3]
[tex3]\boxed{\hspace{2cm}\\[6pt]\hspace{1cm}a+b+c=-13\hspace{1cm}\\\hspace{2cm}}\\[24pt]
[/tex3]
\text{e as raízes de }x^4+13x^2+6=0\text{ são as raízes de }\left\{\begin{array}{}x^2=\dfrac{-13-\sqrt{145}}{2}\\\text{ou}\\x^2=\dfrac{-13+\sqrt{145}}{2}\end{array}\right.\\
\text{ou seja:}
\left\{\begin{array}{m}x=i\sqrt{\dfrac{13-\sqrt{145}}{2}}\\\text{ou}\\x=-i\sqrt{\dfrac{13-\sqrt{145}}{2}}\\\text{ou}\\x=i\sqrt{\dfrac{13+\sqrt{145}}{2}}\\\text{ou}\\x=-i\sqrt{\dfrac{13+\sqrt{145}}{2}}\\\end{array}\right.\text{, todas raízes imaginárias puras.}
[/tex3]
Última edição: rcompany (Sáb 02 Out, 2021 19:33). Total de 1 vez.
Out 2021
02
23:16
Re: (Simulado-Ime/Ita) Equação
- geogebra-export (14).png (52.5 KiB) Exibido 1746 vezes
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