As raízes da equação do 2º grau são, x'=4 e x"= -1.
Acompanhe a solução:
Aplicando a fórmula de Bháskara, temos:
Cálculo:
[tex]\Delta=b^2-4ac\\\\\Delta=(-3)^2-4\cdot1\cdot(-4)\\\\\Delta=9+16\\\\\Large\boxed{\Delta=25}\\\\\\x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\x=\dfrac{-(-3)\pm\sqrt{25}}{2\cdot1}\\\\\Large\boxed{x=\dfrac{3\pm5}{2}}\\\\\\\boxed{\large\begin {array}{l}x'=\dfrac{3+5}{2}\\\\x'=\dfrac{8}{2}\\\\\Large\boxed{\boxed{x'=4}}\Huge\checkmark\end {array}}\quad\quad\boxed{\large\begin {array}{l}x"=\dfrac{3-5}{2}\\\\x"=\dfrac{-2}{2}\\\\\Large\boxed{\boxed{x"=-1}}\Huge\checkmark\end {array}}[/tex]
Resposta:
Portanto, as raízes da equação do 2º grau são, x'=4 e x"= -1.
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