Resposta :

PeH
[tex]a) \ 2x^2 - 7x + 4 = 0 \\\\ \bullet \ a = 2 \\ \bullet \ b = -7 \\ \bullet \ c = 4 \\\\ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\\\ x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 4}}{2 \cdot 2} \\\\ x = \frac{7 \pm \sqrt{17}}{4} \\\\ x' = \frac{7 + \sqrt{17}}{4} \ \text{e} \ x'' = \frac{7 - \sqrt{17}}{4} \rightarrow \small{\boxed{\text{S} = {\{\frac{7 - \sqrt{17}}{4}}, \frac{7 + \sqrt{17}}{4}\}}}[/tex]


[tex]b) \ 9x^2 - 12x + 4 \\\\ \bullet \ a = 9 \\ \bullet \ b = -12 \\ \bullet \ c = 4 \\\\ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\\\ x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 9 \cdot 4}}{2 \cdot 9} \\\\ x = \frac{12 \pm 0}{18} \\\\ x' = x'' = \frac{12}{18} = \frac{2}{3} \rightarrow \small{\boxed{\text{S} = \{\frac{2}{3}\}}}[/tex]


[tex]c) \ 5x^2 + 3x + 5 = 0 \\\\ \bullet \ a = 5 \\ \bullet \ b = 3 \\ \bullet \ c = 5 \\\\ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\\\ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 5 \cdot 5}}{2 \cdot 5} \\\\ x = \frac{-3 \pm \sqrt{-91}}{10} \notin \mathbb{R} \rightarrow {\small{\boxed{\text{S} = \varnothing}}}[/tex]


[tex]d) \ 7x^2 + 13x - 2 = 0 \\\\ \bullet \ a = 7 \\ \bullet \ b = 13 \\ \bullet \ c = -2 \\\\ x = \frac{-13 \pm \sqrt{13^2 - 4 \cdot 7 \cdot (-2)}}{2 \cdot 7} \\\\ x = \frac{-13 \pm \sqrt{13^2 - 4 \cdot 7 \cdot (-2)}}{2 \cdot 7} \\\\ x = \frac{-13 \pm 15}{14} \\\\ x' = \frac{2}{14} = \frac{1}{7} \ \text{e} \ x'' = -2 \rightarrow \small{\boxed{\text{S} = \{-2, \frac{1}{7}\}}}[/tex]

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