Matemática
andersonns20
3

Achar as raízes das equações: a) x2 - x - 20 = 0 b) x2 - 3x -4 = 0 c) x2 - 8x + 7 = 0

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(1) Respostas
josebalddino

a) [latex] x^{2} - x - 20 = 0[/latex] [latex]x' = \frac{-(-1)+\sqrt{(-1)^{2} - 4.1.(-20)}}{2.1}\\ x' = \frac{1+\sqrt{81}}{2}\\ x' = \frac{1+9 }{2} \\ x' = \frac{10 }{2} \\ x'' = 5 \\ \\ x'' = \frac{1-9 }{2} \\ x'' = \frac{-8}{2} \\ x'' = -4[/latex] b) [latex] x^{2} - 3x - 4 = 0[/latex] [latex]x' = \frac{-(-3)+\sqrt{(-3)^{2} - 4.1.(-4)} }{2.1} \\ x' = \frac{3+\sqrt{25} }{2} \\ x' = \frac{1+5 }{2} \\ x' = \frac{6 }{2} \\ x'' = 3 \\ \\ x'' = \frac{1-5 }{2} \\ x'' = \frac{-4}{2} \\ x'' = -2[/latex] c) [latex] x^{2} - 8x + 7 = 0[/latex] [latex]x' = \frac{-(-8)+\sqrt{(-8)^{2} - 4.1.7}}{2.1}\\ x' = \frac{3+\sqrt[64 - 28} }{2} \\ x' = \frac{3+\sqrt[36} }{2} \\ x' = \frac{3+6 }{2} \\ x' = \frac{9 }{2} \\ x'' = \frac{1-6 }{2} \\ x'' = \frac{-5}{2} \\[/latex] [latex]x' = \frac{-(-8)+\sqrt{(-8)^{2} - 4.1.7} }{2.1} \\ x' = \frac{8+\sqrt{36} }{2} \\ x' = \frac{8+6 }{2} \\ x' = \frac{14 }{2} \\ x'' = 7 \\ \\ x'' = \frac{8-6 }{2} \\ x'' = \frac{2}{2} \\ x'' = 1[/latex]

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