Question

equação de 2grau1x2-4x+1=0

Answer 1

Resolvendo a equação do segundo grau descobriu-se que o resultado é:

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• $\large\text{ \underline{\boxed{\tt{2~+~\sqrt{3}}}} }$
• $\large\text{ \underline{\boxed{\tt{2~-~\sqrt{3}}}} }$

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Encontrando os coeficientes da equação (as letras):

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$\large\text{ \tt{1x^{2}~-~4x~+~1~=~0} }\\\\\\\large\tt{a~\rightarrow~1}\\\\\large\tt{b~\rightarrow~-~4}\\\\\large\tt{c~\rightarrow~1}$

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Substituindo na fórmula de Bhaskara:

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$\large\text{ \tt{x~=~\dfrac{-~b~\pm~\sqrt{b^{2}~-~4~.~a~.~c}}{2~.~a}} }\\\\\\\large\text{ \tt{x~=~\dfrac{-~(-4)~\pm~\sqrt{(-4)^{2}~-~4~.~1~.~1}}{2~.~1}} }\\\\\\\large\text{ \tt{x~=~\dfrac{4~\pm~\sqrt{16~-~4}}{2}} }\\\\\\\large\text{ \tt{x~=~\dfrac{4~\pm~\sqrt{12}}{2}} }$

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A fatoração do número 12 fica:

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$\large\text{ \tt{2~.~2~.~3~=~2^{2}~.~3} }$

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Substituindo dentro da raiz quadrada:

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$\large\text{ \tt{x~=~\dfrac{4~\pm~\sqrt{2^{2}~.~3}}{2}} }\\\\\\\large\text{ \tt{x~=~\dfrac{4~\pm~2~\sqrt{3}}{2}} }\\\\\\\large\text{ \tt{x'~=~\dfrac{4~+~2~\sqrt{3}}{2}~=~\dfrac{4}{2}+\dfrac{2\sqrt{3}}{2}~=~\underline{\boxed{\tt{2+\sqrt{3}}}}} }\\\\\\\large\text{ \tt{x''~=~\dfrac{4~-~2~\sqrt{3}}{2}~=~\dfrac{4}{2}-\dfrac{2\sqrt{3}}{2}~=~\underline{\boxed{\tt{2-\sqrt{3}}}}} }$

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Portanto, o resultado da equação é:

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$\large\begin{cases}\tt{x'~=~2~+~\sqrt{3}\\\tt{x''~=~2~-~\sqrt{3}}}\end{cases}$

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