Question

ajuda estou precisando.......​

Answer 1

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⠀⠀☞ Utilizando o método da Substituição de Variáveis podemos transformar e resolver as equação biquadradas através da Fórmula de Bháskara encontrando suas raízes reais.✅ Acompanhe:

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⠀⠀$\Large\gray{\boxed{\sf\blue{~~x^2 = k~~\iff~~x = \sqrt{k}}}}$

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$\large\gray{\blue{\text{ \sf~01)~a)~~\pink{\overbrace{1}^{a}}k^2 + \green{\overbrace{(-13)}^{b}}k + \gray{\overbrace{36}^{c}} = 0 }}}$

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$\Large\blue{\sf \Delta = \green{(-13)}^2 - 4 \cdot \pink{1} \cdot \gray{36} = 25}$  

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$\begin{cases}\large\blue{\text{ \sf k_1 = \dfrac{-(-13) + \sqrt{25}}{2 \cdot 1} = \dfrac{13 + 5}{2} = 9 }}\\\\\\\large\blue{\text{ \sf k_2 = \dfrac{-(-13) - \sqrt{25}}{2 \cdot 1} = \dfrac{13 - 5}{2} = 4 }}\end{cases}$

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$\begin{cases}\large\blue{\sf x_{1} = \pm \sqrt{k_1} = \pm \sqrt{9} = \pm 3}\\\\\\\large\blue{\sf x_{2} = \pm \sqrt{k_2} = \pm \sqrt{4} = \pm 2}\end{cases}$

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$\huge\green{\boxed{\rm~~~\gray{S}~\pink{=}~\blue{ \{\pm 2, \pm 3\} }~~~}}$ ✅

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01) b) k² + 3k - 4 = 0

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$\Large\blue{\sf \Delta = 25}$

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$\begin{cases}\large\blue{\sf k_{1} = 1~~\pink{\Longrightarrow}~~x_1 = \pm 1}\\\\\\\large\blue{\sf k_{2} = -4~~\pink{\Longrightarrow}~~x_2 \notin \mathbb{R}}\end{cases}$

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$\Huge\green{\boxed{\rm~~~\blue{S = \{\pm 1\}~~~}}}$ ✅  

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01) c) k² - 7k + 12 = 0

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$\Large\blue{\sf \Delta = 1}$

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$\begin{cases}\large\blue{\sf k_{1} = 4~~\pink{\Longrightarrow}~~x_1 = \pm 2}\\\\\\\large\blue{\sf k_{2} = 3~~\pink{\Longrightarrow}~~x_2 = \pm \sqrt{3}}\end{cases}$

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$\LARGE\green{\boxed{\rm~~~\blue{S = \{\pm \sqrt{3}, \pm 2\}~~~}}}$ ✅  

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01) d) k² + 5k + 6 = 0

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$\Large\blue{\sf \Delta = 1}$

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$\begin{cases}\large\blue{\sf k_{1} = -2~~\pink{\Longrightarrow}~~x_1 \notin \mathbb{R}}\\\\\\\large\blue{\sf k_{2} = -3~~\pink{\Longrightarrow}~~x_2 \notin \mathbb{R}}\end{cases}$

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$\Huge\green{\boxed{\rm~~~\blue{S = \{\emptyset\}~~~}}}$ ✅

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01) e) k² - 18k + 32 = 0

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$\Large\blue{\sf \Delta = 196}$

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$\begin{cases}\large\blue{\sf k_{1} = 16~~\pink{\Longrightarrow}~~x_1 = \pm 4}\\\\\\\large\blue{\sf k_{2} = 2~~\pink{\Longrightarrow}~~x_2 = \pm \sqrt{2}}\end{cases}$

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$\LARGE\green{\boxed{\rm~~~\blue{S = \{\pm \sqrt{2}, \pm 4\}~~~}}}$ ✅

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02) k² - 50k + 49

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$\Large\blue{\sf \Delta = 2304}$

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$\begin{cases}\large\blue{\sf k_{1} = 49~~\pink{\Longrightarrow}~~x_1 = \pm 7}\\\\\\\large\blue{\sf k_{2} = 1~~\pink{\Longrightarrow}~~x_2 = \pm 1}\end{cases}$

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$\huge\green{\boxed{\rm~~~\blue{S = \{\pm 1, \pm 7\}~~~}}}$ ✅

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03) a) k² - 26k + 25 = 0

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$\Large\blue{\sf \Delta = 576}$

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$\begin{cases}\large\blue{\sf k_{1} = 25~~\pink{\Longrightarrow}~~x_1 = \pm 5}\\\\\\\large\blue{\sf k_{2} = 1~~\pink{\Longrightarrow}~~x_2 = \pm 1}\end{cases}$

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$\huge\green{\boxed{\rm~~~\blue{S = \{\pm 1, \pm 5\}~~~}}}$ ✅

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03) b) k² - 5k + 6 = 0

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$\Large\blue{\sf \Delta = 1}$

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$\begin{cases}\large\blue{\sf k_{1} = 3~~\pink{\Longrightarrow}~~x_1 = \pm \sqrt{3}}\\\\\\\large\blue{\sf k_{2} = 2~~\pink{\Longrightarrow}~~x_2 = \pm \sqrt{2}}\end{cases}$

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$\LARGE\green{\boxed{\rm~~~\blue{S = \{\pm \sqrt{2} ,\pm \sqrt{3}\}~~~}}}$ ✅

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04) x^4 = 4 + 3x²

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$\Large\gray{\blue{\text{ \sf~\pink{\overbrace{1}^{a}}k^2 + \green{\overbrace{(-3)}^{b}}k + \gray{\overbrace{(-4)}^{c}} = 0 }}}$

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$\Large\blue{\sf \Delta = 25}$

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$\begin{cases}\large\blue{\sf k_{1} = 4~~\pink{\Longrightarrow}~~x_1 = \pm 2}\\\\\\\large\blue{\sf k_{2} = -1~~\pink{\Longrightarrow}~~x_2 \notin \mathbb{R}}\end{cases}$

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$\Huge\green{\boxed{\rm~~~\blue{S = \{\pm 2\}~~~}}}$ ✅  

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05) k² + 2k - 1 = 0

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$\Large\blue{\sf \Delta = \green{2}^2 - 4 \cdot \pink{1} \cdot \gray{(-1)} = 8}$

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$\begin{cases}\large\blue{\sf k_{1} \approx 0,414~~\pink{\Longrightarrow}~~x_1 \approx \pm 0,64}\\\\\\\large\blue{\sf k_{2} \approx -2,414~~\pink{\Longrightarrow}~~x_2 \notin \mathbb{R}}\end{cases}$

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$\Huge\green{\boxed{\rm~~~\red{(C)}~\blue{ duas }~~~}}$ ✅

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$\bf\large\red{\underline{\quad\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$

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$\bf\large\red{\underline{\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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