Question

Sendo S a soma e P o produto das raízes da equação 6x ao quadrado − 15x − 21 = 0
A) S − P = 6 .
B) S + P = 2 .
C) S ⋅ P = 4 .
D) S/P= 1

Answer 1

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$\large\green{\boxed{\rm~~~\red{A)}~\gray{S - P}~\pink{=}~\blue{ 6 }~~~}}$

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$\green{\rm\underline{EXPLICAC_{\!\!\!,}\tilde{A}O\ PASSO{-}A{-}PASSO\ \ \ }}$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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☺lá, Ribora, como tens passado nestes tempos de quarentena⁉ E os estudos à distância, como vão⁉ Espero que bem❗ Acompanhe a resolução abaixo, feita através de algumas manipulações algébricas, e após o resultado você encontrará um resumo com mais informações sobre as Relações de Girard que talvez te ajude com exercícios semelhantes no futuro. ✌

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$\large\gray{\boxed{\blue{F(x) = \pink{6}x^2 + \green{(-15)}x + \gray{(-21)} = 0}}}$

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$\rm\large\pink{\Longrightarrow~~a = 6}~~~$

$\rm\large\green{\Longrightarrow~~b = -15}~~~$

$\rm\large\gray{\Longrightarrow~~c = -21}~~~$

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➡ $\large\sf\blue{ S = \dfrac{-b}{a} = \dfrac{15}{6} }$

➡ $\large\sf\blue{ P = \dfrac{c}{a} = \dfrac{-21}{6} }$

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Ⓐ_____________________________✍

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$\large\gray{\boxed{\rm\blue{ S - P }}}$

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$\sf\blue{ = \dfrac{15}{6} - \dfrac{-21}{6} }$

$\sf\blue{ = \dfrac{15}{6} + \dfrac{21}{6} }$

$\sf\blue{ = \dfrac{15 + 21}{6} }$

$\sf\blue{ = \dfrac{36}{6}}$

$\sf\blue{ = 6 }$  ✅

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Ⓑ_____________________________✍

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$\large\gray{\boxed{\rm\blue{ S + P }}}$

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$\sf\blue{ = \dfrac{15}{6} + \dfrac{-21}{6} }$

$\sf\blue{ = \dfrac{15}{6} - \dfrac{21}{6} }$

$\sf\blue{ = \dfrac{15 - 21}{6} }$

$\sf\blue{ = \dfrac{-6}{6}}$

$\sf\blue{ = -1 \neq 2}$  ❌

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Ⓒ_____________________________✍

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$\large\gray{\boxed{\rm\blue{ S \cdot P }}}$

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$\sf\blue{ = \dfrac{15}{6} \cdot \dfrac{-21}{6} }$

$\sf\blue{ = \dfrac{15 \cdot (-21)}{6 \cdot 6}}$

$\sf\blue{ = \dfrac{-315}{36} }$

$\sf\blue{ = \dfrac{-105}{2} \neq 4}$  ❌

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Ⓓ_____________________________✍

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$\large\gray{\boxed{\rm\blue{ S \div P }}}$

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$\sf\blue{ = \dfrac{15}{6} \div \dfrac{-21}{6} }$

$\sf\blue{ = \dfrac{15}{6} \cdot \dfrac{6}{-21} }$

$\sf\blue{ = \dfrac{15 \cdot 6}{6 \cdot (-21)}}$

$\sf\blue{ = \dfrac{-90}{126} }$

$\sf\blue{ = \dfrac{-5}{7} \neq 1}$  ❌

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$\large\green{\boxed{\rm~~~\red{A)}~\gray{S - P}~\pink{=}~\blue{ 6 }~~~}}$ ✅

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_________________________________

$\sf\large\red{COEFICIENTES~E~RA\acute{I}ZES }$

$\sf\large\red{(Relac_{\!\!\!,}\tilde{o}es~de~Girard) }$

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☔ Ao realizarmos nossas manipulações algébricas em busca das raízes de uma equação polinomial de segundo grau dada na forma

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$\sf\large\red{\boxed{\pink{\boxed{\begin{array}{rcl} & & \\ & \orange{f(x) = a \cdot x^2 + b \cdot x + c} & \\ & & \\ \end{array}}}}}$

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através da fórmula de Bháskara, onde

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$\sf\large\red{\boxed{\pink{\boxed{\begin{array}{rcl} & & \\ & \orange{x = \dfrac{-b \pm \sqrt{b^2 - 4 \cdot a \cdot c}}{2 \cdot a}} & \\ & & \\ \end{array}}}}}$

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encontramos as seguintes raízes

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$\sf\large\begin{cases}\orange{x_{1}= \dfrac{-b + \sqrt{b^2 - 4 \cdot a \cdot c}}{2 \cdot a}}\\\\\\ \orange{x_{2}= \dfrac{-b - \sqrt{b^2 - 4 \cdot a \cdot c}}{2 \cdot a}}\end{cases}$

