Question

Equação de segundo grau.

2x² - x - 10 = 0
2x² +3x - 5 = 0

Answer 1

Para calcularmos as raízes de uma equação do 2º grau devemos primeiramente identificarmos o A, B e o C da sua equação. Vamos lá.

$\large\begin{array}{lr} \sf \underline{2}x^{2}\ \underline{-1}x\ \underline{- 10}=0\left\{\begin{array}{ll}\sf a=2\\\sf b=-1\\\sf c=-10 \end{array}\right.\end{array}$

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$\large\begin{array}{lr} \sf \underline{2}x^{2}\ \underline{+3}x\ \underline{- 5}=0\left\{\begin{array}{ll}\sf a=2\\\sf b=3\\\sf c=-5 \end{array}\right.\end{array}$

Agora devemos aplicar a fórmula do discriminante ( $\Delta$ ).

$\large\begin{array}{lr} \sf \underline{2}x^{2}\ \underline{-1}x\ \underline{- 10}=0\left\{\begin{array}{ll}\sf a=2\\\sf b=-1\\\sf c=-10 \end{array}\right.\end{array}$

$\large\begin{array}{lr} \sf \Delta = b^{2} -4\cdot a\cdot c\\\\\sf \Delta = (-1)^{2} -4\cdot 2\cdot (-10)\\\\\sf \Delta = 1 + 80\\\\\sf \Delta = \underline{\boxed{\red{\sf 81}}}\end{array}$

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$\large\begin{array}{lr} \sf \underline{2}x^{2}\ \underline{+3}x\ \underline{- 5}=0\left\{\begin{array}{ll}\sf a=2\\\sf b=3\\\sf c=-5 \end{array}\right.\end{array}$      

$\large\begin{array}{lr} \sf \Delta = b^{2} -4\cdot a\cdot c\\\\\sf \Delta = 3^{2} -4\cdot 2\cdot (-5)\\\\\sf \Delta = 9 + 40\\\\\sf \Delta = \underline{\boxed{\red{\sf 49}}}\end{array}$        

Agora devemos aplicar a fórmula de bhaskara.

$\large\begin{array}{lr} \sf \underline{2}x^{2}\ \underline{-1}x\ \underline{- 10}=0\left\{\begin{array}{ll}\sf a=2\\\sf b=-1\\\sf c=-10 \end{array}\right.\end{array}$

$\large\begin{array}{lr} \sf \Delta = b^{2} -4\cdot a\cdot c\\\\\sf \Delta = (-1)^{2} -4\cdot 2\cdot (-10)\\\\\sf \Delta = 1 + 80\\\\\sf \Delta = \underline{\boxed{\red{\sf 81}}}\end{array}$

$\large\begin{array}{lr} \sf x=\dfrac{-b\pm \sqrt{\Delta} }{2\cdot a} \\\\\sf x=\dfrac{-(-1)\pm \sqrt{81} }{2\cdot 2} \\\\\sf x=\dfrac{1\pm 9}{4}\\\\\sf x'=\dfrac{10}{4}\\\\\sf x'= \underline{\boxed{\red{\sf \frac{5}{2}}}}\\\\\sf x''=\dfrac{-8}{4}\\\\\sf x''= \underline{\boxed{\red{\sf -2}}}\end{array}$

   

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$\large\begin{array}{lr} \sf \underline{2}x^{2}\ \underline{+3}x\ \underline{- 5}=0\left\{\begin{array}{ll}\sf a=2\\\sf b=3\\\sf c=-5 \end{array}\right.\end{array}$      

$\large\begin{array}{lr} \sf \Delta = b^{2} -4\cdot a\cdot c\\\\\sf \Delta = 3^{2} -4\cdot 2\cdot (-5)\\\\\sf \Delta = 9 + 40\\\\\sf \Delta = \underline{\boxed{\red{\sf 49}}}\end{array}$        

$\large\begin{array}{lr} \sf x=\dfrac{-b\pm \sqrt{\Delta} }{2\cdot a} \\\\\sf x=\dfrac{-3\pm \sqrt{41} }{2\cdot 2} \\\\\sf x=\dfrac{-3\pm 7}{4}\\\\\sf x'=\dfrac{4}{4}\\\\\sf x'= \underline{\boxed{\red{\sf 1}}}\\\\\sf x''=\dfrac{-10}{4}\\\\\sf x''= \underline{\boxed{\red{\sf -\frac{5}{2}}}}\end{array}$

Espero ter ajudado.

Bons estudos.

• Att. FireClassis.

Answer 2

Explicação passo-a-passo:

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$2x {}^{2} - x - 10 = 0$

$a = 2$

$b = - 1$

$c = - 10$

$∆ = b {}^{2} - 4ac$

$∆ = ( - 1) {}^{2} - 4 \times 2 \times - 10$

$∆ = 1 + 80$

$∆ = 81$

$\dfrac{ - b \frac{ + }{} \sqrt{∆} }{2a}$

$\dfrac{1 \frac{ + }{} \sqrt{81} }{2 \times 2}$

$\dfrac{1 \frac{ + }{} 9 }{4}$

$x {}^{1} = \dfrac{1 + 9}{4} = \dfrac{10}{4} = \dfrac{10 {}^{ \div 2} }{4 {}^{ \div 2} } = \dfrac{5}{2}$

$x {}^{2} = \dfrac{1 - 9}{4} = - \dfrac{8}{4} = - 2$

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$2x {}^{2} + 3x - 5 = 0$

$a = 2$

$b = 3$

$c = - 5$

$∆ = b {}^{2} - 4ac$

$∆ =3 {}^{2} - 4 \times 2 \times - 5$

$∆ = 9 + 40$

$∆ = 49$

$\dfrac{ - b \frac{ + }{} \sqrt{∆} }{2a}$

$\dfrac{ - 3 \frac{ + }{} \sqrt{49} }{2 \times 2}$

$\dfrac{ - 3 \frac{ + }{} 7}{4}$

$x {}^{1} = \dfrac{ - 3 + 7}{4} = \dfrac{4}{4} = 1$

$x {}^{2} = \dfrac{ - 3 - 7}{4} = - \dfrac{10}{4} = - \dfrac{10 {}^{ \div 2} }{4 {}^{ \div 2} } = - \dfrac{5}{2}$