Matemática, perguntado por cv9126724, 4 meses atrás

2.RESOLVA AS EQUAÇÕES DE 2º GRAU ou Fórmula de Bhaskara a) 9y² – 12y + 4 = 0 b) x² - 3 = 4x + 2 c) -x² - x + 30 = 0 d) (x - 5)² = 1 e) 3x2 + 55 = 0 f) x2 - 6x = 0


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Soluções para a tarefa

Respondido por auditsys
0

Resposta:

\textsf{Leia abaixo}

Explicação passo a passo:

\mathsf{9y^2 - 12y + 4 = 0}

\mathsf{\Delta = b^2 - 4.a.c}

\mathsf{\Delta = (-12)^2 - 4.9.4}

\mathsf{\Delta = 144 - 144}

\mathsf{\Delta = 0}

\mathsf{y = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{12 \pm \sqrt{0}}{18} \rightarrow \begin{cases}\mathsf{y' = \dfrac{12 + 0}{18} = \dfrac{12}{18} = \dfrac{2}{3}}\\\\\mathsf{y'' = \dfrac{12 - 0}{18} = \dfrac{12}{18} = \dfrac{2}{3}}\end{cases}}

\boxed{\boxed{\mathsf{S = \left\{\dfrac{2}{3}\right\}}}}

\mathsf{x^2 - 3 = 4x + 2}

\mathsf{x^2 - 4x - 5 = 0}

\mathsf{\Delta = b^2 - 4.a.c}

\mathsf{\Delta = (-4)^2 - 4.1.(-5)}

\mathsf{\Delta = 16 + 20}

\mathsf{\Delta = 36}

\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{4 \pm \sqrt{36}}{2} \rightarrow \begin{cases}\mathsf{x' = \dfrac{4 + 6}{2} = \dfrac{10}{2} = 5}\\\\\mathsf{x'' = \dfrac{4 - 6}{2} = -\dfrac{2}{2} = -1}\end{cases}}

\boxed{\boxed{\mathsf{S = \left\{5;-1\right\}}}}

\mathsf{-x^2 - x + 30 = 0}

\mathsf{\Delta = b^2 - 4.a.c}

\mathsf{\Delta = (-1)^2 - 4.(-1).30}

\mathsf{\Delta = 1 + 120}

\mathsf{\Delta = 121}

\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{1 \pm \sqrt{121}}{-2} \rightarrow \begin{cases}\mathsf{x' = \dfrac{1 + 11}{-2} = -\dfrac{12}{2} = -6}\\\\\mathsf{x'' = \dfrac{1 - 11}{-2} = \dfrac{-10}{-2} = 5}\end{cases}}

\boxed{\boxed{\mathsf{S = \left\{5;-6\right\}}}}

\mathsf{(x - 5)^2 = 1}

\mathsf{x^2 - 10x + 25 = 1}

\mathsf{x^2 - 10x + 24 = 0}

\mathsf{\Delta = b^2 - 4.a.c}

\mathsf{\Delta = (-10)^2 - 4.1.24}

\mathsf{\Delta =100 - 96}

\mathsf{\Delta =4}

\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{10 \pm \sqrt{4}}{2} \rightarrow \begin{cases}\mathsf{x' = \dfrac{10 + 2}{2} = \dfrac{12}{2} = 6}\\\\\mathsf{x'' = \dfrac{10 - 2}{2} = \dfrac{8}{2} = 4}\end{cases}}

\boxed{\boxed{\mathsf{S = \left\{6;4\right\}}}}

\mathsf{3x^2 + 55 = 0}

\mathsf{\Delta = b^2 - 4.a.c}

\mathsf{\Delta = (0)^2 - 4.3.55}

\mathsf{\Delta = 0 - 660}

\mathsf{\Delta = -660}

\boxed{\boxed{\mathsf{S = \left\{\emptyset\right\}}}}

\mathsf{x^2 - 6x = 0}

\mathsf{\Delta = b^2 - 4.a.c}

\mathsf{\Delta = (-6)^2 - 4.1.0}

\mathsf{\Delta = 36 - 0}

\mathsf{\Delta = 36}

\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{6 \pm \sqrt{36}}{2} \rightarrow \begin{cases}\mathsf{x' = \dfrac{6 + 6}{2} = \dfrac{12}{2} = 6}\\\\\mathsf{x'' = \dfrac{6 - 6}{2} = \dfrac{0}{2} = 0}\end{cases}}

\boxed{\boxed{\mathsf{S = \left\{6;0\right\}}}}

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