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☔ Quando somamos nossas duas raízes nós obtemos

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➡ $\sf\large\orange{ x_1 + x_2 = \dfrac{-b + \sqrt{b^2 - 4 \cdot a \cdot c}}{2 \cdot a} + \dfrac{-b - \sqrt{b^2 - 4 \cdot a \cdot c}}{2 \cdot a} }$

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$\sf\large\orange{ = \dfrac{-b -b + \overbrace{\sqrt{b^2 - 4 \cdot a \cdot c} - \sqrt{b^2 - 4 \cdot a \cdot c}}^{=0} }{2 \cdot a} }$

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$\sf\large\orange{ = \dfrac{-\diagup\!\!\!\!{2}b}{\diagup\!\!\!\!{2}a} = \dfrac{-b}{a} }$

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$\rm\large\red{\boxed{\pink{\boxed{\begin{array}{rcl}&&\\&\orange{ x_1 + x_2 = \dfrac{-b}{a} }&\\&&\\\end{array}}}}}$

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☔ Quando multiplicamos nossas duas raízes nós obtemos

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➡ $\sf\large\orange{ x_1 \cdot x_2 = \dfrac{-b + \sqrt{b^2 - 4 \cdot a \cdot c}}{2 \cdot a} \cdot \dfrac{-b - \sqrt{b^2 - 4 \cdot a \cdot c}}{2 \cdot a} }$

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$\sf\large\orange{ = \dfrac{b^2 \overbrace{- 2 \cdot b \cdot \sqrt{b^2 - 4 \cdot a \cdot c} + 2 \cdot b \cdot \sqrt{b^2 - 4 \cdot a \cdot c}}^{= 0} - b^2 + 4 \cdot a \cdot c}{4 \cdot a^2} }$

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$\sf\large\orange{ = \dfrac{\diagup\!\!\!\!{b}^2 - \diagup\!\!\!\!{b}^2 + 4 \cdot a \cdot c}{4 \cdot a^2} }$

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$\sf\large\orange{ = \dfrac{\diagup\!\!\!\!{4} \cdot \diagup\!\!\!\!{a} \cdot c}{\diagup\!\!\!\!{4} \cdot a^{\diagup\!\!\!\!{2}}} = \dfrac{c}{a}}$

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$\rm\large\red{\boxed{\pink{\boxed{\begin{array}{rcl}&&\\&\orange{ x_1 \cdot x_2 = \dfrac{c}{a} }&\\&&\\\end{array}}}}}$

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_______________________________☁

☕ Bons estudos.

(Dúvidas nos comentários) ☄

___________________________$\LaTeX$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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"Absque sudore et labore nullum opus perfectum est."

Answer 2

Resposta:

Alternativa A.

Explicação passo-a-passo:

Primeiro vamos encontrar os valor de S e P:

$6x^2 - 15x - 21 =0\\\\Soma(S) = x_1 + x_2 =\boxed{-\dfrac{b}{a}}\\\\Produto(P) = x_1\times x_2 = \boxed{\dfrac{c}{a}}\\\\\\S= \dfrac{-b}{a}=-\dfrac{-15}{6}=\dfrac{15}{6}=\dfrac{5}{2} \Longrightarrow \boxed{S=\dfrac{5}{2}}\\\\\\P=\dfrac{c}{a}=\dfrac{-21}{6}=-\dfrac{7}{2}\Longrightarrow \boxed{P=-\dfrac{7}{2}}$

Agora vamos realizar as operações solicitadas:

$A)\ S - P = \dfrac{5}{2}-\left(-\dfrac{7}{2}\right)=\dfrac{5}{2}+\dfrac{7}{2}=\dfrac{12}{2} =\boxed{6} \\\\\left \ \ \ \boxed{S-P=6} \rightarrow\ \bf{Verdadeiro}\right \\\\B)\ S+P=\dfrac{5}{2}+\left(-\dfrac{7}{2}\right)=\dfrac{5}{2}-\dfrac{7}{2}=-\dfrac{2}{2}=\boxed{-1}\\\\\left \ \ \ \boxed{S+P=2} \rightarrow\ Falso \right \\\\\\C)\ S\times P = \dfrac{5}{2}\times-\dfrac{7}{2}= -\dfrac{35}{4}=\boxed{-8,75}\\\\\left \ \ \ \boxed{S\times P=4} \rightarrow\ Falso \right$

$D)\ \dfrac{S}{P} = \dfrac{\dfrac{5}{2}}{-\dfrac{7}{2}}=\dfrac{5}{2}\times-\dfrac{2}{7}=\dfrac{5}{\not{2}^1}\times-\dfrac{\not{2}^1}{7}=\boxed{-\dfrac{5}{7}}\\\\\left \ \ \ \boxed{\dfrac{S}{P} =1} \rightarrow\ Falso \right$

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