Question

Calcule as equações do 2° grau

Answer 1

$\maltese\ \boldsymbol{\boxed{\sf Equac_{\!\!\!,}\tilde{a}o\ do\ 2^o\ Grau}}$

• Para calcular uma Equação do 2º Grau, primeiro vamos achar o valor das letras: { a, b, c}.

    $\large\boldsymbol {\sf} C) x^2+12x+16= 0\left\{\begin{array}{ll}a= 1\\b=12\\c=16\end{array}\right.$

   $\large\boldsymbol {\sf} D) -x^2-10x+20=0\left\{\begin{array}{ll}a= -1\\b=-10\\c=20\end{array}\right.$

Depois disso, poderemos aplicar a fórmula para acharmos o valor do Delta Δ.

$\large \lozenge\ \boldsymbol{\boxed{\sf F\acute{o}rmula\ do\ Delta: }}$

   $\LARGE\boldsymbol {\sf} \Delta = b^2 - 4 \times a\times c$

$\large \lozenge\ \boldsymbol{\boxed{\sf Aplicando\ a\ f\acute{o}rmula\ do\ Delta: }}$

$\large\boldsymbol {\sf} C) x^2+12x+16= 0\left\{\begin{array}{l}a= 1\\b=12\\c=16\end{array}\right.\\\\\\\large\boldsymbol {\sf} \Delta = 12^2 - 4 \times 1\times 16\\\large\boldsymbol {\sf} \Delta = 144 - 4 \times 1\times 16\\\large\boldsymbol {\sf} \Delta = 144 -64\\\large\boldsymbol {\sf} \Delta = 80$

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$\large\boldsymbol {\sf} D)-x^2-10x+20=0\left\{\begin{array}{l}a= -1\\b=-10\\c=20\end{array}\right.\\\\\\\large\boldsymbol {\sf} \Delta = (-10)^2 - 4 \times -1\times 20\\\large\boldsymbol {\sf} \Delta = 100 -4 \times -1 \times20\\\large\boldsymbol {\sf} \Delta = 100 +80\\\large\boldsymbol {\sf} \Delta = 180$

• Depois disso, poderemos aplicar a fórmula de bhaskara.

$\large \lozenge\ \boldsymbol{\boxed{\sf F\acute{o}rmula\ de\ bhaskara: }}$

        $\LARGE\boldsymbol {\sf}\frac{-b\ (+ou-)\ \sqrt{\Delta} } { 2\ \times\ a}$  

$\large \lozenge\ \boldsymbol{\boxed{\sf Aplicando\ o\ Delta\ mais\ a\ f\acute{o}rmula\ de\ bhaskara: }}$

$\large\boldsymbol {\sf} C) x^2+12x+16= 0\left\{\begin{array}{ll}a= 1\\b=12\\c=16\end{array}\right.\\\\\\\large\boldsymbol {\sf} \Delta = 12^2 - 4 \times 1\times 16\\\large\boldsymbol {\sf} \Delta = 144 - 4 \times 1\times 16\\\large\boldsymbol {\sf} \Delta = 144 -64\\\large\boldsymbol {\sf} \Delta = 80$

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$\large\boldsymbol {\sf} C)\ \boldsymbol {\sf}\dfrac{-12\ (+ou-)\ \sqrt{80} } { 2\ \times\ 1}$

Obs: 80 não é um quadrado perfeito, por isso não tem raiz exata. Pra você encontrar uma raiz aproximada, você deve fatorar 80.

√80 = √2·2·2·2·5 = √2²·2²·5 = 4√5

que é aproximadamente 8.94.

        $\large\boldsymbol {\sf} \ \boldsymbol {\sf}\dfrac{-12\ +\ 4\sqrt{5} } { 2}$

        $\large\boldsymbol {\sf} \ \boldsymbol {\sf}\dfrac{-12\ -\ 4\sqrt{5} } { 2}$

         

$\large\boldsymbol{\boxed{{\sf}x_{1}= {-6\ +\ 2\sqrt{5} } }}$

$\large\boldsymbol{\boxed{{\sf}x_{2}= {-6\ -\ 2\sqrt{5} } }}$

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$\large\boldsymbol {\sf} D)-x^2-10x+20=0\left\{\begin{array}{ll}a= -1\\b=-10\\c=20\end{array}\right.\\\\\\\large\boldsymbol {\sf} \Delta = (-10)^2 - 4 \times -1\times 20\\\large\boldsymbol {\sf} \Delta = 100 -4 \times -1 \times20\\\large\boldsymbol {\sf} \Delta = 100 +80\\\large\boldsymbol {\sf} \Delta = 180$

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$\large\boldsymbol {\sf} D)\ \boldsymbol {\sf}\dfrac{-(-10)\ (+ou-)\ \sqrt{180} } { 2\ \times\ -1}$

        $\large\boldsymbol {\sf} \ \boldsymbol {\sf}\dfrac{10\ +\ {6\sqrt5} } {-2}$

        $\large\boldsymbol {\sf} \ \boldsymbol {\sf}\dfrac{10\ -\ 6\sqrt{5} } {-2}$

$\large\boldsymbol{\boxed{{\sf}x_{1}= {-5\ +\ 3\sqrt{5} } }}$

$\large\boldsymbol{\boxed{{\sf}x_{2}= {-5\ -\ 3\sqrt{5} } }}$

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$\LARGE \boxed{\sf Espero\ ter\ ajudado! :)}$

$\LARGE \boxed{\sf Bons\ estudos!}$